Sat10e SSM 12

Chapter 12 Systems of Equations and Inequalities Section 12.1

 3x + 3 y + 2 z = 4  15.  x − y − z = 0  2 y − 3z = −8  Substituting the values of the variables:  3(1) + 3(−1) + 2(2) = 3 − 3 + 4 = 4  1 − (−1) − 2 = 1 + 1 − 2 = 0  2( −1) − 3(2) = −2 − 6 = −8 

1. 3x + 4 = 8 − x 4x = 4 x =1 The solution set is {1} . 3. inconsistent 5. (3, −2)

Each equation is satisfied, so x = 1, y = −1, z = 2 ,

7. b 2 x − y = 5 9.  5 x + 2 y = 8 Substituting the values of the variables: 2(2) − (−1) = 4 + 1 = 5  5(2) + 2(−1) = 10 − 2 = 8 Each equation is satisfied, so x = 2, y = −1 , or (2, −1) , is a solution of the system of equations.  3x − 4 y = 4  11.  1 1  2 x − 3 y = − 2 Substituting the values of the variables:  1  3(2) − 4  2  = 6 − 2 = 4      1 (2) − 3  1  = 1 − 3 = − 1    2 2 2 2

Each equation is satisfied, so x = 2, y = 1 , or 2 1 , is a solution of the system of equations. 2, 2

( )

 x− y = 3  13.  1  2 x + y = 3 Substituting the values of the variables, we obtain: 4 − 1 = 3  1  2 (4) + 1 = 2 + 1 = 3 Each equation is satisfied, so x = 4, y = 1 , or (4, 1) , is a solution of the system of equations.

or (1, −1, 2) , is a solution of the system of equations. 3x + 3 y + 2 z = 4  17.  x − 3 y + z = 10 5 x − 2 y − 3 z = 8  Substituting the values of the variables: 3 ( 2 ) + 3 ( −2 ) + 2 ( 2 ) = 6 − 6 + 4 = 4  2 − 3 ( −2 ) + 2 = 2 + 6 + 2 = 10  5 ( 2 ) − 2 ( −2 ) − 3 ( 2 ) = 10 + 4 − 6 = 8 Each equation is satisfied, so x = 2 , y = −2 , z = 2 , or (2, −2, 2) is a solution of the system of equations.

x + y = 8 19.  x − y = 4 Solve the first equation for y, substitute into the second equation and solve: y = 8− x  x − y = 4

x − (8 − x) = 4 x −8+ x = 4 2 x = 12 x=6 Since x = 6, y = 8 − 6 = 2 . The solution of the system is x = 6, y = 2 or using ordered pairs (6, 2) .

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Chapter 12: Systems of Equations and Inequalities

= 24 3x 23.  x 2 y + = 0  Solve the first equation for x and substitute into the second equation: x=8   x + 2 y = 0 8 + 2y = 0 2y = −8 y = −4 The solution of the system is x = 8, y = − 4 or

using ordered pairs (8, −4) 5 x − y = 21 21.  2 x + 3 y = −12 Multiply each side of the first equation by 3 and add the equations to eliminate y: 15 x − 3 y = 63  2 x + 3 y = −12  = 51 x=3 Substitute and solve for y: 5(3) − y = 21 15 − y = 21 −y = 6 y = −6 The solution of the system is x = 3, y = −6 or 17 x

3x − 6 y = 2 25.  5 x + 4 y = 1 Multiply each side of the first equation by 2 and each side of the second equation by 3, then add to eliminate y:  6 x − 12 y = 4  15 x + 12 y = 3

using ordered an pair ( 3, −6 ) .

=7 1 x= 3 Substitute and solve for y: 3 (1/ 3) − 6 y = 2 21x

1− 6y = 2 −6y = 1 y=−

1 6

1 1 The solution of the system is x = , y = − or 3 6 1 1 using ordered pairs  , −  . 3 6

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Section 12.1: Systems of Linear Equations: Substitution and Elimination 4 x + 2(2 x) = 12 4 x + 4 x = 12 8 x = 12 3 x= 2 3 3 Since x = , y = 2   = 3 2 2

The solution of the system is x =

3 , y = 3 or 2

3  using ordered pairs  ,3  . 2 

 2x + y = 1 27.  4 x + 2 y = 3 Solve the first equation for y, substitute into the second equation and solve:  y = 1− 2x  4 x + 2 y = 3 4 x + 2(1 − 2 x) = 3 4x + 2 − 4x = 3 0 =1 This equation is false, so the system is inconsistent.

 x + 2y = 4 31.  2 x + 4 y = 8 Solve the first equation for x, substitute into the second equation and solve: x = 4 − 2 y  2 x + 4 y = 8 2(4 − 2 y ) + 4 y = 8 8 − 4y + 4y = 8 0=0 These equations are dependent. The solution of the system is either x = 4 − 2 y , where y is any real 4− x , where x is any real number. 2 Using ordered pairs, we write the solution as {( x, y) x = 4 − 2 y, y is any real number} or as

2 x − y = 0 29.  4 x + 2 y = 12 Solve the first equation for y, substitute into the second equation and solve:  y = 2x  4 x + 2 y = 12

number or y =

 4− x  , x is any real number  .  ( x, y ) y = 2  

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Chapter 12: Systems of Equations and Inequalities

Substitute and solve for x:

 2 x − 3 y = −1 33.  10 x + y = 11 Multiply each side of the first equation by –5, and add the equations to eliminate x: −10 x + 15 y = 5  10 x + y = 11 

1 1 x + (3) = 3 2 3 1 x +1 = 3 2 1 x=2 2 x=4

16 y = 16 y =1 Substitute and solve for x: 2 x − 3(1) = −1 2 x − 3 = −1 2x = 2 x =1 The solution of the system is x = 1, y = 1 or using ordered pairs (1, 1).

The solution of the system is x = 4, y = 3 or using ordered pairs (4, 3).  3x − 5 y = 3 39.  15 x + 5 y = 21 Add the equations to eliminate y:  3 x − 5 y = 3  15 x + 5 y = 21 4 3 Substitute and solve for y: 3 ( 4 / 3) − 5 y = 3 x=

4 − 5y = 3 −5 y = −1 y=

1 5

4 1 The solution of the system is x = , y = or 3 5 4 1 using ordered pairs  ,  .  3 5

5y = 5 y =1

Since y = 1, x = 1 +

= 24

18 x

2 x + 3 y = 6  35.  1  x − y = 2 Solve the second equation for x, substitute into the first equation and solve: 2 x + 3 y = 6  1   x = y + 2 1  2 y +  + 3y = 6 2  2y +1+ 3y = 6

1 3 = . The solution of the 2 2

1 1 x + y = 8  41.  3 − 5 = 0  x y

3 system is x = , y = 1 or using ordered pairs 2

1 1  2 x + 3 y = 3 37.   1 x − 2 y = −1  4 3 Multiply each side of the first equation by –6 and each side of the second equation by 12, then add to eliminate x: −3 x − 2 y = −18  3x − 8 y = −12 

Rewrite letting u =

1 1 , v= : x y

 u+ v=8  3u − 5v = 0 Solve the first equation for u, substitute into the second equation and solve: u = 8 − v  3u − 5v = 0

− 10 y = −30 y= 3

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Section 12.1: Systems of Linear Equations: Substitution and Elimination 3(8 − v) − 5v = 0

Multiply each side of the first result by

24 − 3v − 5v = 0

add to the second result to eliminate y: 4 y − 4 z = −8 −4 y + 7 z = 11

−8v = −24 v=3

Since v = 3, u = 8 − 3 = 5 . Thus, x =

1 1 = , u 5

3z = 3 z =1 Substituting and solving for the other variables:

1 1 = . The solution of the system is v 3 1 1 1 1 x = , y = or using ordered pairs  ,  . 5 3 5 3 y=

y −1 = − 2 y = −1

ordered triples (2, −1, 1) .  x − y − z =1  47. 2 x + 3 y + z = 2  3x + 2 y =0 

Multiply each side of the first equation by –2 and add to the second equation to eliminate x: −2 x + 2 y = −12 − 3 z = 16

Add the first and second equations to eliminate z: x − y − z =1

2 y − 3z = 4 Multiply each side of the result by –1 and add to the original third equation to eliminate y: −2 y + 3z = − 4 2y + z = 4

2x + 3y + z = 2 3x + 2 y =3 Multiply each side of the result by –1 and add to the original third equation to eliminate y: −3x − 2 y = −3

4z = 0 z=0 Substituting and solving for the other variables: 2y + 0 = 4 2 x − 3(0) = 16 2y = 4 2 x = 16

3x + 2 y = 0 0 = −3 This equation is false, so the system is inconsistent.

 x− y− z = 1  49. − x + 2 y − 3 z = − 4  3x − 2 y − 7 z = 0  Add the first and second equations to eliminate x; multiply the first equation by –3 and add to the third equation to eliminate x: x− y− z = 1 − x + 2 y − 3z = − 4

y=2 x =8 The solution is x = 8, y = 2, z = 0 or using ordered triples (8, 2, 0).

 x − 2 y + 3z = 7  45.  2 x + y + z = 4 −3x + 2 y − 2 z = −10 Multiply each side of the first equation by –2 and add to the second equation to eliminate x; and multiply each side of the first equation by 3 and add to the third equation to eliminate x: −2 x + 4 y − 6 z = −14 2x + y + z =

x − 2(−1) + 3(1) = 7 x+2+3= 7

x=2 The solution is x = 2, y = −1, z = 1 or using

 x− y = 6  43.  2 x − 3 z = 16 2 y + z = 4 

2x

4 and 5

y − 4z = − 3 −3x + 3 y + 3 z = −3

4

3x − 2 y − 7 z = 0

5 y − 5 z = − 10

y − 4 z = −3

3x − 6 y + 9 z = 21 −3x + 2 y − 2 z = −10

Multiply each side of the first result by –1 and add to the second result to eliminate y:

− 4 y + 7 z = 11

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Chapter 12: Systems of Equations and Inequalities − y + 4z = 3

Multiply each side of the first result by –1 and add to the second result to eliminate y: −4 x + y = −1 7x − y = 4

y − 4 z = −3 0= 0 The system is dependent. If z is any real number, then y = 4 z − 3 . Solving for x in terms of z in the first equation: x − (4 z − 3) − z = 1

3x = 3 x =1 Substituting and solving for the other variables: 4(1) − y = 1 3(1) − 2(3) + z = −5 3 − 6 + z = −5 − y = −3 y=3 z = −2 The solution is x = 1, y = 3, z = − 2 or using ordered triplets (1, 3, −2) .

x − 4z + 3 − z = 1 x − 5z + 3 = 1 x = 5z − 2 The solution is {( x, y , z ) x = 5 z − 2, y = 4 z − 3 ,

z is any real number}.

 x + 2 y − z = −3  55.  2 x − 4 y + z = −7 − 2 x + 2 y − 3z = 4 Add the first and second equations to eliminate z; multiply the second equation by 3 and add to the third equation to eliminate z: x + 2y − z = − 3 2x − 4 y + z = − 7 3x − 2 y = −10

 2 x − 2 y + 3z = 6  51.  4 x − 3 y + 2 z = 0 − 2 x + 3 y − 7 z = 1  Multiply the first equation by –2 and add to the second equation to eliminate x; add the first and third equations to eliminate x: −4 x + 4 y − 6 z = −12 4x − 3y + 2z =

0

y − 4 z = −12

6 x − 12 y + 3z = − 21 − 2 x + 2 y − 3z = 4

2 x − 2 y + 3z = 6 − 2x + 3y − 7z = 1

4 x − 10 y

y − 4z = 7 Multiply each side of the first result by –1 and add to the second result to eliminate y: − y + 4 z = 12

Multiply each side of the first result by –5 and add to the second result to eliminate y: −15 x + 10 y = 50 4 x − 10 y = −17

y−4z = 7

−11x

= 33 x= −3 Substituting and solving for the other variables: 3(−3) − 2 y = −10 − 9 − 2 y = −10 − 2 y = −1 1 y= 2

0 = 19 This result is false, so the system is inconsistent.

 x+ y− z = 6  53. 3 x − 2 y + z = −5  x + 3 y − 2 z = 14  Add the first and second equations to eliminate z; multiply the second equation by 2 and add to the third equation to eliminate z: x+ y− z = 6 3x − 2 y + z = −5 4x − y

= 1

6 x − 4 y + 2 z = −10 x + 3 y − 2 z = 14 7x − y

=

= − 17

4

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Section 12.1: Systems of Linear Equations: Substitution and Elimination 63. Let s = the price of a smartphone and t = the price of a tablet. Then: s + t = 965

1 −3 + 2   − z = −3 2 −3 + 1 − z = −3

340 s + 250t = 270500 Solve the first equation for t: t = 965 − s Solve by substitution: 340 s + 250(965 − s ) = 270500 340 s + 241250 − 250 s = 270500 90 s = 29250 s = 325 t = 965 − 325 = 640 The price of the smartphone is $325.00 and the price of the tablet is $640.00.

− z = −1 z =1 1 The solution is x = −3, y = , z = 1 or using 2 1   ordered triplets  −3, , 1 . 2  

57. Let l be the length of the rectangle and w be the width of the rectangle. Then: l = 2w and 2l + 2w = 90

Solve by substitution: 2(2 w) + 2w = 90 4 w + 2w = 90 6w = 90 w = 15 feet l = 2(15) = 30 feet The floor is 15 feet by 30 feet.

65. Let x = the plane’s average airspeed and y = the average wind speed. Rate

Time Distance

With Wind

x+ y

3

600

Against

x− y

4

600

 ( x + y )(3) = 600  ( x − y )(4) = 600

59. Let x = the number of commercial launches and y = the number of noncommercial launches. Then: x + y = 81 and y = 2 x + 12

Multiply each side of the first equation by

1 , 3

multiply each side of the second equation by

Solve by substitution: y = 2(23) + 12 x + (2 x + 12) = 81 3 x = 69 y = 46 + 12 x = 23 y = 58 In 2013 there were 23 commercial launches and 58 noncommercial launches.

1 , 4

and add the result to eliminate y x + y = 200 x − y = 150 2 x = 350 x = 175 175 + y = 200 y = 25 The average airspeed of the plane is 175 mph, and the average wind speed is 25 mph.

61. Let x = the number of pounds of cashews. Let y = is the number of pounds in the mixture. The value of the cashews is 5x . The value of the peanuts is 1.50(30) = 45. The value of the mixture is 3y . Then x + 30 = y represents the amount of mixture. 5 x + 45 = 3 y represents the value of the mixture.

67. Let x = the number of $25-design. Let y = the number of $45-design. Then x + y = the total number of sets of dishes. 25 x + 45 y = the cost of the dishes.

Solve by substitution: 5 x + 45 = 3( x + 30) 2 x = 45 x = 22.5 So, 22.5 pounds of cashews should be used in the mixture.

Setting up the equations and solving by substitution:  x + y = 200  25 x + 45 y = 7400 Solve the first equation for y, the solve by substitution: y = 200 − x

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Chapter 12: Systems of Equations and Inequalities 2 ( 50 ) + 4 y = 400

25 x + 45(200 − x ) = 7400 25 x + 9000 − 45 x = 7400

100 + 4 y = 400

− 20 x = −1600

4 y = 300

x = 80 y = 200 − 80 = 120 Thus, 80 sets of the $25 dishes and 120 sets of the $45 dishes should be ordered.

300 = 75 4 So 50 mg of compound 1 should be mixed with 75 mg of compound 2. y=

69. Let x = the cost per package of bacon. Let y = the cost of a carton of eggs. Set up a system of equations for the problem: 3x + 2 y = 13.45  2 x + 3 y = 11.45 Multiply each side of the first equation by 3 and each side of the second equation by –2 and solve by elimination: 9 x + 6 y = 40.35 − 4 x − 6 y = −22.90 5x

=

73. y = ax 2 + bx + c At (–1, 4) the equation becomes: 4 = a (–1) 2 + b(−1) + c 4 = a −b+c

At (2, 3) the equation becomes: 3 = a(2) 2 + b(2) + c 3 = 4a + 2b + c At (0, 1) the equation becomes: 1 = a(0) 2 + b(0) + c 1= c

17.45

x = 3.49

The system of equations is:  a− b+c = 4  4a + 2b + c = 3  c=1 

Substitute and solve for y: 3(3.49) + 2 y = 13.45 10.47 + 2 y = 13.45 2 y = 2.98 y = 1.49 A package of bacon costs $3.49 and a carton of eggs cost $1.49. The refund for 2 packages of bacon and 2 cartons of eggs will be 2($3.49) + 2($1.49) = $9.96.

Substitute c = 1 into the first and second equations and simplify: 4a + 2b + 1 = 3 a − b +1 = 4 a− b =3 4a + 2b = 2 a = b+3 Solve the first result for a, substitute into the second result and solve: 4(b + 3) + 2b = 2 4b + 12 + 2b = 2 6b = −10 5 b=− 3 5 4 a = − +3= 3 3 4 5 The solution is a = , b = − , c = 1 . The 3 3 4 5 equation is y = x 2 − x + 1 . 3 3

71. Let x = the # of mg of compound 1. Let y = the # of mg of compound 2. Setting up the equations and solving by substitution: 0.2 x + 0.4 y = 40 vitamin C   0.3 x + 0.2 y = 30 vitamin D

Multiplying each equation by 10 yields 2 x + 4 y = 400  6 x + 4 y = 600 Subtracting the bottom equation from the top equation yields 2 x + 4 y − ( 6 x + 4 y ) = 400 − 600 2 x − 6 x = −200 −4 x = −200 x = 50

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Section 12.1: Systems of Linear Equations: Substitution and Elimination

0.06Y − 5000r = 240 75.  0.06Y + 6000r = 900 Multiply the first equation by −1 , the add the result to the second equation to eliminate Y. −0.06Y + 5000r = −240 0.06Y + 6000r = 900

 10  55 65 I2 =   + =  71  71 71 10 65 55 The solution is I1 = , I 2 = , I 3 = . 71 71 71

79. Let x = the number of orchestra seats. Let y = the number of main seats. Let z = the number of balcony seats. Since the total number of seats is 500, x + y + z = 500 . Since the total revenue is $64,250 if all seats are sold, 150 x + 135 y + 110 z = 64, 250 . If only half of the orchestra seats are sold, the revenue is $56,750. 1 So, 150 x + 135 y + 110 z = 56, 750 . 2 Thus, we have the following system: x + y + z = 500 150 x + 135 y + 110 z = 64, 250 75 x + 135 y + 110 z = 56, 750

11000r = 660

r = 0.06 Substitute this result into the first equation to find Y. 0.06Y − 5000(0.06) = 240 0.06Y − 300 = 240 0.06Y = 540 Y = 9000 The equilibrium level of income and interest rates is $9000 million and 6%. I 2 = I1 + I 3   77.  5 − 3I1 − 5I 2 = 0 10 − 5I − 7 I = 0 2 3 

Substitute the expression for I 2 into the second and third equations and simplify: 5 − 3I1 − 5( I1 + I 3 ) = 0

Multiply each side of the first equation by –110 and add to the second equation to eliminate z; multiply each side of the third equation by –1 and add to the second equation to eliminate z:

−8 I1 − 5I 3 = −5

−110 x − 110 y − 110 z = −55, 000

10 − 5( I1 + I 3 ) − 7 I 3 = 0

150 x + 135 y + 110 z = 64, 250

−5I1 − 12 I 3 = −10

40x + 25 y

Multiply both sides of the first result by 5 and multiply both sides of the second result by –8 to eliminate I1 :

=

9250

150 x + 135 y + 110 z = 64, 250

−75 x − 135 y − 110 z = −56, 750

−40 I1 − 25I 3 = −25

75 x

40 I1 + 96 I 3 = 80

= 7500 x = 100

71I 3 = 55

Substituting and solving for the other variables: 40(100) + 25 y = 9250 100 + 210 + z = 500 4000 + 25 y = 9250 310 + z = 500 25 y = 5250 z = 190 y = 210

55 I3 = 71

Substituting and solving for the other variables:  55  − 8I1 − 5   = −5  71  275 − 8I1 − = −5 71 80 −8 I1 = − 71 10 I1 = 71

There are 100 orchestra seats, 210 main seats, and 190 balcony seats.

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Chapter 12: Systems of Equations and Inequalities

A system involving only 2 equations that contain 3 or more unknowns cannot be solved uniquely. 1 Multiply the first equation by − and the 2 1 second equation by , then add to eliminate y: 2 −4 x − 3 y − 3 z = −13.05 5 x + 3 y + 4 z = 15.80

81. Let x = the number of servings of chicken. Let y = the number of servings of corn. Let z = the number of servings of 2% milk.

Protein equation: 30 x + 3 y + 9 z = 66 Carbohydrate equation: 35 x + 16 y + 13 z = 94.5 Calcium equation: 200 x + 10 y + 300 z = 910 Multiply each side of the first equation by –16 and multiply each side of the second equation by 3 and add them to eliminate y; multiply each side of the second equation by –5 and multiply each side of the third equation by 8 and add to eliminate y: −480 x − 48 y − 144 z = −1056 105 x + 48 y + 39 z = 283.5 − 375 x

x = 2.75 − z

Substitute and solve for y in terms of z: 5 ( 2.75 − z ) + 3 y + 4 z = 15.80 13.75 + 3 y − z = 15.80 3 y = z + 2.05

− 105 z = − 772.5

−175 x − 80 y −

1 41 z+ 3 60 Solutions of the system are: x = 2.75 − z , 1 41 y = z+ . 3 60 Since we are given that 0.60 ≤ z ≤ 0.90 , we choose values of z that give two-decimal-place values of x and y with 1.75 ≤ x ≤ 2.25 and 0.75 ≤ y ≤ 1.00 . The possible values of x, y, and z are shown in the table.

65 z = − 472.5

y=

1600 x + 80 y + 2400 z = 7280 1425 x

+ z = 2.75

x

+ 2335 z = 6807.5

Multiply each side of the first result by 19 and multiply each side of the second result by 5 to eliminate x: −7125 x − 1995 z = −14, 677.5 7125 x + 11, 675 z = 34, 037.5 9680 z = 19,360 z=2

Substituting and solving for the other variables: −375 x − 105(2) = −772.5 −375 x − 210 = −772.5 −375 x = −562.5 x = 1.5 30(1.5) + 3 y + 9(2) = 66 45 + 3 y + 18 = 66 3y = 3 y =1

The dietitian should serve 1.5 servings of chicken, 1 serving of corn, and 2 servings of 2% milk.

x

y

z

2.13

0.89

0.62

2.10

0.90

0.65

2.07

0.91

0.68

2.04

0.92

0.71

2.01

0.93

0.74

1.98

0.94

0.77

1.95

0.95

0.80

1.92

0.96

0.83

1.89

0.97

0.86

1.86

0.98

0.89

83. Let x = the price of 1 hamburger. Let y = the price of 1 order of fries. Let z = the price of 1 drink.

We can construct the system  8 x + 6 y + 6 z = 26.10  10 x + 6 y + 8 z = 31.60 732 Copyright © 2016 Pearson Education, Inc.

Section 12.1: Systems of Linear Equations: Substitution and Elimination 85. Let x = Beth’s time working alone. Let y = Bill’s time working alone. Let z = Edie’s time working alone.

1 1 1 1 + + = x y z 10 1 1 1 + = y z 15

We can use the following tables to organize our work:

1 x

Beth Bill Edie Hours to do job Part of job done in 1 hour

x

y

z

1 x

1 y

1 z

Substitute x = 30 into the third equation: 12 12 4 + + =1 30 y z . 12 4 3 + = y z 5 Now consider the system consisting of the last result and the second original equation. Multiply the second original equation by –12 and add it to the last result to eliminate y:

In 10 hours they complete 1 entire job, so 1 1 1 10  + +  = 1 x y z 1 1 1 1 + + = x y z 10

Bill Edie Hours to do job

y

z

Part of job done

1 y

1 z

in 1 hour

−12 + y 12 + y

In 15 hours they complete 1 entire job, so 1 1 15  +  = 1 .  y z

in 1 hour

y

z

1 x

1 y

1 z

−12 15 3 5

8 3 =− z 15 z = 40

Plugging z = 40 to find y: 12 4 3 + = y z 5 12 4 3 + = y 40 5 12 1 = y 2 y = 24 Working alone, it would take Beth 30 hours, Bill 24 hours, and Edie 40 hours to complete the job.

Beth Bill Edie x

−12 = z 4 = z

−

1 1 1 + = y z 15 Hours to do job Part of job done

1 30 x = 30 =

With all 3 working for 4 hours and Beth and Bill working for an additional 8 hours, they complete 1 1 1 1 1 1 entire job, so 4  + +  + 8  +  = 1 x y z x y 12 12 4 + + =1 x y z

87.

Answers will vary.

We have the system  1 1 1 1  + + =  x y z 10  1 1 1 + =  y z 15  12 12 4  + + =1 y z x Subtract the second equation from the first equation: 733 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities 7. Writing the augmented matrix for the system of equations:  x − 5y = 5  1 −5 5 →    4 x 3 y 6 + =  4 3 6

89.

91. sin −1 sin −

10π 9

(

 2x + 3y − 6 = 0 9.  4 x − 6 y + 2 = 0 Write the system in standard form and then write the augmented matrix for the system of equations: 2x + 3 y = 6  3 6 → 2   4 x − 6 y = − 2 4 − 6 − 2

follows the form of the

)

(

11. Writing the augmented matrix for the system of equations:  0.01x − 0.03 y = 0.06  0.01 − 0.03 0.06  →   0.10 0.20  0.13x + 0.10 y = 0.20 0.13

)

equation f −1 f ( x ) = sin −1 sin ( x ) = x , but we cannot use the formula directly since −

10π 9

 π π is not in the interval  − ,  . We need to find  2 2  π π an angle θ in the interval  − ,  for which  2 2

13. Writing the augmented matrix for the system of equations:  x − y + z = 10  1 −1 1 10   = 5 → 3 3 0 5 3x + 3 y  x + y + 2z = 2   1 1 2 2 

10π 10π = sin θ . The angle − is in 9 9 quadrant II so sine is positive. The reference 10π π angle of − is and we want θ to be in 9 9 quadrant I so sine will still be positive. Thus, we 10π π π = sin . Since is in the have sin − 9 9 9 sin −

15. Writing the augmented matrix for the system of equations: 1 −1 2   x+ y−z = 2 1   3 2 2 3 2 0 2  = →  −  x− y 5 x + 3 y − z = 1 3 −1 1  5 17. Writing the augmented matrix for the system of equations:  1 −1 −1 10   x − y − z = 10   2 x + y + 2 z = −1   2 1 2 −1 →   −3 4 0 5   −3 x + 4 y = 5    4 x − 5 y + z = 0  4 −5 1 0 

 π π interval  − ,  , we can apply the equation  2 2 above and get sin −1 sin −

10π 9

= sin −1 sin

π 9

=

π 9

.

 1 −3 −2   x − 3 y = −2 19.  →  2 −5 5   2 x − 5 y = 5

Section 12.2

R2 = − 2r1 + r2  1 −3 −2   1 −3 −2   →  2 5 5 2(1) 2 2( 3) 5 2( 2) 5 − − + − − − − +    

1. matrix 3. third; fifth

1 −3 −2  →  0 1 9 

5. b

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Section 12.2: Systems of Linear Equations: Matrices R3 = 3r1 + r3

 1 −3 4 3   x − 3 y + 4 z = 3  21.  3 −5 6 6  →  3 x − 5 y + 6 z = 6  −5 3 4 6  −5 x + 3 y + 4 z = 6

 1 −3 2 −6   0 1 −1 8     −3 −6 4 6  −3 2 −6   1   0 1 8 → −1   3(1) − 3 3(−3) − 6 3(2) + 4 3(−6) + 6 

R2 = − 3r1 + r2  1 −3 4  3 −5 6   −5 3 4  1 →  −3(1) + 3   −5

3 6  6 

 1 −3 2 −6  → 0 1 −1 8    0 −15 10 −12 

4 3 −3  −3(−3) − 5 −3(4) + 6 −3(3) + 6   3 4 6 

 5 −3 1 −2   5 x − 3 y + z = −2  25.  2 −5 6 −2  →  2 x − 5 y + 6 z = −2   −4 1 4 6  −4 x + y + 4 z = 6

 1 −3 4 3  →  0 4 −6 −3    −5 3 4 6 

R3 = 5r1 + r3

R1 = − 2r2 + r1

 1 −3 4 3   0 4 −6 −3     −5 3 4 6  −3 4 3   1 → 0 −6 −3  4   5(1) − 5 5(−3) + 3 5(4) + 4 5(3) + 6   1 −3 4 3  → 0 4 −6 −3   0 −12 24 21

 5 −3 1  2 −5 6   −4 1 4  −2(2) + 5 2 →   −4

 1 7 −11 2  →  2 −5 6 −2    4 6   −4 1

R3 = 2r2 + r3

 1 −3 2 −6   x − 3 y + 2 z = −6  23.  2 −5 3 −4  →  2 x − 5 y + 3z = −4   −3 −6 4 6  −3x − 6 y + 4 z = 6

R2 = − 2r1 + r2  1 −3 2  2 −5 3   −3 −6 4 1  →  −2(1) + 2   −3

−2  −2   6  −2( −5) − 3 −2(6) + 1 −2(−2) − 2   6 −5 −2  1 4 6 

−6  −4   6  −3 −6 2  −2(−3) − 5 −2(2) + 3 −2(−6) − 4   −6 4 6 

 1 −3 2 −6  →  0 1 −1 8     −3 −6 4 6 

 1 7 −11 2   2 −5 6 −2    4 6   −4 1 −11 1 7 2   → −5 −2  2 6    2(2) + (−4) 2( −5) + 1 2(6) + 4 2(−2) + 6  1 7 −11 2  →  2 −5 6 −2    2   0 −9 16

x = 5 27.   y = −1 Consistent; x = 5, y = −1, or using ordered pairs (5, −1) . x = 1  29.  y = 2 0 = 3  Inconsistent

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Chapter 12: Systems of Equations and Inequalities

x + y = 8 39.  x − y = 4 Write the augmented matrix: 1 1 8 1 8 1  →  ( R2 = −r1 + r2 ) 1 −1 4  0 − 2 − 4   1 1 8 → ( R2 = − 12 r2 )  0 1 2  0 6 → 1 ( R1 = −r2 + r1 )  0 1 2  The solution is x = 6, y = 2 or using ordered pairs (6, 2).

 x + 2 z = −1  31.  y − 4 z = − 2  0=0  Consistent;  x = −1 − 2 z   y = −2 + 4 z  z is any real number 

or {( x, y, z ) | x = −1 − 2 z , y = − 2 + 4 z, z is any real number} x1 = 1   33.  x2 + x4 = 2 x + 2x = 3 4  3 Consistent;  x1 = 1   x2 = 2 − x4   x3 = 3 − 2 x4  x4 is any real number

2 x − 4 y = − 2 41.  3x + 2 y = 3 Write the augmented matrix:  2 − 4 − 2   1 − 2 −1 1  →  ( R1 = 2 r1 ) 3 3 3 2 2 3   →  1 − 2 −1 ( R2 = −3r1 + r2 ) 8 6 0  1 − 2 −1 R2 = 18 r2 → 3  0 1 4   1 0 12   ( R1 = 2r2 + r1 ) →  0 1 34  1 3 The solution is x = , y = or using ordered pairs 2 4 1 3  , . 2 4

or {( x1 , x2 , x3 , x4 ) | x1 = 1, x2 = 2 − x4 , x3 = 3 − 2 x4 , x4 is any real number}

(

x1 + 4 x4 = 2   35.  x2 + x3 + 3x4 = 3  0=0  Consistent;  x1 = 2 − 4 x4   x2 = 3 − x3 − 3 x4  x , x are any real numbers  3 4 or {( x1 , x2 , x3 , x4 ) | x1 = 2 − 4 x4 , x2 = 3 − x3 − 3x4 , x3 and x4 are any real numbers}

)

 x + 2y = 4 43.  2 x + 4 y = 8 Write the augmented matrix:  1 2 4  1 2 4 ( R2 = − 2r1 + r2 )  →   2 4 8  0 0 0  This is a dependent system. x + 2y = 4 x = 4 − 2y The solution is x = 4 − 2 y, y is any real number or {( x, y ) | x = 4 − 2 y, y is any real number}

 x1 + x4 = −2   x + 2 x4 = 2 37.  2  x3 − x4 = 0  0=0 Consistent;  x1 = −2 − x4   x2 = 2 − 2 x4   x3 = x4  x4 is any real number

or {( x1 , x2 , x3 , x4 ) | x1 = −2 − x4 , x2 = 2 − 2 x4 , x3 = x4 , x4 is any real number} 736 Copyright © 2016 Pearson Education, Inc.

Section 12.2: Systems of Linear Equations: Matrices

2 x + 3 y = 6  45.  1  x − y = 2 Write the augmented matrix:  2 3 6  1 3 3 2   →  1 −1 12  1 −1 1  2

 1 −1 0 6     2 0 −3 16  0 2 1 4    1 −1 0 6    →  0 2 −3 4  0 2 1 4  

( R1 = 12 r1 )

3 1 3 2  ( R2 = − r1 + r2 ) → 0 − 5 − 5  2 2  1 3 3 2 R2 = − 52 r2 →  0 1 1 3  R1 = − 32 r2 + r1 → 1 0 2 0 1 1

(

1  → 0  0 1  → 0  0

)

(

)

3 3  The solution is x = , y = 1 or  , 1 . 2 2   3x − 5 y = 3 47.  15 x + 5 y = 21 Write the augmented matrix:  1 −5  3 −5 3 1 3  → 15 5 21 5 21 15

0 1 0

0 6  − 32 2   1 4  − 32 8  − 32 2   4 0 − 32 8  − 32 2   1 0

( R2 = 12 r2 )  R1 = r2 + r1    = − + R 2 r r 2 3  3

( R3 = 14 r3 )

 1 0 0 8  R1 = 32 r3 + r1      → 0 1 0 2  R2 = 3 r3 + r2  0 0 1 0 2     The solution is x = 8, y = 2, z = 0 or (8, 2, 0).

 − 53 1 → 1  ( R2 = −15r1 + r2 )  0 30 6   − 53 1 1  R2 = 30 r2 → 1 1  0 1 5  0 4 3 R1 = 53 r2 + r1 → 1 1  0 1  5

(

1 2

 0 1 → 0 1  0 0

( R1 = 13 r1 )

(

−1

( R2 = − 2r1 + r2 )

 x − 2 y + 3z = 7  51.  2 x + y + z = 4 −3x + 2 y − 2 z = −10  Write the augmented matrix:  1 −2 3 7   1 1 4  2  −3 2 − 2 −10   1 −2 3 7    R2 = − 2r1 + r2  5 −5 −10  → 0    R3 = 3r1 + r3  0 − 4 7  11   1 −2 3 7   R2 = 15 r2 → 0 1 −1 − 2  0 − 4 7 11  

)

)

4 1 4 1 The solution is x = , y = or  ,  . 3 5 3 5 = 6  x− y  49. 2 x − 3 z = 16  2y + z = 4  Write the augmented matrix:

(

1 0 1 3   → 0 1 −1 − 2  0 0 3 3 

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)

 R1 = 2r2 + r1     R3 = 4r2 + r3 

Chapter 12: Systems of Equations and Inequalities

1 0 1 3   R3 = 13 r3 → 0 1 −1 − 2  0 0 1 1   1 0 0 2    R1 = − r3 + r1  → 0 1 0 −1    R2 = r3 + r2  0 0 1 1  

(

 −1 1 1 −1   2 −3 − 4   −1  3 − 2 −7 0    1 −1 −1 1   2 −3 − 4  ( R1 = −r1 ) →  −1  3 − 2 −7 0    1 −1 −1 1   R = r +r  → 0 1 − 4 −3  2 1 2  R = −3r1 + r3  0 1 − 4 −3  3  

)

The solution is x = 2, y = −1, z = 1 or (2, −1, 1) .  2x − 2 y − 2z = 2  53. 2 x + 3 y + z = 2  3x + 2 y =0  Write the augmented matrix: 2 − 2 − 2 2   3 1 2 2 3 0 0 2    1 −1 −1 1   →  2 3 1 2 ( R1 = 12 r1 )  3 2 0 0    1 −1 −1 1    R2 = − 2r1 + r2  → 0 5 3 0    R3 = −3r1 + r3  0 5 3 −3  

 1 0 −5 − 2    R = r +r  → 0 1 − 4 −3  1 2 1  R = − r2 + r3  0 0 0 0   3  The matrix in the last step represents the system  x − 5z = − 2  x = 5z − 2    y − 4 z = −3 or, equivalently,  y = 4 z − 3  0 = 0 0=0  

The solution is x = 5 z − 2 , y = 4 z − 3 , z is any real number or {( x, y, z ) | x = 5 z − 2, y = 4 z − 3, z is any real number}.  2 x − 2 y + 3z = 6  57.  4 x − 3 y + 2 z = 0 − 2 x + 3 y − 7 z = 1  Write the augmented matrix:  2 −2 3 6    4 −3 2 0  − 2 3 −7 1    1 −1 3 3 2   R1 = 12 r1 ) →  4 −3 2 0  (    − 2 3 −7 1 3  1 −1 3 2    R2 = − 4r1 + r2  → 0 1 − 4 −12       R3 = 2r1 + r3  4 − 0 7 1  

 1 −1 −1 1   → 0 5 3 0 ( R3 = −r2 + r3 ) 0 0 0 −3   There is no solution. The system is inconsistent. − x + y + z = − 1  55.  − x + 2 y − 3 z = − 4  3x − 2 y − 7 z = 0  Write the augmented matrix:

1 0 − 5 −9  2    R1 = r2 + r1  → 0 1 − 4 −12       R3 = −r2 + r3  0 0 0 19   There is no solution. The system is inconsistent.

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Section 12.2: Systems of Linear Equations: Matrices

 x+ y− z = 6  59. 3 x − 2 y + z = −5  x + 3 y − 2 z = 14  Write the augmented matrix: 1 1 −1 6    1 −5 3 − 2 3 − 2 14   1 1 1  → 0 −5 0 2 1 1  → 0 1  0 2  1 0 → 0 1  0 0

6 −1  4 − 23 8 −1 −1 6   − 54 23 5   −1 8 7 − 15 5 23  4 −5 5  3 − 6 5 5

7 1 0 − 1 5 5  23  4  − → 0 1 5 5   − 2 0 0 1 1 0 0 1   → 0 1 0 3 0 0 1 − 2  The solution is x = 1,

 1 0 − 1 − 13  4 4  3 1  − → 0 1 8 8   11 −  0 0 − 11  4 4  1 0 − 1 − 13  4 4  3 1 → 0 1 − 8 8   0 0 1 1  1 0 0 −3   1 → 0 1 0 2    0 0 1 1

 R2 = − 3r1 + r2     R3 = − r1 + r3 

( R2 =

− 15 r2

( R3 = − 114 r3 )  R1 = 14 r3 + r1    3    R2 = 8 r3 + r2  1

 

1

 

The solution is x = −3, y = , z = 1 or  −3, , 1 . 2 2

)

2   3x + y − z = 3  63. 2 x − y + z = 1  8  4x + 2 y = 3  Write the augmented matrix:  3 1 −1 2  3   2 −1 1 1  8  4 2 0 3 

 R1 = − r2 + r1     R3 = − 2r2 + r3 

( R3 = 53 r3 )  R1 = 15 r3 + r1     R2 = 4 r3 + r2  5  

1 1 − 1 2 3 3 9  →  2 −1 1 1  8 0 3  4 2 

y = 3, z = − 2 , or (1, 3, −2) .

 x + 2 y − z = −3  61.  2 x − 4 y + z = −7 − 2 x + 2 y − 3z = 4  Write the augmented matrix:  1 2 −1 −3   1 −7   2 −4 − 2 2 −3 4  

 1 2 −1 −3   → 0 − 8 3 −1 0 6 −5 − 2    1 2 −1 −3   3 1 → 0 1 − 8 8   0 6 −5 − 2 

 R1 = − 2r2 + r1     R3 = − 6r2 + r3 

1  → 0  0  1  → 0  0  1  → 0  0

 R2 = − 2r1 + r2     R3 = 2r1 + r3 

( R2 = − 18 r2 )

1 3 5 −3 2 3

− 13 5 3 4 3

1 3

− 13

1

−1

2 3

4 3

0

0

1 −1 0 2

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2 9 5 9  16  9

 2 9 − 13   16  9  1 3 − 13   2

( R1 = 13 r1 )  R2 = − 2r1 + r2     R3 = − 4r1 + r3 

( R2 = − 53 r2 )  R1 = − 13 r2 + r1     R3 = − 2 r2 + r3  3  

Chapter 12: Systems of Equations and Inequalities

1 0 0  → 0 1 −1  0 0 1 1 0 0  → 0 1 0  0 0 1

1 3 − 13 

 1

1 3 2 3

 1

( R3 = 12 r3 ) ( R2 = r3 + r2 )

1 2 1 2  The solution is x = , y = , z = 1 or  , , 1 . 3 3 3 3 

0 −1 −3 − 2   6 1 2 4 0 1 2 2  0 3 13 13

1  0 → 0  0

0 0 −1 0   1 0 0 2 0 1 2 2  0 0 7 7 0 0 −1 0   1 0 0 2 0 1 2 2  0 0 1 1

1  0 → 0  0

 x+ y+ z+ w= 4  2x − y + z = 0  65.  3x + 2 y + z − w = 6  x − 2 y − 2 z + 2 w = −1 Write the augmented matrix: 1 1 1 1 4   1 0 0  2 −1 3 2 1 −1 6     1 − 2 − 2 2 −1 1 1 1 1 4   0 −3 −1 − 2 − 8 → 0 −1 − 2 − 4 − 6    1 −5 0 −3 −3 1 1 1 1 4   0 −1 − 2 − 4 − 6  →  0 −3 −1 − 2 − 8   1 −5  0 −3 −3 1 1 1 1 4   0 6 1 2 4 →  0 −3 −1 − 2 − 8   1 −5  0 −3 −3  1 0 −1 −3 − 2    0 1 2 4 6 → 0 0 5 10 10    0 0 3 13 13

1  0 → 0  0

( R3 = 15 r3 )  R1 = r3 + r1     R2 = − 2 r3 + r2   R = −3 r + r  3 4   4

( R4 = 17 r4 )

 1 0 0 0 1    0 1 0 0 2   R1 = r4 + r1 → 0 0 1 0 0   R3 = − 2 r4 + r3    0 0 0 1 1 The solution is x = 1, y = 2, z = 0, w = 1 or (1, 2, 0, 1).  x + 2y + z =1  67. 2 x − y + 2 z = 2  3x + y + 3z = 3 

 R2 = − 2r1 + r2     R3 = − 3r1 + r3  R = −r + r   4 1 4 

Write the augmented matrix:  1 2 1 1    2 −1 2 2   3 1 3 3    1 2 1 1    R2 = − 2r1 + r2  → 0 −5 0 0     R3 = −3r1 + r3  0 −5 0 0   

 Interchange     r2 and r3 

( R2 = −r2 )

 1 2 1 1   → 0 −5 0 0  ( R3 = −r2 + r3 ) 0 0 0 0    The matrix in the last step represents the system x + 2y + z = 1   − 5y = 0  0=0  Substitute and solve: x + 2(0) + z = 1 −5 y = 0 y=0 z = 1− x The solution is y = 0, z = 1 − x, x is any real

 R1 = − r2 + r1     R3 = 3 r2 + r3   R = 3r + r  2 4  4

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Section 12.2: Systems of Linear Equations: Matrices

number or {( x, y, z ) | y = 0, z = 1 − x, x is any real number}.

1 −1 −1 0    0 1 −7 −7  →  ( R2 = −2r3 + r2 ) 0 2 4 5    0 0 0 0  1 0 −8 −7    0 1 −7 −7   R1 = r2 + r1   → 0 0 18 19   R3 = −2r2 + r3    0 0 0 0 

x− y+ z= 5  69.  3x + 2 y − 2 z = 0 Write the augmented matrix: 1 −1 1 5  3 2 −2 0    1 −1 1 5  → ( R2 = − 3r1 + r2 )   0 5 −5 −15 1 −1 1 5  → R2 = 15 r2   0 1 −1 −3 1 0 0 2  → ( R1 = r2 + r1 )   0 1 −1 −3

(

0 −8 −7   1 −7 −7  1 R3 = 18 r3  0 1 19 18  0 0 0 The matrix in the last step represents the system  x − 8 z = −7  y − 7 z = −7   z = 19  18 Substitute and solve:  19   19  y − 7  = 7 x − 8   = −7  18   18  7 13 y= x= 18 9 13 7 19 , y= , z= or Thus, the solution is x = 9 18 18  13 7 19   , , .  9 18 18  1  0 → 0  0

)

The matrix in the last step represents the system x=2 x = 2  or, equivalently,    y − z = −3 y = z −3 Thus, the solution is x = 2 , y = z − 3 , z is any real number or {( x, y, z ) | x = 2, y = z − 3, z is any real number}. 2 x + 3 y − z = 3  x− y−z =0  71.   −x + y + z = 0  x + y + 3 z = 5 Write the augmented matrix:  2 3 −1 3     1 −1 −1 0   −1 1 1 0     1 1 3 5  1 −1 −1 0   2 3 −1 3 interchange    →   −1 1 1 0   r1 and r2   1 1 3 5  

(

)

 4x + y + z − w = 4 73.   x − y + 2 z + 3w = 3 Write the augmented matrix:  4 1 1 −1 4    1 −1 2 3 3 

1 −1 2 3 3  →    4 1 1 −1 4 

1 −1 −1 0   0 5 1 3   R2 = −2r1 + r2    R3 = r1 + r3  →    0 0 0 0     0 2 4 5   R4 = −r1 + r4    1 −1 −1 0    0 5 1 3  interchange  →  0 2 4 5   r3 and r4     0 0 0 0 

 interchange     r1 and r2 

1 −1 2 3 3 → ( R2 = −4r1 + r2 )  0 5 −7 −13 −8 The matrix in the last step represents the system  x − y + 2 z + 3w = 3   5 y − 7 z − 13w = −8 The second equation yields

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Chapter 12: Systems of Equations and Inequalities 5 y − 7 z − 13w = −8

1 0 12 − 12    → 0 1 1 5  → ( R3 = − 12 r3 ) 2 2   0 0 1 3  1 0 0 −2   R1 = − 12 r3 + r1      → 0 1 0 1   R2 = − 1 r3 + r2    2 0 0 1 3 

5 y = 7 z + 13w − 8 7 13 8 z + w− 5 5 5 The first equation yields x − y + 2 z + 3w = 3 x = 3 + y − 2 z − 3w Substituting for y: 13   8 7 x = 3 +  − + z + w  − 2 z − 3w 5 5 5   3 2 7 x = − z − w+ 5 5 5 3 2 7 Thus, the solution is x = − z − w + , 5 5 5 7 13 8 y = z + w − , z and w are any real numbers or 5 5 5  3 2 7 7 13 8 ( x, y, z , w) x = − z − w + , y = z + w − , 5 5 5 5 5 5  y=

The solution is a = − 2, b = 1, c = 3 ; so the equation is y = −2 x 2 + x + 3 . 77. Each of the points must satisfy the equation f ( x) = ax3 + bx 2 + cx + d . f (−3) = −112 : −27 a + 9b − 3c + d = −112 f (−1) = −2 : −a + b − c + d = −2 f (1) = 4 : a+b+c+d = 4 f (2) = 13 : 8a + 4b + 2c + d = 13 Set up a matrix and solve:  − 27 9 −3 1 −112    − 2  −1 1 −1 1  1 1 1 1 4   13  8 4 2 1  1 1 1 1 4   − 2   Interchange  −1 1 −1 1 →   − 27 9 −3 1 −112   r3 and r1    13  8 4 2 1 1 1 1 1 4     R2 = r1 + r2   0 0 2 2 2   R = 27 r1 + r3  → 0 36 24 28 − 4   3    R4 = − 8 r1 + r4  0 − 4 − 6 −7 −19 

 z and w are any real numbers  .  75. Each of the points must satisfy the equation y = ax 2 + bx + c . (1, 2) : 2 = a+b+c (−2, −7) : − 7 = 4a − 2b + c − 3 = 4a + 2b + c (2, −3) : Set up a matrix and solve: 1 1 1 2    4 − 2 1 −7   4 2 1 −3 1 1 1 2    R2 = − 4r1 + r2  →  0 − 6 −3 −15  R = − 4r + r  1 3   3 0 − 2 −3 −11 1 1 1 2   5 1 R2 = − 16 r2 → 0 1 2 2    0 − 2 −3 −11 1 1 0 − 12  2    R1 = − r2 + r1  5 1 → 0 1 2 2      R3 = 2 r2 + r3  0 0 − 2 − 6

(

1 1 1 1 4   0 0 1 1 1 → 0 36 24 28 − 4    0 − 4 − 6 −7 −19 

)

1  0 → 0  0

0

1  0 →  0  0

0

( R2 = 12 r2 )

3  1 1 1 0 24 − 8 − 40   0 − 6 −3 −15

 R1 = − r2 + r1     R3 = −36 r2 + r3     R4 = 4 r2 + r4 

3  1 1 1 5 5 0 1 − 3 − 3 0 − 6 −3 −15

( R3 = 241 r3 )

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1 0

1 0

0

0

Section 12.2: Systems of Linear Equations: Matrices

1  0 → 0  0 1  0 → 0  0

0 0

1 3

14  3

1 5 1 0 1 − 3 − 3  0 0 −5 − 25 1 0

1

0 0

1 3

14  3

1  5 1 0 1 − 3 − 3  0 0 5 1 1 0

1

Substitute z = 1 and solve: −198 y − 295(1) = − 691 −198 y = −396 y=2

 R1 = − r3 + r1     R4 = 6 r3 + r4 

x + 10(2) + 16(1) = 37.5 x + 36 = 37.5 x = 1.5

( R4 = − 15 r4 )

The dietitian should serve 1.5 servings of salmon steak, 2 servings of baked eggs, and 1 serving of acorn squash.

1 0 0 0 3  R = − 1 r + r  1 1 3 4    0 1 0 0 − 4  →  R2 = −r4 + r2  0 0 1 0 0      R3 = 13 r4 + r3  5  0 0 0 1 The solution is a = 3, b = − 4, c = 0, d = 5 ; so the

81. Let x = the amount invested in Treasury bills. Let y = the amount invested in Treasury bonds. Let z = the amount invested in corporate bonds. Total investment equation: x + y + z = 10, 000 Annual income equation: 0.06 x + 0.07 y + 0.08 z = 680 Condition on investment equation: z = 0.5 x x − 2z = 0 Set up a matrix and solve:

equation is f ( x) = 3 x 3 − 4 x 2 + 5 . 79. Let x = the number of servings of salmon steak. Let y = the number of servings of baked eggs. Let z = the number of servings of acorn squash. Protein equation: 30 x + 15 y + 3 z = 78 Carbohydrate equation: 20 x + 2 y + 25 z = 59 Vitamin A equation: 2 x + 20 y + 32 z = 75 Set up a matrix and solve:  30 15 3 78    20 2 25 59   2 20 32 75  2 20 32 75    Interchange  →  20 2 25 59    r3 and r1   30 15 3 78   1 10 16 37.5   →  20 2 25 59  ( R1 = 12 r1 )  30 15 3 78 1 10 16 37.5    R2 = −20r1 + r2  → 0 −198 −295 −691   0 −285 −477 −1047   R3 = −30r1 + r3  1 10 16 37.5   R3 = − 95 r + r3 → 0 −198 −295 −691 66 2 3457 3457   0 − 66 − 66  0 1 10 16 37.5   66 → 0 −198 −295 −691 R3 = − 3457 r3 0  0 1 1 

(

(

 1 1 1 10,000    0.06 0.07 0.08 680    1 0 −2 0  

1 1 1 10,000    → 0 0.01 0.02 80  0 −3 −10,000  −1 

 R2 = − 0.06 r1 + r2     R3 = − r1 + r3 

 1 1 1 10,000    → 0 1 2 8000  0 −1 −3 −10,000   

( R2 = 100 r2 )

1  → 0 0  1  → 0 0 

0 −1 2000   8000  1 2 0 −1 − 2000 

0 −1 2000   1 2 8000  0 1 2000   1 0 0 4000    → 0 1 0 4000  0 0 1 2000   

)

 R1 = − r2 + r1     R3 = r2 + r3 

( R3 = − r3 )  R1 = r3 + r1    = − + R r r 2 3 2  2

Carletta should invest $4000 in Treasury bills, $4000 in Treasury bonds, and $2000 in corporate bonds.

)

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Chapter 12: Systems of Equations and Inequalities 83. Let x = the number of Deltas produced. Let y = the number of Betas produced. Let z = the number of Sigmas produced. Painting equation: 10 x + 16 y + 8 z = 240 Drying equation: 3x + 5 y + 2 z = 69 Polishing equation: 2 x + 3 y + z = 41 Set up a matrix and solve: 10 16 8 240   3 5 2 69     2 3 1 41   1 1 2 33 →  3 5 2 69   2 3 1 41 1 1 2 33  → 0 2 −4 −30  0 1 −3 −25 1 1 2 33  → 0 1 −2 −15 0 1 −3 −25 1 → 0 0 1 → 0 0

0  1 1  1

( R1 = −3r2 + r1 )  R2 = −3r1 + r2     R3 = −2r1 + r3 

( R2 = 12 r2 )

0 4 48  1 −2 −15  0 −1 −10 

 R1 = r1 − r2     R3 = r3 − r2 

0 4 48  1 −2 −15 0 1 10 

( R3 = −r3 )

1 0 0 8   R1 = −4r3 + r1   → 0 1 0 5     R2 = 2r3 + r2   0 0 1 10  The company should produce 8 Deltas, 5 Betas, and 10 Sigmas.

2 0 0 0 0 5 0 3 0 0 −1 −1

4  8 4  0

1  0 → 1  1

0

1  0 → 0  0

0

5 8  0 1 2 0 −5 − 4   0 −1 − 6 − 8

 R2 = 12 r2     R3 = − r1 + r3   R = − r + r  1 4  4

1  0 → 0  0

5 8  0 1 2 0 −1 − 6 − 8  0 3 −5 − 4 

 Interchange     r3 and r4 

1  0 → 0  0

0 0 5 8  0 1 0 2 0 1 6 8  0 0 − 23 − 28

1  0 → 0  0

0 0 5 1 0 0 0 1 6

1  0 → 0  0

5 8  0 4 2 0 0 4  0 −1 −1 0 

0

0 0 3

0 0 3

0 0

8  2 8  28 0 0 1 23  0 0 0 1 0 0

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 R3 = − r3    3 R r r = − + 3 4  4

( R4 = − 231 r4 )

44  23

 2 16  0 1 0 23  28  0 0 1 23  44 The solution is I1 = , 23 28 I4 = . 23

85. Rewrite the system to set up the matrix and solve: 2I2 = 4 − 4 + 8 − 2 I 2 = 0    8 = 5 I 4 + I1   I1 + 5I 4 = 8 →  4 = 3I 3 + I1   I1 + 3I 3 = 4   I1 − I 3 − I 4 = 0 I 3 + I 4 = I1

 Interchange     r2 and r1 

 R1 = −5 r4 + r1     R3 = − 6 r4 + r3 

I 2 = 2 , I3 =

16 , 23

Section 12.2: Systems of Linear Equations: Matrices 87. Let x = the amount invested in Treasury bills. Let y = the amount invested in corporate bonds. Let z = the amount invested in junk bonds. a.

 1 1 1 25, 000     0.07 0.09 0.11 2000 

Total investment equation: x + y + z = 20, 000 Annual income equation: 0.07 x + 0.09 y + 0.11z = 2000 Set up a matrix and solve:  1 1 1 20, 000     0.07 0.09 0.11 2000  1 → 7 1 → 0

1 1 20, 000   9 11 200, 000  1 1 20, 000   2 4 60, 000 

1 1 1 20, 000  →   0 1 2 30, 000 

( R2 = 100r2 )

1 → 7 1 → 0

1 1 25, 000   9 11 200, 000  1 1 25, 000   2 4 25, 000 

1 → 0 1 → 0

1 1 25, 000   1 2 12,500  0 −1 12,500   1 2 12,500 

( R2 = 100r2 ) ( R2 = r2 − 7r1 )

( R2 = 12 r2 ) ( R1 = r1 − r2 )

The matrix in the last step represents the  x − z = 12,500 system   y + 2 z = 12,500 Thus, the solution is x = z + 12,500 , y = −2 z + 12,500 , z is any real number.

( R2 = r2 − 7r1 )

( R2 = 12 r2 )

1 0 −1 −10, 000  →  ( R1 = r1 − r2 )  0 1 2 30, 000  The matrix in the last step represents the  x − z = −10, 000 system   y + 2 z = 30, 000 Therefore the solution is x = −10, 000 + z , y = 30, 000 − 2 z , z is any real number.

Possible investment strategies:

Possible investment strategies:

Amount Invested At

Amount Invested At

7%

9%

11%

0

10,000

10,000

1000

8000

11,000

2000

6000

12,000

3000

4000

13,000

4000

2000

14,000

5000

0

15,000

c.

b. Total investment equation: x + y + z = 25, 000 Annual income equation: 0.07 x + 0.09 y + 0.11z = 2000 Set up a matrix and solve:

7%

9%

11%

12,500

12,500

0

14,500

8500

2000

16,500

4500

4000

18,750

0

6250

Total investment equation: x + y + z = 30, 000 Annual income equation: 0.07 x + 0.09 y + 0.11z = 2000 Set up a matrix and solve:  1 1 1 30, 000     0.07 0.09 0.11 2000  1 → 7 1 → 0

1 1 30, 000   9 11 200, 000 

( R2 = 100r2 )

1 1 30, 000   2 4 −10, 000 

( R1 = r2 − 7r1 )

1 → 0 1 → 0

1 1 30, 000   1 2 −5000  0 −1 35, 000   1 2 −5000 

( R2 = 12 r2 ) ( R1 = r1 − r2 )

The matrix in the last step represents the 745 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities

 x − z = 35, 000 system   y + 2 z = −5000 Thus, the solution is x = z + 35, 000 , y = −2 z − 5000 , z is any real number. However, y and z cannot be negative. From y = −2 z − 5000 , we must have y = z = 0.

Therefore the solution is x = 50 −

1 y = 75 + z , z is any real number. 8

Possible combinations:

One possible investment strategy Amount Invested At

7%

9%

11%

30,000

0

0

This will yield ($30,000)(0.07) = $2100, which is more than the required income.

Supplement 1

Supplement 2

Supplement 3

50mg

75mg

0mg

36mg

76mg

8mg

22mg

77mg

16mg

8mg

78mg

24mg

91 – 93. Answers will vary.

d. Answers will vary. 89. Let x = the amount of supplement 1. Let y = the amount of supplement 2. Let z = the amount of supplement 3. 0.20 x + 0.40 y + 0.30 z = 40 Vitamin C  0.30 x + 0.20 y + 0.50 z = 30 Vitamin D Multiplying each equation by 10 yields 2 x + 4 y + 3z = 400  3x + 2 y + 5 z = 300

Set up a matrix and solve:  2 4 3 400   3 2 5 300    3 1 2 2 200  R1 = 12 r1 →  3 2 5 300   3 1 2 2 200   ( R2 = r2 − 3r1 ) → 0 −4 1 −300    2

(

1 2 3 200  2 →  1 0 1 − 75  8 

7 z, 4

)

( R2 = − 14 r2 )

1 0 7 50  4 →  ( R1 = r1 − 2r2 ) 1 0 1 − 8 75 The matrix in the last step represents the system  x + 74 z = 50  1  y − 8 z = 75

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Section 12.3: Systems of Linear Equations: Determinants

95.

R( x) =

2 x2 − x − 1 x2 + 2x + 1

=

(2 x + 1)( x − 1) ( x + 1)( x + 1)

Domain: { x x ≠ −1} . R ( x) =

p ( x) = 2 x 2 − x − 1; q ( x) = x 2 + 2 x + 1;

2 x2 − x − 1

is in lowest terms. x2 + 2 x + 1 2 ⋅ 02 − 0 − 1 −1 = = −1 . Plot the point ( 0, −1) . The y-intercept is f (0) = 2 0 + 2⋅0 +1 1 1 The x-intercepts are the zeros of p ( x) : 1 and − . 2 2 x2 − x − 1 R( x) = 2 is in lowest terms. The vertical asymptotes are the zeros of q( x) : x + 2x + 1 x = −1 . Graph this asymptote with dashed lines. 2 Since n = m , the line y = = 2 is the horizontal asymptote. Solve to find intersection points: 1 Plot the line y = 2 using dashes. Graph:

97. cos −1 ( −0.75) = 2.42 radians

Section 12.3 9.

1. ad − bc

−3 −1 4

2

3. False; If ad=bc, the the det = 0. 5. False; See Thm (14) 7.

6 4 = 6(3) − (−1)(4) = 18 + 4 = 22 −1 3

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= −3(2) − 4(−1) = − 6 + 4 = −2

Chapter 12: Systems of Equations and Inequalities

3

11.

4 2 5 5 5 = 3 −1 1 −1 −4 1 + 2 1 −1 2 2 − − 2 1 1 2 1 2 −2 = 3 ( −1) (− 2) − 2(5)  − 4 [1(− 2) − 1(5) ] + 2 [1(2) − 1( −1) ] = 3( − 8) − 4( −7) + 2(3) = − 24 + 28 + 6 = 10

13.

= 4 [ −1(4) − 0(−3) ] + 1[ 6(4) − 1(0)]

+ 2[ 6( −3) − 1(−1) ]

= 4(− 4) + 1(24) + 2(−17) = −16 + 24 − 34 = − 26

= 6−0 = 6

1 2

24 0 = 48 − 0 = 48 0 2

3 24 = 0 − 24 = − 24 1 0 Find the solutions by Cramer's Rule: D y − 24 D 48 x= x = =8 y= = = −4 6 6 D D The solution is (8, −4) .

3 −6 5

4

= 12 − (−30) = 42

−6 Dx = 24 = 96 − (−72) = 168 12 4

8

1 = − 8 − 4 = −12 4 −1

3 24 = 36 − 120 = − 84 5 12 Find the solutions by Cramer's Rule: Dy − 84 D 168 =4 = = −2 x= x = y= 42 42 D D The solution is (4, −2) . Dy =

8 Dy = 1 = 4−8 = −4 1 4 Find the solutions by Cramer's Rule: Dy − 4 D −12 x= x = y= =6 = =2 D D −2 −2 The solution is (6, 2).

 5 x − y = 13 17.  2 x + 3 y = 12

Dy =

3 0

Dy =

D=

D = 1 1 = −1 − 1 = − 2 1 −1

Dx =

D=

 3 x − 6 y = 24 21.  5 x + 4 y = 12

x + y = 8 15.  x − y = 4

D=

= 24 3 x 19.   x + 2y = 0

Dx =

4 −1 2 0 6 0 6 −1 − (−1) +2 6 −1 0 = 4 −1 −3 4 1 4 1 −3 1 −3 4

Dx =

Dy 34 Dx 51 = =3 y= = =2 D 17 D 17 The solution is (3, 2). x=

3x − 2 y = 4 23.  6 x − 4 y = 0 3 −2 = −12 − (−12) = 0 6 −4 Since D = 0 , Cramer's Rule does not apply. D=

5 −1 = 15 + 2 = 17 2 3 13 −1 = 39 + 12 = 51 12 3 5 13

= 60 − 26 = 34 2 12 Find the solutions by Cramer's Rule:

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Section 12.3: Systems of Linear Equations: Determinants

2x − 4 y = − 2 25.  3 x + 2 y = 3

15 Dx − 2 3 = = x= −5 D 2

−4 = 4 + 12 = 16 D= 2 3 2 Dx =

−2 −4 3

2

= −4 + 12 = 8

D= Dx =

4 1

 x+ y− z = 6  33. 3 x − 2 y + z = −5  x + 3 y − 2 z = 14 

2 −1 =10 − (−10) = 20 10 5 Find the solutions by Cramer's Rule: Dy 20 2 D 5 1 x= x = = y= = = D 50 10 D 50 5  1 2 The solution is  ,  .  10 5 

1 1 −1 D = 3 −2 1 − 2 3 1 1 −1 3 1 + (−1) 3 − 2 =1 −2 3 −2 3 1 −2 1 = 1(4 − 3) − 1(− 6 − 1) − 1(9 + 2) = 1 + 7 − 11 = −3

2 x + 3 y = 6  29.  1  x − y = 2

6 1 −1 Dx = −5 − 2 1 3 −2 14

D = 2 3 = − 2 − 3 = −5 1 −1

Dy =

−1

= −6 −

3 −5 = 15 − (−105) = 120 21 5

The solution is  ,  .  3 5

Dy =

3

3 −5 = 15 − (−75) = 90 15 5

3 3 = 63 − 45 = 18 15 21 Find the solutions by Cramer's Rule: Dy 18 1 D 120 4 x= x = = y= = = D D 90 5 90 3

−3 = − 10 − (−15) = 5 Dx = −1 5 10

1 2

−5 =1 −5

Dy =

2 −3 = 20 − (−30) = 50 10 10

6

D

=

 3x − 5 y = 3 31.  15 x + 5 y = 21

 2 x − 3 y = −1 27.  10 x + 10 y = 5

Dx =

Dy

3  The solution is  , 1 . 2 

−2 = 6 + 6 = 12 Dy = 2 3 3 Find the solutions by Cramer's Rule: Dy 12 3 D 8 1 x= x = = y= = = D 16 2 D 16 4 1 3 The solution is  ,  . 2 4

D=

y=

1 − 1 −5 1 + (−1) −5 − 2 = 6 −2 3 −2 3 14 − 2 14 = 6(4 − 3) − 1(10 − 14) − 1(−15 + 28) = 6 + 4 − 13 = −3

3 15 =− 2 2

2 6

= 1 − 6 = −5 1 1 2 Find the solutions by Cramer's Rule:

749 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities

1 6 Dy = 3 −5

1 −3 −1 Dy = 2 −7 1 −2 4 −3

−1

1 1 14 − 2

1 −6 3 1 + (−1) 3 −5 14 − 2 1 −2 1 14 = 1(10 − 14) − 6(− 6 − 1) − 1(42 + 5)

=1

−5

1 − (−3) 2 1 + (−1) 2 −7 − 2 −3 −2 4 4 −3 = 1(21 − 4) + 3(− 6 + 2) − 1(8 − 14) =1

−7

= −4 + 42 − 47

= 17 − 12 + 6

= −9

= 11

1 1 6 2 − Dz = 3 −5 1 =1

Dz =

3 14

− 2 −5

−1

3 −5

+6

3 −2

3 14 3 1 14 1 = 1(− 28 + 15) − 1(42 + 5) + 6(9 + 2) = −13 − 47 + 66 =6 Find the solutions by Cramer's Rule: Dy −9 D −3 x= x = =1 y= = =3 D −3 D −3 D 6 z= z = = −2 D −3 The solution is (1, 3, −2) .

1 2 −3 2 − 4 −7 −2 2 4

= 1 − 4 −7 − 2 2 −7 + (−3) 2 − 4 −2 −2 2 4 4 2 = 1(−16 + 14) − 2(8 − 14) − 3(4 − 8) = − 2 + 12 + 12 = 22 Find the solutions by Cramer's Rule: Dy 11 1 D − 66 x= x = = −3 y= = = D D 22 2 22 D 22 z= z = =1 D 22 1   The solution is  −3, , 1 . 2  

 x + 2 y − z = −3  35.  2 x − 4 y + z = −7 − 2 x + 2 y − 3z = 4 

 x − 2 y + 3z = 1  37.  3x + y − 2 z = 0  2x − 4 y + 6z = 2 

1 2 −1 1 D = 2 −4 −2 2 −3

3 1 −2 D= 3 1 −2 6 2 −4

1 −2 2 1 + (−1) 2 − 4 =1 −4 2 − −2 3 3 − − 2 2 = 1(12 − 2) − 2(− 6 + 2) − 1(4 − 8) = 10 + 8 + 4 = 22 −3 2 −1 Dx = −7 − 4 1 4 2 −3

1 − 2 − (− 2) 3 − 2 + 3 3 1 6 6 −4 2 2 −4 = 1(6 − 8) + 2(18 + 4) + 3(−12 − 2) = − 2 + 44 − 42 =0 Since D = 0 , Cramer's Rule does not apply. =1

1 − 2 −7 1 + (−1) −7 − 4 = −3 − 4 2 −3 4 −3 4 2 = −3(12 − 2) − 2(21 − 4) − 1(−14 + 16) = −30 − 34 − 2 = − 66

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Section 12.3: Systems of Linear Equations: Determinants

 x + 2y − z = 0  39.  2 x − 4 y + z = 0 − 2 x + 2 y − 3z = 0 

43.

By Theorem (11), the value of a determinant changes sign if any two rows are interchanged. 1 2 3 Thus, u v w = − 4 . x y z

1 2 −1 D = 2 −4 1 −2 2 −3 1 −2 2 1 + (−1) 2 − 4 =1 −4 − 2 −3 −2 2 −3 2 = 1(12 − 2) − 2(− 6 + 2) − 1(4 − 8) = 10 + 8 + 4 = 22 0 2 −1 Dx = 0 − 4 1 = 0 [By Theorem (12)] 0 2 −3 1 0 −1 Dy = 2 0 1 =0 − 2 0 −3 1 2 0 Dz = 2 − 4 0 = 0 −2 2 0

x y z u v w =4 1 2 3

x y z 45. Let u v w = 4 . 1 2 3 x y z x y z −3 − 6 −9 = − 3 1 2 3 u v w u v w

[Theorem (14)]

x y z = −3( −1) u v w [Theorem (11)] 1 2 3

[By Theorem (12)]

= 3(4) = 12

x y z 47. Let u v w = 4 1 2 3

[By Theorem (12)]

Find the solutions by Cramer's Rule: Dy D 0 0 x= x = =0 y= = =0 D 22 D 22 D 0 z= z = =0 D 22 The solution is (0, 0, 0).

1 2 3 x −3 y −6 z −9 2u 2v 2w 1 2 3 = 2 x −3 y −6 z −9 u v w

 x − 2 y + 3z = 0  41.  3x + y − 2 z = 0  2x − 4 y + 6z = 0 

= 2(−1)

3 1 −2 2 − D= 3 1 6 2 −4

x −3 y −6 z −9 1 2 3 u v w

[Theorem (14)]

[Theorem (11)]

x−3 y −6 z −9 = 2(−1)(−1) u v w [Theorem (11)] 1 2 3

1 − 2 − (− 2) 3 − 2 + 3 3 1 −4 6 6 2 2 −4 = 1(6 − 8) + 2(18 + 4) + 3(−12 − 2) = − 2 + 44 − 42 =0 Since D = 0 , Cramer's Rule does not apply. =1

x y z = 2(−1)(−1) u v w 1 2 3 = 2(−1)(−1)(4) = 8

751 Copyright © 2016 Pearson Education, Inc.

[Theorem (15)] ( R1 = −3r3 + r1 )

Chapter 12: Systems of Equations and Inequalities 55. Solve for x:

x y z 49. Let u v w = 4 1 2 3

x 2 3 0 =7 1 x 6 1 −2

1 2 3 2x 2y 2z u −1 v − 2 w − 3 1 2 3 =2 x y z u −1 v − 2 w − 3 x y z = 2(−1) 1 2 3 u −1 v − 2 w − 3

0 −2 1 0 +3 1 x = 7 x x 6 −2 6 1 1 −2 x ( − 2 x ) − 2 ( − 2 ) + 3 (1 − 6 x ) = 7 − 2 x 2 − 18 x = 0 − 2x ( x + 9) = 0 x = 0 or x = −9

[Theorem (11)]

x y z = 2(−1)(−1) u − 1 v − 2 w − 3 [Theorem (11)] 1 2 3 x y z = 2(−1)(−1) u v w 1 2 3 = 2(−1)(−1)(4) =8

− 2 x 2 + 4 + 3 − 18 x = 7

[Theorem (14)]

[Theorem (15)] ( R2 = −r3 + r2 )

57. Expanding the determinant: x x1 x2 x

y1 1 x 1 x −y 1 +1 1 y2 1 x2 1 x2

y1 =0 y2

x( y1 − y2 ) − y ( x1 − x2 ) + ( x1 y2 − x2 y1 ) = 0

x( y1 − y2 ) + y ( x2 − x1 ) = x2 y1 − x1 y2

y ( x2 − x1 ) = x2 y1 − x1 y2 + x( y2 − y1 )

y ( x2 − x1 ) − y1 ( x2 − x1 )

51. Solve for x: x

x =5 3 4 3x − 4 x = 5 −x = 5 x = −5

= x2 y1 − x1 y2 + x( y2 − y1 ) − y1 ( x2 − x1 )

( x2 − x1 ) ( y − y1 ) = x( y2 − y1 ) + x2 y1 − x1 y2 − y1 x2 + y1 x1

( x2 − x1 ) ( y − y1 ) = ( y2 − y1 ) x − ( y2 − y1 ) x1 ( x2 − x1 ) ( y − y1 ) = ( y2 − y1 )( x − x1 )

53. Solve for x: x

1 1 4 3 2 =2 −1 2 5 x

y 1 y1 1 = 0 y2 1

( y2 − y1 ) ( x − x1 ) ( x2 − x1 ) This is the 2-point form of the equation for a line. ( y − y1 ) =

3 2 4 2 4 3 −1 +1 =2 2 5 −1 5 −1 2 x (15 − 4 ) − ( 20 + 2 ) + ( 8 + 3) = 2 11x − 22 + 11 = 2 11x = 13 13 x= 11

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Section 12.3: Systems of Linear Equations: Determinants 59.

If the vertices of a triangle are (2, 3), (5, 2), and (6, 5), then: 2 5 6 1 D= 3 2 5 2 1 1 1

−240 − 120 x + 80 y + 20 x 2 + 20 y 2 = 0 20 x 2 − 120 x + 20 y 2 + 80 y = 240 x 2 − 6 x + y 2 + 4 y = 12 x 2 − 6 x + 9 + y 2 + 4 y + 4 = 12 + 9 + 4

( x − 3)2 + ( y + 2)2 = 25

3 5 3 2  1 2 5 = 2 −5 +6  1 1 1 1  2 1 1 1 = [ 2(2 − 5) − 5(3 − 5) + 6(3 − 2) ] 2 1 = [ 2(−3) − 5(−2) + 6(1) ] 2 1 = [ −6 + 10 + 6] 2 =5 The area of the triangle is 5 = 5 square units.

61. A =

1 2

(

6 8 1 6 −1 3 6 −2 8 4 + + + + −1 3 1 6 6 −2 8 4 6 8

65. If a = 0, then b ≠ 0 and c ≠ 0 since

by = s  ad − bc ≠ 0 , and the system is  . cx + dy = t s The solution of the system is y = , b

t − dy t − d ( b ) tb − sd = = . Using Cramer’s c c bc 0 b Rule, we get D = = −bc , c d s b Dx = = sd − tb , t d 0 s Dy = = 0 − sc = − sc , so c t D ds − tb td − sd x= x = = and D −bc bc Dy − sc s y= = = , which is the solution. Note D −bc b that these solutions agree if d = 0. s

x=

)

1 [(36−8)+(3+ 6)+(2−18)+(24+16)+(64− 24)] 2 1 = [ 28 + 9 − 16 + 40 + 40] = 50.5 square units 2 =

1

63.

2

2

2

x

7

3

6

y

−5

3

2

x2 + y2 7 = −5

74 18 40

3

6

1

3

2 − x −5

74 18 40

1

1

3

2

If b = 0, then a ≠ 0 and d ≠ 0 since =s ax ad − bc ≠ 0 , and the system is  . cx + dy = t

74 18 40

1

1

1

+y 7

3

6 − x2 + y2

74 18 40

(

)

(

s , a

1

1 1

The solution of the system is x =

7

3 6

t − cx at − cs . Using Cramer’s Rule, we = d ad a 0 s 0 get D = = ad , Dx = = sd , and c d t d a s D sd s = Dy = = at − cs , so x = x = c t D ad a Dy at − cs = , which is the solution. and y = D ad Note that these solutions agree if c = 0. y=

−5 3 2

)

= ( −240) − x (120) + y (80) − x 2 + y 2 ( −20) = −240 − 120 x + 80 y + 20 x 2 + 20 y 2

Now set this expression equal to 0. Then complete the square to obtain the standard form.

If c = 0, then a ≠ 0 and d ≠ 0 since ax + by = s ad − bc ≠ 0 , and the system is  . dy = t  753 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities

The solution of the system is y =

t , d

s − by sd − tb = . Using Cramer’s Rule, we a ad a b s b get D = = ad , Dx = = sd − tb , 0 d t d a s D sd − tb = at , so x = x = and Dy = and 0 t D ad Dy at t y= = = , which is the solution. Note D ad d that these solutions agree if b = 0. x=

If d = 0, then b ≠ 0 and c ≠ 0 since ax + by = s . ad − bc ≠ 0 , and the system is  =t cx t The solution of the system is x = , c s − ax cs − at y= = . Using Cramer’s Rule, we b bc a b = 0 − bc = −bc , get D = c 0 Dx = Dy =

s b t

0

a s c

t

= 0 − tb = −tb , and = at − cs , so x =

Dy

at − cs cs − at = = , which is the D −bc bc solution. Note that these solutions agree if a = 0. y=

67. Evaluating the determinant to show the relationship: a11 a12 ka21 ka22 a31 a32 = a11

ka22 a32

a13 ka23 a33 ka23 ka ka23 ka ka22 − a12 21 + a13 21 a33 a31 a33 a31 a32

= a11 ( ka22a33 − ka23a32 ) − a12 (ka21a33 − ka23a31 ) + a13 ( ka21a32 − ka22a31 ) = ka11 (a22a33 − a23a32 ) − ka12 (a21a33 − a23a31 ) + ka13 (a21a32 − a22a31 ) = k ( a11 (a22a33 − a23a32 ) − a12 ( a21a33 − a23a31 ) + a13 (a21a32 − a22a31 ) )

 = k  a11  a11 = k a21 a31

a22 a32 a12 a22 a32

Dx −tb t = = and D −bc c

754 Copyright © 2016 Pearson Education, Inc.

a23 a a a a  − a12 21 23 + a13 21 22  a33 a31 a33 a31 a32  a13 a23 a33

Section 12.4: Matrix Algebra 69. Evaluating the determinant to show the relationship: a11 + ka21 a12 + ka22 a21 a22 a31 a32 = (a11 + ka21 )

a22 a32

+ (a13 + ka23 )

Section 12.4 1. square

a13 + ka23 a23 a33

3. false

a23 a a − ( a12 + ka22 ) 21 23 a33 a31 a33

5. True

a21 a22 a31 a32

0 0 3 −5  4 1 9. A + B =   +  − 2 3 − 2 1 2 6     −5 + 0   0 + 4 3 +1 = 1 ( 2) 2 3 6 (− 2)  + − + + 

7. a

= (a11 + ka21 )(a22a33 − a23a32 ) − (a12 + ka22 )( a21a33 − a23a31 ) + (a13 + ka23 )(a21a32 − a22a31 )

 4 4 −5  =   −1 5 4 

= a11 ( a22a33 − a23a32 ) + ka21 (a22a33 − a23a32 ) − a12 (a21a33 − a23a31 ) − ka22 ( a21a33 − a23a31 )

0 11. 4 A = 4  1 4 ⋅ 0 =  4 ⋅1

+ a13 (a21a32 − a22a31 ) + ka23 ( a21a32 − a22a31 ) = a11 ( a22a33 − a23a32 ) + ka21a22a33 − ka21a23a32 − a12 ( a21a33 − a23a31 ) − ka22a21a33 + ka22a23a31 + a13 (a21a32 − a22a31 ) + ka23a21a32

 0 12 − 20  = 24  4 8

− ka23a22a31 = a11 ( a22a33 − a23a32 ) − a12 (a21a33 − a23a31 )

0  0 3 −5  4 1 13. 3 A − 2 B = 3  − 2    1 2 6  − 2 3 − 2 0  0 9 −15  8 2 = −    3 6 18  − 4 6 − 4 

+ a13 (a21a32 − a22a31 ) = a11

a22 a32

a11 a12 = a21 a22 a31 a32

71.

73.

3 −5  2 6  4 ⋅ 3 4(−5)  4⋅2 4 ⋅ 6 

a23 a a a a − a12 21 23 + a13 21 22 a33 a31 a33 a31 a32 a13 a23 a33

 − 8 7 −15 =   7 0 22   4 1  0 3 −5    ⋅   6 2 1 2 6    − 2 3   0(4) + 3(6) + (−5)(− 2) 0(1) + 3(2) + ( −5)(3)  = 1(1) + 2(2) + 6(3)   1(4) + 2(6) + 6(− 2)

f ( x) = 2 x3 − 5 x 2 + x − 10 p must be a factor of 10: p = ±1, ± 2, ±5, ±10 q must be a factor of 2: q = ±1, ±2 The possible rational zeros are: p 1 5 = ± , ± , ±1, ±2, ±5, ±10 2 2 q

15. AC = 

 28 − 9  =   4 23 

tan 42° − cot 48° = cot(90° − 42°) − cot 48° = cot(48°) − cot 48° = 0

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Chapter 12: Systems of Equations and Inequalities

 4 1  0 3 −5 17. CA =  6 2  ⋅    1 2 6  − 2 3 4(3) + 1(2) 4(−5) + 1(6)   4(0) + 1(1) 6(3) + 2(2) 6( −5) + 2(6)  =  6(0) + 2(1)  − 2(0) + 3(1) − 2(3) + 3(2) − 2(−5) + 3(6)  1 14 −14  =  2 22 −18   3 0 28   4 19. C ( A + B) =  6  − 2  4 =  6  − 2

25. a11 = 2(2) + (− 2)(3) = −2 a12 = 2(1) + (− 2)(−1) = 4 a13 = 2(4) + (− 2)(3) = 2 a14 = 2(6) + (− 2)(2) = 8 a21 = 1(2) + 0(3) = 2 a22 = 1(1) + 0(−1) = 1 a23 = 1(4) + 0(3) = 4 a24 = 1(6) + 0(2) = 6

1 0   0 3 −5  4 1 2     + − 2 3 − 2  1 2 6    3   1  4 4 −5 2  ⋅  −1 5 4  3 

 15 21 −16  =  22 34 − 22   −11 7 22 

 1 2 1 2 3    27.    −1 0  0 −1 4   2 4    1(2) + 2(0) + 3(4)   1(1) + 2(−1) + 3(2) = 0(1) ( 1)( 1) 4(2) 0(2) + − − + + (−1)(0) + 4(4)   5 14  =  9 16 

 4 1  0 3 −5   1 0  6 2  − 3  ⋅     1 2 6   − 2 3  0 1     28 − 9   3 0  = −   4 23  0 3 

21. AC − 3I 2 = 

 25 − 9  =   4 20 

 1 0 1  1

3 2   3 6 1  8 −1

29.  2 4 1 6

 1(1) + 0(6) + 1(8) 1(3) + 0(2) + 1(−1)  =  2(1) + 4(6) + 1(8) 2(3) + 4(2) + 1( −1)   3(1) + 6(6) + 1(8) 3(3) + 6(2) + 1(−1)   9 2 =  34 13  47 20 

23. CA − CB  4 1  4 1 0 0 3 −5   4 1   =  6 2 ⋅  −  6 2  ⋅   1 2 6 − 2 3 − 2   − 2 3   − 2 3   1 14 −14   14 7 − 2  =  2 22 −18 −  20 12 − 4   3 0 28  −14 7 − 6   −13 7 −12  =  −18 10 −14   17 −7 34 

 2 − 2   2 1 4 6   − 2 4 2 8 = 1 0   3 −1 3 2   2 1 4 6  

 2 1 31. A =    1 1 Augment the matrix with the identity and use row operations to find the inverse: 2 1 1 0  1 1 0 1    1 1 0 1  Interchange  →     2 1 1 0   r1 and r2  1 1 1 0 →  ( R2 = − 2r1 + r2  0 −1 1 − 2 

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)

Section 12.4: Matrix Algebra

1 → 0 1 → 0

 2 1 1 0  a a 0 1  

0 1 ( R2 = −r2 ) 1 −1 2  0 1 −1 ( R1 = −r2 + r1 ) 1 −1 2   1 −1 Thus, A−1 =  .  −1 2  1

 1 12 12 0  →   a a 0 1 1 →  0

1 0  1 −3

 1 1 0 12  →   0 1 −1 3

a

− 12

0  a 1 1 2

1  1 12 0 2 → 2 −  0 1 1 a 

 6 5 33. A =   2 2 Augment the matrix with the identity and use row operations to find the inverse: 6 5 1 0  2 2 0 1   2 2 0 → 6 5 1 2 2 0 →  0 −1 1

1 2

1 2

( R1 = 12 r1 )

 1 0 1 − 1a  →  2  0 1 −1 a 

( R2 = − a r1 + r2 )

( R2 = a2 r2 ) ( R1 = − 12 r2 + r1 )

 1 − 1a  Thus, A−1 =  . 2  −1 a 

 Interchange     r1 and r2 

 1 −1 1 37. A =  0 − 2 1  − 2 −3 0  Augment the matrix with the identity and use row operations to find the inverse:  1 −1 1 1 0 0   0 − 2 1 0 1 0    − 2 −3 0 0 0 1

( R2 = − 3r1 + r2 )  R1 = 12 r1     R2 = −r2 

 1 0 1 − 52  →  ( R1 = − r2 + r1 ) 3  0 1 −1  1 − 52  Thus, A−1 =  . 3  −1

 1 −1 1 1 0 0  →  0 − 2 1 0 1 0   0 −5 2 2 0 1 1 1 0  1 −1  1 → 0 1 − 2 0 − 12  0 −5 2 2 0

 2 1 35. A =   where a ≠ 0. a a  Augment the matrix with the identity and use row operations to find the inverse:

1 1 0 2  → 0 1 − 12 0 0 − 1  2

1 − 12 0 − 12

2 − 52

( R3 = 2 r1 + r3 ) 0 0  1

0  0 1

( R2 = − 12 r2 )  R1 = r2 + r1     R3 = 5 r2 + r3 

1 1 0 1 − 12 0 2   1 1 → 0 1 − 2 0 −2 0 0 0 1 −4 5 − 2   3 −3 1 1 0 0  →  0 1 0 − 2 2 −1  0 0 1 − 4 5 − 2  1  3 −3  −1 Thus, A =  − 2 2 −1 .  − 4 5 − 2 

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( R3 = − 2 r3 )  R1 = − 12 r3 + r1     R2 = 1 r3 + r2    2

Chapter 12: Systems of Equations and Inequalities

 1 1 1 39. A = 3 2 −1 3 1 2  Augment the matrix with the identity and use row operations to find the inverse:  1 1 1 1 0 0  3 2 −1 0 1 0    3 1 2 0 0 1 1 1 1 1  → 0 −1 − 4 −3 0 − 2 −1 −3 1 1 1 1  → 0 1 4 3 0 − 2 −1 −3

0 0 1 0  0 1 0 0 −1 0  0 1

 R2 = −3 r1 + r2     R3 = −3 r1 + r3 

( R2 = − r2 )

 1 0 −3 − 2 1 0   → 0 1 4 3 −1 0  R3 = 17 r3 3 2 1 0 0 − 7 7  1 7  3 1  1 0 0 − 75 7 7    R1 = 3r3 + r1  9 1 → 0 1 0 − 74    7 7 R = − 4r3 + r2   3 2 1  2 −7 0 0 1 7 7  3 1  − 75 7 7   1 4 . Thus, A−1 =  97 − 7 7  3 2 1  7  7 − 7

(

)

2 x + y = 8 41.   x+ y =5 Rewrite the system of equations in matrix form:  2 1  x 8  A= , X =  , B =     1 1  y 5 

Find the inverse of A and solve X = A−1 B :  1 −1 From Problem 29, A−1 =   , so  −1 2   1 −1 8  3 X = A−1 B =    =  .  −1 2  5  2  The solution is x = 3, y = 2 or (3, 2) . 2 x + y = 0 43.   x+ y =5 Rewrite the system of equations in matrix form:  2 1  x 0  , X =  , B =   A=   1 1  y  5

Find the inverse of A and solve X = A−1 B :  1 −1 From Problem 29, A−1 =   , so  −1 2   1 −1 0   −5 X = A−1 B =    =   .  −1 2   5  10  The solution is x = −5, y = 10 or (−5, 10) .  6x + 5 y = 7 45.  2 x + 2 y = 2 Rewrite the system of equations in matrix form:  6 5  x 7  , X =  , B =   A=  2 2  y 2

Find the inverse of A and solve X = A−1 B :  1 − 52  From Problem 31, A−1 =   , so 3  −1  1 − 52   7   2  X = A−1 B =    =   . 3  2   −1  −1 The solution is x = 2, y = −1 or (2, −1) .

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Section 12.4: Matrix Algebra

 6 x + 5 y = 13 47.  2 x + 2 y = 5 Rewrite the system of equations in matrix form:  6 5  x 13 A= , X =  , B =    2 2  y  5

 x− y+z = 0  53.  − 2 y + z = −1 − 2 x − 3 y = −5  Rewrite the system of equations in matrix form:  1 −1 1  x  0     A =  0 − 2 1 , X =  y  , B =  −1  − 2 −3 0   z   −5

Find the inverse of A and solve X = A−1 B :  1 − 52  From Problem 31, A−1 =   , so 3  −1  1 − 52  13  12  X = A−1 B =    =   . 3  5   2   −1

Find the inverse of A and solve X = A−1 B : 1  3 −3  −1 From Problem 35, A =  − 2 2 −1 , so  − 4 5 − 2 

1 1  The solution is x = , y = 2 or  , 2  . 2 2 

1  0   − 2   3 −3 X = A−1 B =  − 2 2 −1  −1 =  3 .  − 4 5 − 2   −5  5 The solution is x = − 2, y = 3, z = 5 or (−2, 3, 5) .

2 x + y = −3 a≠0 49.   ax + ay = − a Rewrite the system of equations in matrix form:  2 1  x  −3 A=  , X =  y  , B =  −a  a a      

 x− y+z = 2 55.  −2y + z = 2  − 2 x − 3 y = 1  2 Rewrite the system of equations in matrix form:  2  1 −1 1  x       A =  0 − 2 1 , X =  y  , B =  2  1  − 2 −3 0   z  2

Find the inverse of A and solve X = A−1 B :  1 − 1a  From Problem 33, A−1 =   , so 2  −1 a   1 − 1a   −3  − 2  X = A−1 B =  =  .  2  −a  a  −1     1 The solution is x = −2, y = 1 or (−2, 1) .

Find the inverse of A and solve X = A−1 B : 1  3 −3  −1 From Problem 35, A =  − 2 2 −1 , so  − 4 5 − 2 

7  2 x + y = 51.  a a≠0  ax + ay = 5 Rewrite the system of equations in matrix form:  7a   2 1  x A= , X = , B =     y  5 a a   

1  2   12   3 −3     X = A−1 B =  − 2 2 −1  2  =  − 12  1  − 4 5 − 2   2   1

Find the inverse of A and solve X = A−1 B :  1 − 1a  From Problem 33, A−1 =   , so 2  −1 a 

1 1 The solution is x = , y = − , z = 1 or 2 2 1 1   , − , 1 . 2 2 

 1 − 1a   7   a2  X = A−1 B =   a = 3 . 2    −1 a   5  a  2 3 2 3 The solution is x = , y = or  ,  . a a a a

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Chapter 12: Systems of Equations and Inequalities

 x+ y+ z =9  57. 3 x + 2 y − z = 8 3x + y + 2 z = 1 

Rewrite the system of equations in matrix form:  1 1 1  x 9      A = 3 2 −1 , X =  y  , B = 8 3 1 2   z   1 Find the inverse of A and solve X = A−1 B : 3 1  − 75 7 7   1 From Problem 37, A−1 =  97 − 74  , so 7  3 2 1  7 − 7 7  3 34  1  − 75  − 7 7 9  7  9   85  −1 1 4   − 7  8 =  7  . X =A B= 7 7  3  12  2 1   1  7 − 7 7     7  The solution is x = −

34 85 12 or ,y= ,z= 7 7 7

 34 85 12  − , ,  .  7 7 7

 x+ y+ z = 2  7 59. 3 x + 2 y − z =  3  10 3x + y + 2 z = 3  Rewrite the system of equations in matrix form:  2  1 1 1  x       A = 3 2 −1 , X =  y  , B =  73   10  3 1 2   z   3 

Find the inverse of A and solve X = A−1 B : 3 1  − 75 7 7   1 From Problem 37, A−1 =  97 − 74  , so 7  3 2 1 7  7 − 7  5 3 1 1 − 7   2  3  7 7  9     −1 1 4 − 7   73  =  1 . X =A B= 7 7  3 2 2 1   10    3   3  7  7 − 7 1 2 1 The solution is x = , y = 1, z = or  , 1, 3 3 3

4 2 61. A =    2 1 Augment the matrix with the identity and use row operations to find the inverse: 4 2 1 0  2 1 0 1   1 0 4 2 1 →  ( R2 = − 2 r1 + r2 ) 1 0 0 1 − 2   1 1 1 2 0 1 4 →  ( R1 = 4 r1 ) 1 0 0 − 1  2 There is no way to obtain the identity matrix on the left. Thus, this matrix has no inverse. 15 3 63. A =   10 2  Augment the matrix with the identity and use row operations to find the inverse: 15 3 1 0  10 2 0 1   1 0 15 3 → 2 0 0 − 1 3 

( R2 = − 23 r1 + r2 )

 1 15 151 0  → R1 = 151 r1  2  0 0 − 3 1 There is no way to obtain the identity matrix on the left; thus, there is no inverse.

(

)

1 −1  −3  65. A =  1 − 4 −7  2 5  1 Augment the matrix with the identity and use row operations to find the inverse: 1 −1 1 0 0   −3  1 − 4 −7 0 1 0     1 2 5 0 0 1

2 . 3

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Section 12.4: Matrix Algebra

2 5  1  →  1 − 4 −7  −3 1 −1 2 5 1 →  0 − 6 −12  0 7 14

0 0 0

1

1 0 0 0 0

1

1 0

1 0  0  1 −1 3

1 2 5 0 0 1   →  0 1 2 0 − 16 16   0 7 14 1 0 3 

 Interchange     r1 and r3 

 25

61

 3

4

71. A = 18 −12

−12  10  7  ; B =  −9  −1  12 

Enter the matrices into a graphing utility and use A−1B to solve the system. The result is shown below:

 R2 = − r1 + r2     R3 = 3 r1 + r3 

( R2 = − 16 r2 ) Thus, the solution to the system is x ≈ 4.57 , y ≈ − 6.44 , z ≈ − 24.07 or (4.57, −6.44, −24.07) .

1 2 1 0 1 0 3 3    R1 = − 2r2 + r1  →  0 1 2 0 − 16 16     R3 = −7 r2 + r3   7 11  0 0 0 1  6 6  There is no way to obtain the identity matrix on the left; thus, there is no inverse.

 25 61 −12   21   73. A = 18 −12 7  ; B =  7   3  −2  −1  4

 25 61 −12  67. A =  18 −2 4   8 35 21

Thus, the solution to the system is x ≈ −1.19 , y ≈ 2.46 , z ≈ 8.27 or (−1.19, 2.46, 8.27) . 75.

Thus, A

−1

0.05 − 0.01  0.01  ≈  0.01 − 0.02 0.01  − 0.02 0.01 0.03

21 18  44  −2 10 15 69. A =   21 12 −12  4  −8 −16

6 5 4  9

 2 x + 3 y = 11  5 x + 7 y = 24 Multiply each side of the first equation by 5, and each side of the second equation by −2 . Then add the equations to eliminate x:  10 x + 15 y = 55  −10 x − 14 y = −48 y=7 Substitute and solve for x: 2 x + 3 ( 7 ) = 11 2 x + 21 = 11 2 x = −10 x = −5 The solution of the system is x = −5, y = 7 or

0.01  0.02 − 0.04 − 0.01  − 0.02  0.05 0.03 − 0.03 . Thus, A−1 ≈   0.02 0.01 − 0.04 0.00    0.06 0.07 0.06   − 0.02

using ordered pairs ( −5, 7 ) .

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Chapter 12: Systems of Equations and Inequalities

 

x − 2 y + 4z = 2

 1 −5 4 −3   2 17  0 1 − 3 24  R2 = r2 / 24   0 48 −32 38

77. −3x + 5 y − 2 z = 17  4x − 3y 

= −22

Write the augmented matrix:

 1 −5 4 −3  2 17  0 1 − 3 24    0 4  R3 = −48r2 + r3 0 0

 1 −2 4 2    −3 5 −2 17   4 −3 0 −22   

 1 −2 4 2  0 −1 10 23 R2 = 3r1 + r2  0 5 −16 −30  R3 = −4r1 + r3  1 −2 4 2 0 1 −10 −23 R2 = −r2  0 5 −16 −30   1 −2 4 2 0 1 −10 −23  0 0 34 85 R = −5r + r 3 2 3 1 0  0 1 0  0 1 0  0

−2 4 2 1 −10 −23 5 0 1 2  R3 = r3 / 34 −2 0 −8 R1 = −4r3 + r1 1 0 2  R2 = 10r3 + r2 5 0 1 2 0 0 −4  R1 = 2r2 + r1 1 0 2 5 0 1 2

The solution is x = −4, y = 2, z =  5x − y + 4z = 2  79.  − x + 5 y − 4 z = 3 7 x + 13 y − 4 z = 17 

Write the augmented matrix:  5 −1 4 2     −1 5 −4 3   7 13 −4 17   

 1 −5 4 −3 R1 = −r2    5 −1 4 2  R2 = r1 7 13 −4 17   1 −5 4 −3    0 24 −16 17  R2 = −5r1 + r2 0 48 −32 38 R = −7 r + r 3 1 3

5 5  or  −4,2,  . 2 2 

The last row of our matrix is a contradiction. Therefore, the system is inconsistent. The solution set is { } , or ∅ .  2x − 3y + z = 4  81. −3x + 2 y − z = −3  − 5y + z = 6 

Write the augmented matrix:  2 −3 1 4     −3 2 −1 −3  0 −5 1 6    3 1   −2 2  R1 = r1 / 2 2  1  −3 2 −1 −3    0 −5 1 6   −3 1  2 2 2 1  0 − 52 12 3 R2 = 3r1 + r2   0 −5 1 6  1  −3  2 2 2 1 6 1 2   1 − 5 − 5  R2 = − 5 r2 0 0 −5 6  1  1 0  0  1 0  0

 2 − 65  1  0 0 0  R3 = 5r2 + r3 1 1 0 5 5 R1 = 3 r2 + r1  2 1 − 6 − 5 5 1  0 0 0  − 32

1 2 − 15

Since the last row yields an identity, and no contradictions exist in the other rows, there are an infinite number of solutions. The solution is 1 1 1 6 x = − z + , y = z − , and z is any real 5 5 5 5

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Section 12.4: Matrix Algebra

number. That is, 1 1 1 6  ( x, y, z ) | x = − z + , y = z − , 5 5 5 5 

87. a.

z is any real number}

83. a. b.

6

A=

9

3 12

AB = =

6

9

; B=

121.00 340.13

121.00

3 12 340.13 6(121.00) + 9(340.13) 3(121.00) + 12(340.13)

=

3787.17 4444.56

Nikki’s total tuition is $3787.17, and Joe’s total tuition is $4444.56. 85. a.

The rows of the 2 by 3 matrix represent stainless steel and aluminum. The columns represent 10-gallon, 5-gallon, and 1-gallon. 500 350 400  The 2 by 3 matrix is:  . 700 500 850   500 700  The 3 by 2 matrix is:  350 500  .  400 850 

1 → 2   1 1 → 0  0

1 0 0 1 0 1 1 1 0 0  1 1 0 0 1 1 0 0 1 0 −1 1 1 −2 0   0 1 0 −1 1

1 → 0 0 1 → 0 0

1

0

1 −1 −1 0

1

1 0

0 0

1 0 −1 0

Thus, K

The days usage of materials is: 15  500 350 400     11,500  ⋅  700 500 850   8 = 17, 050     3    

b.

−1

M = E⋅K

Thus, 11,500 pounds of stainless steel and 17,050 pounds of aluminum were used that day. d. The 1 by 2 matrix representing cost is: [0.10 0.05] . e.

0

1

0

1 0 0 1 → 0 1 0 −1 0 0 1 0

b. The 3 by 1 matrix representing the amount of 15 material is:  8 .  3 c.

 2 1 1 K =  1 1 0    1 1 1 Augment the matrix with the identity and use row operations to find the inverse: 2 1 1 1 0 0  1 1 0 0 1 0    1 1 1 0 0 1

The total cost of the day’s production was:  11,500  [0.10 0.05] ⋅   = [ 2002.50] . 17, 050  The total cost of the day’s production was $2002.50.

763 Copyright © 2016 Pearson Education, Inc.

1 0 2 0  −1 1 1 0 1 1 −1 1 0 −1 1 1 −1 1

 Interchange   r and r  1  2  R2 = −2 r1 + r2  R = −r + r  1 3   3

( R2 = −r2 )

( R2 = r2 + r3 ) ( R1 = r1 − r2 )

 1 0 −1 =  −1 1 1 .  0 −1 1

−1

 47 34 33  1 0 −1 =  44 36 27   −1 1 1  47 41 20   0 −1 1 13 1 20  =  8 9 19   6 21 14 

Chapter 12: Systems of Equations and Inequalities

because a11 = 47(1) + 34(−1) + 33(0) = 13

é0 ê ê1 ê 91. A = ê1 ê ê0 ê ê0 ë

a12 = 47(0) + 34(1) + 33(−1) = 1 a13 = 47(−1) + 34(1) + 33(1) = 20 a21 = 44(1) + 36(−1) + 27(0) = 8 a22 = 44(0) + 36(1) + 27(−1) = 9

é2 ê ê0 ê 2 A = ê0 ê ê1 ê ê1 ë

a31 = 47(1) + 41(−1) + 20(0) = 6 a32 = 47(0) + 41(1) + 20(−1) = 21 a33 = 47(−1) + 41(1) + 20(1) = 14 13 → M ; 1 → A; 20 → T ; 8 → H ; 9 → I ; 19 → S ; 6 → F ; 21 → U ; 14 → N The message: Math is fun.

0  0 1  1 a

b 1 a → bc d 0 − a 

1 → 0

−

b a ad −bc a

− ac

1 a −c ad −bc

1 0 → 0 1 

1 a

1 0 → 0 1 

d ad −bc −c ad −bc

0   a  ad −bc 

bc a ( ad −bc ) −c ad −bc

1 0 d D → 0 1 − c D 

−b  ad −bc  a  ad −bc 

( R2 = −c ⋅ r1 + r2 )

( R2 = ad a−bc ⋅ r2 )

é ê ê ê ê 93. a. XR = êê ê ê ê ê ëê

( R1 =

− Db   a  D 

⋅ r2 + r1

(3 − 2

)

b.

− Db  1  d −b  =   a   D  −c a  D 

where D = ad − bc .

2 0 0 1 1

é0 1ù ê ú ê4 ú 1 ú 3 ê 1ú A = ê 3 ê ú ê1 0úú ê ê0 1úû ë 4 3 2 2 3

5 2 2 3 2

2 5 4 2 1

3 1 1 1 2

4 0 0 2 2

0 4 3 1 0

2ù ú 2ú ú 1ú ú 2úú 1úû

3ù ú 4ú ú 2ú ú 3úú 2úû

ù 1 3 0úú 2 2 ú é 6ù éê 3 - 2 3 ùú úê ú ê 3 1 ú 0úú ê 4ú = ê3 3 + 2ú ê ú 2 2 ú ê 1ú êê 1 úú 0 0 1úú ë û ëê ûú ú ûú

The coordinates would be − ba

−b  ad − bc  a  ad − bc 

 d Thus, A−1 =  Dc  − D

0 2 2 0 0

b. The largest number in row 1 (Page 1) is 5 which corresponds to Page 3.

0  1 

1 a

1 b a → 0 1 

+

0  1 

c a

0 2 1 1 0

Yes, all pages can reach every other page within 3 clicks.

( R1 = 1a ⋅ r1 )

1 a

0ù ú 1ú ú 0ú ú 1úú 0úû

0 1 1 0 0

é2 ê ê5 ê 2 3 A + A + A = ê4 ê ê2 ê ê1 ë

a b  89. A =   c d  If D = ad − bc ≠ 0 , then a ≠ 0 and d ≠ 0 , or b ≠ 0 and c ≠ 0 . Assuming the former, then a b 1 0  c d 0 1   1 ba → c d

1 0 0 1 0

a.

a23 = 44(−1) + 36(1) + 27(1) = 19

c.

1 0 0 0 1

R −1

3,3 3 + 2

1 2 3 = − 2 0

)

3 0 2 1 0 ; This is the rotation 2 0 1

matrix needed to get the translated coordinates back to the original coordinates.

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Section 12.5: Partial Fraction Decomposition 95. Since the product is found by multiplying the components from the columns of the first matrix by the components in the rows of the second matrix and then adding those products, then the number of columns in the first must equal the number of rows in the second.

7. The rational expression

perform the division: 1

9 The proper rational expression is: x2 + 5 9 = 1+ 2 2 x −4 x −4

−1

A AX = A 0 IX = 0 X =0

If the inverse of A does not exist, then A is singular and you would not be able to multiply A by its inverse to give the identity inverse and X would have no solution.

9. The rational expression

x 2 − 4 5 x3

101. Let θ = csc u so that csc θ = u , −

π 2

)

is improper,

− 20 x

22 x − 1 The proper rational expression is:

≤θ ≤

π 2

5 x3 + 2 x − 1

,

2

x −4

u ≥ 1 . Then, cos csc−1 u = cos θ = cos θ ⋅

x2 − 4

+ 2x −1

5 x3

θ = cos −1 ( −1) = 180° −1

5 x3 + 2 x − 1

so perform the division: 5x

99. v ⋅ w = ( −2)(2) + ( −1)(1) = −4 − 1 = −5 v⋅w −5 −5 cos θ = = = = −1 v w 5 5 5

=

is improper, so

x2 − 4

AX = 0

(

x2 − 4

x2 − 4 x2 + 5

97. If the inverse of A exists: −1

x2 + 5

= 5x +

22 x − 1 x2 − 4

11. The rational expression x( x − 1) x2 − x is improper, so = 2 ( x + 4)( x − 3) x + x − 12 perform the division: 1

sin θ = cot θ sin θ sin θ

cot θ cot 2 θ csc2 θ − 1 = = csc θ csc θ csc θ

x 2 + x − 12 x 2 − x + 0

u2 −1 = u

x 2 + x − 12 − 2 x + 12 The proper rational expression is: x( x − 1) − 2 x + 12 − 2( x − 6) =1+ 2 =1+ ( x + 4)( x − 3) ( x + 4)( x − 3) x + x − 12

Section 12.5

13. Find the partial fraction decomposition: 4 A B = + x( x − 1) x x − 1

1. True

(

)



4  B  A  = x( x − 1)  +  − −1  x ( x 1) x x   

3. 3x 4 + 6 x3 + 3x 2 = 3x 2 x 2 + 2 x + 1 = 3x 2 ( x + 1)

5. The rational expression

x( x − 1) 

2

4 = A( x − 1) + Bx

x

is proper, since x −1 the degree of the numerator is less than the degree of the denominator. 2

765 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities

Let x = 1 , then 4 = A(0) + B B=4 Let x = 0 , then 4 = A(−1) + B (0) A = −4 4 4 −4 = + x( x − 1) x x −1

15. Find the partial fraction decomposition: 1 2

x( x + 1) 

=

A Bx + C + x x2 + 1

  A Bx + C  2  = x( x + 1)  + 2  x ( x + 1)  x x +1   

x( x 2 + 1) 

1

2

2

1 = A( x + 1) + ( Bx + C ) x Let x = 0 , then 1 = A(02 + 1) + ( B (0) + C )(0) A =1

Let x = 1 , then

2

1 = A(1 + 1) + ( B (1) + C )(1) 1 = 2A + B + C 1 = 2(1) + B + C B + C = −1

Let x = −1 , then

1 = A((−1) 2 + 1) + ( B(−1) + C )(−1) 1 = A(1 + 1) + (− B + C )(−1) 1 = 2A + B − C 1 = 2(1) + B − C B − C = −1

Solve the system of equations: B + C = −1 B − C = −1 2B = −2 B = −1

1 x( x + 1)

=

19. Find the partial fraction decomposition: x2 A B C = + + 2 2 x +1 ( x − 1) ( x + 1) x − 1 ( x − 1)

Multiplying both sides by ( x − 1) 2 ( x + 1) , we obtain: x 2 = A( x − 1)( x + 1) + B ( x + 1) + C ( x − 1) 2 Let x = 1 , then 12 = A(1 − 1)(1 + 1) + B (1 + 1) + C (1 − 1) 2 1 = A(0)(2) + B (2) + C (0) 2 1 = 2B 1 B= 2

Let x = −1 , then (−1) 2 = A(−1 − 1)(−1 + 1) + B (−1 + 1) + C (−1 − 1) 2 1 = A(−2)(0) + B (0) + C (−2) 2 1 = 4C 1 C= 4

Let x = 0 , then 02 = A(0 − 1)(0 + 1) + B (0 + 1) + C (0 − 1) 2 0 = −A + B + C A = B+C 1 1 3 A= + = 2 4 4

−1 + C = −1 C=0 2

Let x = 1 , then 1 = A(1 − 2) + B (1 − 1) 1 = −A A = −1 Let x = 2 , then 2 = A(2 − 2) + B(2 − 1) 2=B x −1 2 = + ( x − 1)( x − 2) x − 1 x − 2

1 −x + 2 x x +1

x2

17. Find the partial fraction decomposition: x A B = + ( x − 1)( x − 2) x − 1 x − 2

( x − 1) 2 ( x + 1)

Multiplying both sides by ( x − 1)( x − 2) , we obtain: x = A( x − 2) + B( x − 1)

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=

3 4

+

1 2

x − 1 ( x − 1) 2

+

1 4

x +1

Section 12.5: Partial Fraction Decomposition

Let x = 1 , then

21. Find the partial fraction decomposition: 1 3

x −8

1

=

12 = A(1 − 1)(1 + 1) 2 + B(1 + 1) 2

2

( x − 2)( x + 2 x + 4) A Bx + C = + 2 2 ( x − 2)( x + 2 x + 4) x − 2 x + 2 x + 4

+ C (1 − 1) 2 (1 + 1) + D(1 − 1) 2

1

1 = 4B

we obtain: 1 = A( x 2 + 2 x + 4) + ( Bx + C )( x − 2)

1 4 Let x = −1 , then

Let x = 2 , then

(−1) 2 = A(−1 − 1)(−1 + 1) 2 + B (−1 + 1) 2

B=

Multiplying both sides by ( x − 2)( x 2 + 2 x + 4) ,

(

)

1 = A 22 + 2(2) + 4 + ( B (2) + C )(2 − 2)

+ C (−1 − 1) 2 (−1 + 1) + D(−1 − 1) 2 1 = 4D

1 = 12 A

1 4 Let x = 0 , then

1 12 Let x = 0 , then

D=

A=

(

02 = A(0 − 1)(0 + 1) 2 + B (0 + 1) 2

)

1 = A 02 + 2(0) + 4 + ( B (0) + C )(0 − 2)

+ C (0 − 1) 2 (0 + 1) + D(0 − 1) 2 0 = −A + B + C + D

1 = 4 A − 2C 1 = 4 (1/12 ) − 2C

A−C = B + D

−2C = 23

A−C =

C = −1 3

Let x = 2 , then

Let x = 1 , then

(

22 = A(2 − 1)(2 + 1) 2 + B (2 + 1) 2

)

1 = A 12 + 2(1) + 4 + ( B (1) + C )(1 − 2)

+ C (2 − 1) 2 (2 + 1) + D(2 − 1) 2 4 = 9 A + 9 B + 3C + D 9 A + 3C = 4 − 9 B − D

1= 7A − B − C 1 = 7 (1/12 ) − B + B=−

1 3

1 1 3 9 A + 3C = 4 − 9   − = 4 4 2 1 3A + C = 2 Solve the system of equations: A−C = 1 2 1 3A + C = 2 4A =1 A= 1 4 3 +C = 1 4 2 C =−1 4 1 1 1 2 −1 x 4 4 = 4 + + 4 + 2 2 2 x − 1 ( x − 1) x + 1 ( x + 1) 2 ( x − 1) ( x + 1)

1 12

1 x3 − 8

= =

1 12

x−2 1 12

x−2

+ +

1 x−1 − 12 3

x2 + 2x + 4 1 x+4 − 12 ( )

x2 + 2x + 4

23. Find the partial fraction decomposition: x2 2

( x − 1) ( x + 1)2

=

1 1 1 + = 4 4 2

A B C D + + + x − 1 ( x − 1)2 x + 1 ( x + 1) 2

Multiplying both sides by ( x − 1)2 ( x + 1)2 , we obtain: x 2 = A( x − 1)( x + 1)2 + B( x + 1)2 + C ( x − 1) 2 ( x + 1) + D ( x − 1) 2

767 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities 25. Find the partial fraction decomposition: x −3 A B C = + + 2 x + 2 x + 1 ( x + 1)2 ( x + 2)( x + 1)

Multiplying both sides by ( x + 2)( x + 1) 2 , we obtain: x − 3 = A( x + 1) 2 + B ( x + 2)( x + 1) + C ( x + 2) Let x = − 2 , then − 2 − 3 = A(− 2 + 1) 2 + B( − 2 + 2)( − 2 + 1) + C ( − 2 + 2) −5 = A A = −5

Let x = −1 , then 2

−1 − 3 = A(−1 + 1) + B(−1 + 2)(−1 + 1) + C (−1 + 2) −4 = C C = −4 Let x = 0 , then 0 − 3 = A(0 + 1) 2 + B (0 + 2)(0 + 1) + C (0 + 2) −3 = A + 2 B + 2C −3 = −5 + 2 B + 2(− 4) 2 B = 10 B=5 x −3

( x + 2)( x + 1)

2

=

5 −5 −4 + + x + 2 x + 1 ( x + 1) 2

Let x = 2 , then 2 + 4 = A(2)(22 + 4) + B(22 + 4) + (C (2) + D)(2)2 6 = 16 A + 8B + 8C + 4 D 6 = 16 A + 8 + 8C + 4 D 16 A + 8C + 4 D = − 2 Solve the system of equations: 5A + C + D = 0 −5 A − C + D = − 2 2D = − 2 D = −1 5A + C −1 = 0 C = 1− 5A

16 A + 8(1 − 5 A) + 4(−1) = − 2 16 A + 8 − 40 A − 4 = − 2 − 24 A = − 6 1 A= 4 5 1 1 C = 1− 5  = 1− = − 4 4 4   1 1 1 − x −1 x+4 = 4 + 2 + 42 2 2 x ( x + 4) x x x +4 =

27. Find the partial fraction decomposition: x+4 2

2

x ( x + 4)

=

A B Cx + D + + x x2 x2 + 4

Multiplying both sides by x 2 ( x 2 + 4) , we obtain: x + 4 = Ax( x 2 + 4) + B( x 2 + 4) + (Cx + D) x 2

Let x = 0 , then 0 + 4 = A(0)(02 + 4) + B (02 + 4) + ( C (0) + D ) (0) 2 4 = 4B B =1

Let x = 1 , then 1 + 4 = A(1)(12 + 4) + B (12 + 4) + (C (1) + D)(1) 2 5 = 5 A + 5B + C + D 5 = 5A + 5 + C + D 5A + C + D = 0 Let x = −1 , then −1 + 4 = A(−1)((−1) 2 + 4) + B ((−1) 2 + 4)

1 4

x

+

1 x

2

+

− 14 ( x + 4 ) x2 + 4

29. Find the partial fraction decomposition: x2 + 2 x + 3 A Bx + C = + 2 2 x + 1 ( x + 1)( x + 2 x + 4) x + 2x + 4

Multiplying both sides by ( x + 1)( x 2 + 2 x + 4) , we obtain: x 2 + 2 x + 3 = A( x 2 + 2 x + 4) + ( Bx + C )( x + 1) Let x = −1 , then (−1) 2 + 2(−1) + 3 = A((−1) 2 + 2(−1) + 4) + ( B(−1) + C )(−1 + 1) 2 = 3A 2 A= 3 Let x = 0 , then 02 + 2(0) + 3 = A(02 + 2(0) + 4) + ( B(0) + C )(0 + 1)

+ (C ( −1) + D)(−1) 2 3 = −5 A + 5 B − C + D 3 = −5 A + 5 − C + D −5 A − C + D = − 2

768 Copyright © 2016 Pearson Education, Inc.

3 = 4A + C 3 = 4 ( 2 / 3) + C C=

1 3

Section 12.5: Partial Fraction Decomposition

Let x = 1 , then 2

Let x = −3 , then −3 = A(−3 − 1) + B (−3 + 3) −3 = − 4 A 3 A= 4

2

1 + 2(1) + 3 = A(1 + 2(1) + 4) + ( B (1) + C )(1 + 1) 6 = 7 A + 2 B + 2C 6 = 7 ( 2 / 3) + 2 B + 2 (1/ 3) 2 B = 6 − 143 − 23 =

x

2 3

x2 + 2 x − 3

B = 13 x2 + 2 x + 3 ( x + 1)( x 2 + 2 x + 4)

= =

2 3

x +1 2 3

x +1

+ +

1 3

=

3 4

x+3

+

1 4

x −1

35. Find the partial fraction decomposition: x 2 + 2 x + 3 Ax + B Cx + D = 2 + ( x 2 + 4) 2 x + 4 ( x 2 + 4) 2

x + 13

x2 + 2 x + 4 1 ( x + 1) 3 2

Multiplying both sides by ( x 2 + 4) 2 , we obtain:

x + 2x + 4

x 2 + 2 x + 3 = ( Ax + B )( x 2 + 4) + Cx + D

31. Find the partial fraction decomposition: x A B = + (3x − 2)(2 x + 1) 3x − 2 2 x + 1

x 2 + 2 x + 3 = Ax3 + Bx 2 + 4 Ax + 4 B + Cx + D x 2 + 2 x + 3 = Ax3 + Bx 2 + (4 A + C ) x + 4 B + D

Multiplying both sides by (3 x − 2)(2 x + 1) , we obtain: x = A(2 x + 1) + B (3 x − 2) Let x = − 1 , then 2 1 − = A ( 2 ( −1/ 2 ) + 1) + B ( 3 ( −1/ 2 ) − 2 ) 2 1 7 − =− B 2 2 1 B= 7 2 Let x = , then 3 2 = A ( 2 ( 2 / 3) + 1) + B ( 3 ( 2 / 3) − 2 ) 3 2 7 = A 3 3 2 A= 7

A=0;

B =1;

4A + C = 2 4(0) + C = 2 C=2 2

x + 2x + 3 2

( x + 4)

2

=

4B + D = 3 4(1) + D = 3 D = −1 1 2

x +4

+

2x −1 ( x 2 + 4) 2

37. Find the partial fraction decomposition: 7x + 3 7x + 3 = 3 2 x − 2 x − 3 x x( x − 3)( x + 1) A B C = + + x x − 3 x +1 Multiplying both sides by x( x − 3)( x + 1) , we obtain: 7 x + 3 = A( x − 3)( x + 1) + Bx( x + 1) + Cx( x − 3)

Let x = 0 , then

7(0) + 3 = A(0 − 3)(0 + 1) + B(0)(0 + 1) + C (0)(0 − 3)

2 1 x = 7 + 7 (3x − 2)(2 x + 1) 3x − 2 2 x + 1

3 = −3 A A = −1

Let x = 3 , then

33. Find the partial fraction decomposition: x x A B = = + 2 x + 2 x − 3 ( x + 3)( x − 1) x + 3 x − 1

7(3) + 3 = A(3 − 3)(3 + 1) + B (3)(3 + 1) + C (3)(3 − 3) 24 = 12 B B=2

Multiplying both sides by ( x + 3)( x − 1) , we obtain: x = A( x − 1) + B( x + 3)

Let x = −1 , then 7(−1) + 3 = A(−1 − 3)(−1 + 1) + B(−1)(−1 + 1) + C (−1)(−1 − 3)

Let x = 1 , then 1 = A(1 − 1) + B(1 + 3) 1 = 4B 1 B= 4

−4 = 4C C = −1

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Chapter 12: Systems of Equations and Inequalities

7x + 3 3

2

x − 2 x − 3x

=

2 −1 −1 + + x x − 3 x +1

39. Perform synthetic division to find a factor: 2 1 −4 5 −2 2 −4 2 1 −2

1

Multiplying both sides by ( x 2 + 16)3 , we obtain: x3 = ( Ax + B )( x 2 + 16) 2 + (Cx + D)( x 2 + 16) + Ex + F x3 = ( Ax + B )( x 4 + 32 x 2 + 256) + Cx3 + Dx 2 + 16Cx + 16 D + Ex + F x3 = Ax5 + Bx 4 + 32 Ax3 + 32 Bx 2 + 256 Ax

0

+ 256 B + Cx3 + Dx 2 + 16Cx + 16 D + Ex + F

x3 − 4 x 2 + 5 x − 2 = ( x − 2)( x 2 − 2 x + 1) = ( x − 2)( x − 1) 2

Find the partial fraction decomposition: x2 x3 − 4 x 2 + 5 x − 2

=

x2

( x − 2)( x − 1) 2 A B C = + + x − 2 x − 1 ( x − 1) 2

Multiplying both sides by ( x − 2)( x − 1) 2 , we obtain: x 2 = A( x − 1) 2 + B ( x − 2)( x − 1) + C ( x − 2) Let x = 2 , then 22 = A(2 − 1) 2 + B (2 − 2)(2 − 1) + C (2 − 2) 4= A Let x = 1 , then 12 = A(1 − 1) 2 + B (1 − 2)(1 − 1) + C (1 − 2) 1 = −C C = −1 Let x = 0 , then

x3 = Ax5 + Bx 4 + (32 A + C ) x3 + (32 B + D) x 2 + (256 A + 16C + E ) x + (256 B + 16 D + F )

32 B + D = 0 32(0) + D = 0 D=0

x2 2

x − 4 x + 5x − 2

=

256 A + 16C + E = 0 256(0) + 16(1) + E = 0 E = −16

256 B + 16 D + F = 0 256(0) + 16(0) + F = 0 F =0 x3 2

( x + 16)

3

02 = A(0 − 1) 2 + B (0 − 2)(0 − 1) + C (0 − 2) 0 = A + 2 B − 2C 0 = 4 + 2 B − 2(−1) −2 B = 6 B = −3 3

32 A + C = 1 32(0) + C = 1 C =1

A = 0; B = 0 ;

4 −3 −1 + + x − 2 x − 1 ( x − 1) 2

41. Find the partial fraction decomposition: x3 Ax + B Cx + D Ex + F = 2 + 2 + 2 2 3 2 x + 16 ( x + 16) ( x + 16) ( x + 16)3

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=

x 2

( x + 16)

2

+

−16 x 2

( x + 16)3

Section 12.5: Partial Fraction Decomposition

Let x = −3 , then 2(−3) + 3 = A(−3)(−3 − 3)(−3 + 3) + B(−3 − 3)(−3 + 3)

43. Find the partial fraction decomposition: 4 4 A B = = + 2 x 2 − 5 x − 3 ( x − 3)(2 x + 1) x − 3 2 x + 1

Multiplying both sides by ( x − 3)(2 x + 1) , we obtain: 4 = A(2 x + 1) + B( x − 3)

+ C (−3) 2 (−3 + 3) + D(−3)2 (−3 − 3) −3 = −54 D 1 D= 18

1 , then 2   1   1  4 = A  2  −  + 1 + B  − − 3  2 2       7 4=− B 2 8 B=− 7 Let x = 3 , then 4 = A(2(3) + 1) + B (3 − 3) 4 = 7A 4 A= 7 4 − 87 4 7 = + 2 x2 − 5x − 3 x − 3 2 x + 1

Let x = −

Let x = 1 , then 2 ⋅1 + 3 = A ⋅1(1 − 3)(1 + 3) + B (1 − 3)(1 + 3) + C ⋅12 (1 + 3) + D ⋅12 (1 − 3) 5 = − 8 A − 8B + 4C − 2 D 5 = − 8 A − 8 ( −1/ 3) + 4 (1/ 6 ) − 2 (1/18 ) 8 2 1 5 = −8A + + − 3 3 9 16 −8A = 9 2 A=− 9 1 1 −2 −1 2x + 3 = 9 + 23 + 6 + 18 4 2 x x−3 x+3 x − 9x x

45. Find the partial fraction decomposition: 2x + 3 2x + 3 = 2 4 2 x − 9x x ( x − 3)( x + 3) A B C D = + 2+ + x x x −3 x +3

47.

2

Multiplying both sides by x ( x − 3)( x + 3) , we obtain: 2 x + 3 = Ax( x − 3)( x + 3) + B( x − 3)( x + 3)

x3 + x 2 − 3 x 2 + 3x − 4 Dividing: x−2 x3 + x 2 + 0 x − 3

2

x + 3x − 4

(

− x3 + 3x 2 − 4 x

+ Cx 2 ( x + 3) + Dx 2 ( x − 3)

− 2x + 4x − 3 − −2 x 2 − 6 x + 8

Let x = 0 , then 2 ⋅ 0 + 3 = A ⋅ 0(0 − 3)(0 + 3) + B (0 − 3)(0 + 3) 2

)

2

(

2

+ C ⋅ 0 (0 + 3) + D ⋅ 0 (0 − 3)

3

2

x + x −3

3 = −9 B 1 B=− 3 Let x = 3 , then 2 ⋅ 3 + 3 = A ⋅ 3(3 − 3)(3 + 3) + B(3 − 3)(3 + 3)

)

10 x − 11 10 x − 11

, x ≠ −4,1 x + 3x − 4 x 2 + 3x − 4 Find the partial fraction decomposition: 10 x − 11 10 x − 11 A B = = + x 2 + 3x − 4 ( x + 4)( x − 1) x + 4 x − 1 2

= x−2+

Multiplying both sides by ( x + 4)( x − 1) , we obtain: 10 x − 11 = A( x − 1) + B( x + 4) Let x = 1 , then

+ C ⋅ 32 (3 + 3) + D ⋅ 32 (3 − 3) 9 = 54C 1 C= 6

771 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities 10 (1) − 11 = A (1 − 1) + B (1 + 4 )

−11x − 32 x2 + 4 x + 4

−1 = 5 B 1 B=− 5 Let x = −4 , then 10(−4) − 11 = A((− 4) − 1) + B (− 4 + 4) −51 = −5 A 51 A= 5 51 − 15 10 x − 11 5 = + x 2 + 3x − 4 x + 4 x − 1

Thus,

49.

x3 + x 2 − 3 x 2 + 3x − 4

= x−2+

51 5

x+4

+

− 15 x −1

Thus, x 4 − 5 x3 + x − 4 2

x + 4x + 4

x2 + 1 Dividing: x x 2 + 0 x + 1 x3 + 0 x 2 + 0 x

(

− x3 + 0 x 2 + x x 2

x +1

−11x − 32 A B = + ( x + 2)( x + 2) x + 2 ( x + 2) 2

Multiplying both sides by ( x + 2) 2 , we obtain: −11x − 32 = A( x + 2) + B −11x − 32 = Ax + 2 A + B −11x − 32 = Ax + (2 A + B) Since the coefficient of x is A then A = −11 . Let A = −11 , then −32 = 2 A + B −32 = 2(−11) + B −10 = B −11x − 32 −11 −10 = + 2 x + 4 x + 4 x + 2 ( x + 2) 2

x3

3

=

= x+

−x

53.

)

= x2 − 4 x + 7 +

x5 + x 4 − x 2 + 2 x4 − 2 x2 + 1 Dividing: x +1 x 4 + 0 x3 − 2 x 2 + 0 x + 1 x5 + x 4 + 0 x3 − x 2 + 0 x + 2

−x

(

2

− x5 + 0 x 4 − 2 x3 + 0 x 2 + x

x +1

4

)

3

2x − x

x2 + 4 x + 4 Dividing:

x5 + x 4 − x 2 + 2

(

)

− 4 x3 − 9 x 2 + x − −4 x3 − 16 x 2 − 16 x

(

)

7x 2 + 17 x − 4 − 7 x 2 + 28 x + 28

(

= x +1+

2

− x +1

2 x3 + x 2 − x + 1

, x4 − 2x2 + 1 x4 − 2 x2 + 1 x ≠ 1, −1 Find the partial fraction decomposition: 2 x3 + x 2 − x + 1 A B C D = + + + ( x + 1) 2 ( x − 1)2 x + 1 ( x + 1) 2 x − 1 ( x − 1) 2

x2 − 4 x + 7 x 4 + 0 x3 − 5 x 2 + x − 4

− x 4 + 4 x3 + 4 x 2

)

2

(

x 4 − 5 x3 + x − 4

x2 + 4 x + 4

3

x + 2x − x − x + 2 − x 4 + 0 x3 − 2 x 2 + 0 x + 1

Since x 2 + 1 is irreducible then we cannot go any further. 51.

−11 −10 + x + 2 ( x + 2) 2

Multiplying both sides by ( x + 1) 2 ( x − 1) 2 , we obtain:

)

− 11x − 32 x 4 − 5 x3 + x − 4 −11x − 32 , = x2 − 4 x + 7 + 2 2 x + 4x + 4 x + 4x + 4 x ≠ −2 Find the partial fraction decomposition:

2 x3 + x 2 − x + 1 = A( x + 1)( x − 1) 2 + B( x − 1) 2 + C ( x + 1) 2 ( x − 1) + D ( x + 1) 2

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Section 12.6: Systems of Nonlinear Equations

Let x = −1 , then 3

2

2(−1) + (−1) − (−1) + 1 = A(−1 + 1)(−1 − 1)

55. A = P 1 +

2

+ B (−1 − 1) 2 + C (−1 + 1) 2 (−1 − 1) + D(−1 + 1) 2 1 = 4B 1 =B 4

365t

2 = (1.000493)

365t

ln 2 = ln (1.000493)

2

ln 2 = 365t ln (1.000493)

+ B (1 − 1) 2 + C (1 + 1) 2 (1 − 1) + D(1 + 1) 2 3 = 4D 3 =D 4

t=

ln 2 365ln (1.000493)

= 3.85 years

1 cos 308° cos 52° 1 cos 308° = cos(360° − 52°) 1 = cos 308° = 1 cos 308°

Let x = 0 , then 2(0)3 + (0) 2 − (0) + 1 = A(0 + 1)(0 − 1) 2 1 3 + (0 − 1) 2 + C (0 + 1) 2 (0 − 1) + (0 + 1) 2 4 4 1 3 1= A+ −C + 4 4 0 = A−C Let x = 2 , then 2(2)3 + (2) 2 − (2) + 1 = A(2 + 1)(2 − 1) 2 1 3 + (2 − 1)2 + C (2 + 1) 2 (2 − 1) + (2 + 1) 2 4 4 1 27 19 = 3 A + + 9C + 4 4 12 = 3 A + 9C

57. sec 52° cos 308° =

Section 12.6 1. y = 3x + 2 The graph is a line. x-intercept: 0 = 3x + 2 3x = −2 2 x=− 3

0 = A−C 12 = 3 A + 9C A=C 12 = 3 A + 9 A 12 = 12 A 1 = A and 1 = C

y-intercept: y = 3 ( 0 ) + 2 = 2 y

2

1 3 2 x3 + x 2 − x + 1 1 1 4 = + + + 4 2 4 2 2 x + 1 x − 1 x − 2x +1 ( x + 1) ( x − 1) x5 + x 4 − x 2 + 2 x4 − 2 x2 + 1

0.18 365

365t

2

2(1) + (1) − (1) + 1 = A(1 + 1)(1 − 1)

Thus,

nt

8400 = 4200 1 +

Let x = 1 , then 3

r n

(0, 2)

 2   − 3 ,0   

2

−2

= x +1

−2

1 3 1 4 + + + 4 2 2 x − 1 ( x + 1) ( x − 1)

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x

Chapter 12: Systems of Equations and Inequalities

 y = 36 − x 2 7.   y = 8 − x

y 2 = x2 − 1

3. 2

2

x − y =1 x2 y 2 − =1 12 12 The graph is a hyperbola with center (0, 0), transverse axis along the x-axis, and vertices at (−1, 0) and (1, 0) . The asymptotes are y = − x and y = x . y

5 (−1, 0) −5

(1, 0) 5

x

(2.59, 5.41) and (5.41, 2.59) are the intersection points.

−5

Solve by substitution: 36 − x 2 = 8 − x

2  y = x + 1 5.   y = x + 1

36 − x 2 = 64 − 16 x + x 2 2 x 2 − 16 x + 28 = 0 x 2 − 8 x + 14 = 0 8 ± 64 − 56 2 8± 2 2 = 2 = 4± 2

x=

( ) 2, y = 8 − ( 4 − 2 ) = 4 + 2 ( 4 + 2, 4 − 2 ) and ( 4 − 2, 4 + 2 )

If x = 4 + 2, y = 8 − 4 + 2 = 4 − 2 If x = 4 −

(0, 1) and (1, 2) are the intersection points. Solve by substitution: x2 + 1 = x + 1 2

x −x=0 x( x − 1) = 0

Solutions:  y = x 9.   y = 2 − x

x = 0 or x = 1 y =1 y=2 Solutions: (0, 1) and (1, 2)

(1, 1) is the intersection point.

774 Copyright © 2016 Pearson Education, Inc.

Section 12.6: Systems of Nonlinear Equations Solve by substitution: x = 2− x

Substitute 4 for x 2 + y 2 in the second equation. 2x + 4 = 0 2x = − 4 x = −2

x = 4 − 4 x + x2 x2 − 5x + 4 = 0 ( x − 4)( x − 1) = 0 x=4

y = 4 − (− 2) 2 = 0

or x = 1

Solution: (–2, 0)

y = − 2 or y =1 Eliminate (4, –2); we must have y ≥ 0 . Solution: (1, 1)

y = 3x − 5  15.  2 2  x + y = 5

 x = 2 y 11.  2  x = y − 2 y

(1, –2) and (2, 1) are the intersection points. Solve by substitution: x 2 + (3 x − 5) 2 = 5 x 2 + 9 x 2 − 30 x + 25 = 5

(0, 0) and (8, 4) are the intersection points.

10 x 2 − 30 x + 20 = 0

Solve by substitution: 2 y = y2 − 2 y

x 2 − 3x + 2 = 0

y2 − 4 y = 0

( x − 1)( x − 2) = 0 x =1 or x = 2 y = −2 y =1 Solutions: (1, –2) and (2, 1)

y ( y − 4) = 0 y = 0 or y =4 x = 0 or x =8 Solutions: (0, 0) and (8, 4)

 x 2 + y 2 = 4 17.  2  y − x = 4

 x +y =4 13.  2 2  x + 2 x + y = 0 2

2

(–1, 1.73), (–1, –1.73), (0, 2), and (0, –2) are the intersection points.

(–2, 0) is the intersection point.

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Chapter 12: Systems of Equations and Inequalities

 x 2 + y 2 = 4 21.  y = x2 − 9 

Substitute x + 4 for y 2 in the first equation:

x2 + x + 4 = 4 x2 + x = 0 x ( x + 1) = 0 x=0

or x = −1

2

y =4

y2 = 3

y = ±2

y=± 3

(

)(

Solutions: (0, – 2), (0, 2), −1, 3 , −1, − 3 xy = 4  19.  2 2  x + y = 8

) No solution; Inconsistent. Solve by substitution: x 2 + ( x 2 − 9) 2 = 4 x 2 + x 4 − 18 x 2 + 81 = 4 x 4 − 17 x 2 + 77 = 0 17 ± 289 − 4(77) 2 17 ± −19 = 2 There are no real solutions to this expression. Inconsistent. x2 =

(–2, –2) and (2, 2) are the intersection points. Solve by substitution: 2

4 x2 +   = 8 x 16 x2 + 2 = 8 x 4 x + 16 = 8 x 2 x 4 − 8 x 2 + 16 = 0

 y = x 2 − 4 23.   y = 6 x − 13

( x 2 − 4) 2 = 0 x2 − 4 = 0 x2 = 4 x = 2 or x = −2 y = 2 or y = −2 Solutions: (–2, –2) and (2, 2)

(3, 5) is the intersection point. Solve by substitution: x 2 − 4 = 6 x − 13 x2 − 6 x + 9 = 0 ( x − 3) 2 = 0 x−3 = 0 x=3 y = (3) 2 − 4 = 5 Solution: (3,5)

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Section 12.6: Systems of Nonlinear Equations 29. Solve the first equation for y, substitute into the second equation and solve: x + y +1 = 0  y = −x −1   2 2  x + y + 6 y − x = −5

25. Solve the second equation for y, substitute into the first equation and solve: 2 x 2 + y 2 = 18   4 xy = 4  y =  x 

x 2 + (− x − 1)2 + 6(− x − 1) − x = −5

2

x 2 + x 2 + 2 x + 1 − 6 x − 6 − x = −5

4 2 x 2 +   = 18  x 16 2 x 2 + 2 = 18 x 4 2 x + 16 = 18 x 2

2 x2 − 5x = 0 x(2 x − 5) = 0 x = 0 or x = 5 2 y = −(0) − 1 = −1 If x = 0 : 5 5 7 y = − −1 = − If x = : 2 2 2 5 7 Solutions: (0, −1), ( 2 , − 2 )

2 x 4 − 18 x 2 + 16 = 0

(x

x4 − 9 x2 + 8 = 0 2

)(

)

− 8 x2 − 1 = 0

x2 = 8 x = ± 8= ± 2 2

or

If x = 2 2 :

y=

If x = − 2 2 :

y=

If x = 1:

y=

(2

4 2 2 4

= 2

−2 2

2

=− 2

5  2 5  2 4 x 2 − 3x  − x +  + 9  − x +  = 15 3  3 3  3

4 =4 1 4 y= = −4 −1

If x = −1:

Solutions:

31. Solve the second equation for y, substitute into the first equation and solve: 4 x 2 − 3xy + 9 y 2 = 15   2 5 2x + 3y = 5  y = − x +  3 3 

x2 = 1 x = ±1

or

)(

4 x 2 + 2 x 2 − 5 x + 4 x 2 − 20 x + 25 = 15 10 x 2 − 25 x + 10 = 0 2 x2 − 5x + 2 = 0 (2 x − 1)( x − 2) = 0

)

2, 2 , − 2 2, − 2 , (1, 4), (−1, − 4)

1 or x = 2 2 1 21 5 4 If x = : y = −  + = 2 32 3 3 2 5 1 If x = 2 : y = − (2) + = 3 3 3 1 4  1 Solutions:  ,  ,  2,   2 3  3 x=

27. Substitute the first equation into the second equation and solve: y = 2x + 1   2 2 2 x + y = 1 2

2 x 2 + ( 2 x + 1) = 1 2 x2 + 4 x2 + 4 x + 1 = 1 6 x2 + 4x = 0

33. Multiply each side of the second equation by 4 and add the equations to eliminate y:  x 2 − 4 y 2 = −7 ⎯⎯ → x2 − 4 y 2 = − 7  2 4 2 → 12 x 2 + 4 y 2 = 124 3x + y = 31 ⎯⎯ = 117 13x 2

2 x ( 3x + 2 ) = 0 2 x = 0 or 3 x + 2 = 0 2 x = 0 or x = − 3 y = 2(0) + 1 = 1 If x = 0 :

x2 = 9 x = ±3 If x = 3 : 3(3) 2 + y 2 = 31  y 2 = 4  y = ±2

4 1  2 y = 2 −  +1 = − +1 = − 3 3  3  2 1 Solutions: (0, 1),  − , −   3 3 2 If x = − : 3

777 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities

If x = −3 : 3(−3) 2 + y 2 = 31  y 2 = 4  y = ±2 Solutions: (3, 2), (3, –2), (–3, 2), (–3, –2) 35. 7 x 2 − 3 y 2 + 5 = 0  2 2  3x + 5 y = 12

9 x 2 + 15 y 2 = 36 1 4 1 2

If x = 1 : 2 2

9 3 1 3   + 5 y 2 = 12  y 2 =  y = ± 4 2 2 If x = − 1 : 2 2

 1 3  −  + 5 y 2 = 12   2 Solutions: 1 3 1 3  1  , ,  , − ,  − , 2 2 2 2  2

( 2)

2

− 2⋅y = 2 4 2

 y=2 2

If x = − 2 :

(

) − (− 2 ) y = 2



2 ⋅ y = −4  y =

3 − 2

2

Solutions:

(

)(

−4 2

 y = −2 2

2, 2 2 , − 2, − 2 2

)

39.  2 x 2 + y 2 = 2  2 2  x − 2 y + 8 = 0 2 2 2 x + y = 2  2 2  x − 2 y = −8 Multiply each side of the first equation by 2 and add the equations to eliminate y: 4x2 + 2 y 2 = 4

44 x 2 = 11

x=±

3

 − 2 ⋅ y = −4  y =

2 2 7 x − 3 y = −5  2 2 3x + 5 y = 12 Multiply each side of the first equation by 5 and each side of the second equation by 3 and add the equations to eliminate y: 35 x 2 − 15 y 2 = −25

x2 =

If x = 2 :

y2 =

9 3  y=± 4 2

3  1 3 ,  − , −  2  2 2

37. Multiply each side of the second equation by 2 and add the equations to eliminate xy: 2 → x 2 + 2 xy = 10  x + 2 xy = 10 ⎯⎯  2 2 → 6 x 2 − 2 xy = 4 3x − xy = 2 ⎯⎯ 7 x2

= 14 2

x =2 x=± 2

x 2 − 2 y 2 = −8 5 x 2 = −4 4 x2 = − 5 No real solution. The system is inconsistent.

41.  x 2 + 2 y 2 = 16  2 2  4 x − y = 24 Multiply each side of the second equation by 2 and add the equations to eliminate y: x 2 + 2 y 2 = 16 8 x 2 − 2 y 2 = 48 9 x 2 = 64 64 x2 = 9 8 x=± 3 8 If x = : 3 2 8 80   2 2   + 2 y = 16  2 y = 9 3

 y2 =

40 2 10  y=± 9 3

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Section 12.6: Systems of Nonlinear Equations

6 1 45.  4 + 4 = 6 x y    2 − 2 = 19  x 4 y 4

8 If x = − : 3 2 80  8 2 2  −  + 2 y = 16  2 y = 3 9   40 2 10  y=± 9 3 Solutions:  8 2 10   8 2 10   8 2 10   , ,  , − ,  − , , 3   3 3   3 3  3

Multiply each side of the first equation by –2 and add the equations to eliminate x: −2 12 − = −12 x4 y4 2 2 − = 19 x4 y4 14 − 4 =7 y

 y2 =

 8 2 10   − , −  3   3

y 4 = −2 There are no real solutions. The system is inconsistent.

2 5 43.  2 − 2 + 3 = 0 y x  3 1  + =7  x 2 y 2 2 5  x 2 − y 2 = −3    3 + 1 =7  x 2 y 2 Multiply each side of the second equation by 2 and add the equations to eliminate y: 5 2 − 2 = −3 2 x y 6 2 + 2 = 14 2 x y

47.  x 2 − 3 xy + 2 y 2 = 0  x 2 + xy = 6  Subtract the second equation from the first to eliminate the x 2 term. −4 xy + 2 y 2 = −6

2 xy − y 2 = 3 Since y ≠ 0 , we can solve for x in this equation to get y2 + 3 , y≠0 x= 2y Now substitute for x in the second equation and solve for y. x 2 + xy = 6

11

= 11 x2 x2 = 1 x = ±1 If x = 1: 3 1 1 1 + 2 = 7  2 = 4  y2 = 2 4 y y (1) 1  y=± 2 If x = −1: 3 1 1 1 + 2 = 7  2 = 4  y2 = 2 4 y y (−1) 1  y=± 2 1  1  1  1  Solutions: 1,  , 1, −  ,  −1,  ,  −1, −  2  2  2  2 

2

 y2 + 3   y2 + 3    + y=6  2y   2y  y4 + 6 y2 + 9 y2 + 3 + =6 2 4 y2 y 4 + 6 y 2 + 9 + 2 y 4 + 6 y 2 = 24 y 2 3 y 4 − 12 y 2 + 9 = 0

(y

y4 − 4 y2 + 3 = 0 2

)(

)

− 3 y2 −1 = 0

Thus, y = ± 3 or y = ±1 . If y = 1: x = 2 ⋅1 = 2 If y = −1: x = 2(−1) = − 2

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Chapter 12: Systems of Equations and Inequalities

If y = 3 :

x= 3

If y = − 3 :

x=− 3

Solutions: (2, 1), (–2, −1),

(

)(

3, 3 , − 3, − 3

)

49.  y 2 + y + x 2 − x − 2 = 0  x−2  =0 y +1+  y  Multiply each side of the second equation by –y and add the equations to eliminate y: y 2 + y + x2 − x − 2 = 0 2

−y − y

y 2 (3 + y )(3 − y ) = 0 y = 0 or y = −3 or y = 3 Since ln y is undefined when y ≤ 0 , the only solution is y = 3 . If y = 3 :

Solution:

−x+2=0 x ( x − 2) = 0

x = 0 or x = 2 If x = 0 : 2

x = y 4  x = 34 = 81

(81, 3)

 x2 + x + y 2 − 3 y + 2 = 0  55.  y2 − y x +1+ =0  x 

x2 − 2 x = 0

2

 x = y 4 So we have the system  2  x = 9 y Therefore we have : 9 y 2 = y 4  9 y 2 − y 4 = 0  y 2 (9 − y 2 ) = 0

2

y + y+0 −0−2 = 0  y + y−2 = 0  ( y + 2)( y − 1) = 0  y = − 2 or y = 1

( x + 1 )2 + y − 3 2 = ( 2) 2   2 2 ( x + 12 ) + ( y − 12 ) = 

1 2 1 2

If x = 2 : y 2 + y + 22 − 2 − 2 = 0  y 2 + y = 0  y ( y + 1) = 0  y = 0 or y = −1 Note: y ≠ 0 because of division by zero. Solutions: (0, –2), (0, 1), (2, –1)

51. Rewrite each equation in exponential form:  log x y = 3 → y = x3  5 log x (4 y ) = 5 → 4 y = x

Substitute the first equation into the second and solve: 4 x3 = x5 x5 − 4 x3 = 0

57. Graph: y1 = x ∧ (2 / 3); y2 = e ∧ (− x) Use INTERSECT to solve: 3.1

–4.7

x3 ( x 2 − 4) = 0 x3 = 0 or x 2 = 4  x = 0 or x = ±2 The base of a logarithm must be positive, thus x ≠ 0 and x ≠ − 2 .

4.7

-3.1 Solution: x = 0.48, y = 0.62 or (0.48, 0.62)

If x = 2 : y = 23 = 8 Solution: (2, 8)

53. Rewrite each equation in exponential form: 4

ln x = 4 ln y  x = e4ln y = eln y = y 4 log3 x = 2 + 2 log 3 y 2

x = 32 + 2log3 y = 32 ⋅ 32log3 y = 32 ⋅ 3log3 y = 9 y 2

780 Copyright © 2016 Pearson Education, Inc.

Section 12.6: Systems of Nonlinear Equations 65. Solve the first equation for x, substitute into the second equation and solve: x + 2y = 0  x = − 2y   2 2 ( x − 1) + ( y − 1) = 5

59. Graph: y1 = 3 2 − x 2 ; y2 = 4 / x3 Use INTERSECT to solve: 3.1

–4.7

(− 2 y − 1) 2 + ( y − 1) 2 = 5

4.7

4 y2 + 4 y + 1 + y2 − 2 y + 1 = 5  5 y2 + 2 y − 3 = 0 (5 y − 3)( y + 1) = 0 3 y = =0.6 or y = −1 5 6 x = − = − 1.2 or x = 2 5  6 3 The points of intersection are  − ,  , (2, –1) .  5 5

–3.1 Solution: x = −1.65, y = −0.89 or (–1.65, –0.89)

61. Graph: y1 = 4 12 − x 4 ; y2 = − 4 12 − x 4 ; y3 = 2 / x ; y4 = − 2 / x Use INTERSECT to solve:

67. Complete the square on the second equation. y2 + 4 y + 4 = x −1+ 4 ( y + 2)2 = x + 3 Substitute this result into the first equation. ( x − 1) 2 + x + 3 = 4

Solutions: x = 0.58, y = 1.86; x = 1.81, y = 1.05; x = 1.81, y = −1.05; ; x = 0.58, y = −1.86 or (0.58, 1.86), (1.81, 1.05), (1.81, –1.05), (0.58, –1.86)

x2 − 2 x + 1 + x + 3 = 4 x2 − x = 0 x( x − 1) = 0 x = 0 or x = 1

63. Graph: y1 = 2 / x; y2 = ln x Use INTERSECT to solve: 3.1

If x = 0 :

( y + 2) 2 = 0 + 3

y + 2 = ± 3  y = −2± 3

–4.7

If x = 1: ( y + 2) 2 = 1 + 3 y + 2 = ±2  y = − 2 ± 2 The points of intersection are:

4.7

( 0, − 2 − 3 ) , ( 0, − 2 + 3 ) , (1, – 4) , (1, 0) .

–3.1 Solution: x = 2.35, y = 0.85 or (2.35, 0.85)

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Chapter 12: Systems of Equations and Inequalities

The points of intersection are: (1, –2), (5, 2).

69. Solve the first equation for x, substitute into the second equation and solve: 4  y = x −3   x2 − 6x + y2 + 1 = 0  4 x−3 4 x −3 = y 4 x = +3 y y=

( y + 2 )2 + y 2 = 10 y 2 + 4 y + 4 + y 2 = 10 y2 + 2 y − 3 = 0

2

4  4  2  + 3 − 6  + 3 + y +1 = 0 y  y  16 24 24 + + 9 − − 18 + y 2 + 1 = 0 y y2 y 16 y 4

71. Let x and y be the two numbers. The system of equations is:  x = y+2  x − y = 2  2 2  x + y = 10 Solve the first equation for x, substitute into the second equation and solve:

2

+ y2 − 8 = 0

2

16 + y − 8 y = 0 y 4 − 8 y 2 + 16 = 0 ( y 2 − 4) 2 = 0 y2 − 4 = 0 y2 = 4 y = ±2 4 x = +3 = 5 If y = 2 : 2 4 If y = − 2 : x = +3 =1 −2

( y + 3)( y − 1) = 0

 y = −3 or y = 1

If y = −3 : x = −3 + 2 = −1 x = 1+ 2 = 3 If y = 1: The two numbers are 1 and 3 or –1 and –3.

73. Let x and y be the two numbers. The system of equations is: 4  xy = 4  x =  y   x2 + y 2 = 8  Solve the first equation for x, substitute into the second equation and solve: 2

4 2   + y =8 y   16 + y2 = 8 y2 16 + y 4 = 8 y 2 y 4 − 8 y 2 + 16 = 0

(y

2

−4

)

2

=0

y2 = 4 y = ±2

782 Copyright © 2016 Pearson Education, Inc.

Section 12.6: Systems of Nonlinear Equations

Solve the first equation for y, substitute into the second equation and solve. 2 x + 2 y = 16 2 y = 16 − 2 x y = 8− x

4 4 = 2; If y = − 2 : x = = −2 2 −2 The two numbers are 2 and 2 or –2 and –2. If y = 2 : x =

75. Let x and y be the two numbers. The system of equations is:  x − y = xy  1 1 x + y = 5  Solve the first equation for x, substitute into the second equation and solve: x − xy = y y x (1 − y ) = y  x = 1− y

x ( 8 − x ) = 15 8 x − x 2 = 15 x 2 − 8 x + 15 = 0

( x − 5 )( x − 3) = 0 x = 5 or x = 3 The dimensions of the rectangle are 3 inches by 5 inches.

81. Let x = the radius of the first circle. Let y = the radius of the second circle.

1

1 y + y =5

2π x + 2π y = 12π  2 2  π x + π y = 20π Solve the first equation for y, substitute into the second equation and solve: 2π x + 2π y = 12π x+ y =6 y = 6− x

1− y

1− y 1 + =5 y y 2− y =5 y 2 − y = 5y 6y = 2 y=

1 3

1

1

1  x= 3 = 3 = 1 2 1− 3 3 2

π x 2 + π y 2 = 20π x 2 + y 2 = 20

The two numbers are 1 and 1 . 2 3

x 2 + (6 − x) 2 = 20 x 2 + 36 − 12 x + x 2 = 20  2 x 2 − 12 x + 16 = 0

 a 2 =  77.  b 3 a + b = 10  a = 10 − b Solve the second equation for a , substitute into the first equation and solve: 10 − b 2 = 3 b 3(10 − b) = 2b 30 − 3b = 2b 30 = 5b b=6a=4 a + b = 10; b − a = 2 The ratio of a + b to b − a is 10 = 5 . 2

x 2 − 6 x + 8 = 0  ( x − 4)( x − 2) = 0 x = 4 or x = 2 y=2 y=4 The radii of the circles are 2 centimeters and 4 centimeters.

83. The tortoise takes 9 + 3 = 12 minutes or 0.2 hour longer to complete the race than the hare. Let r = the rate of the hare. Let t = the time for the hare to complete the race. Then t + 0.2 = the time for the tortoise and r − 0.5 = the rate for the tortoise. Since the length of the race is 21 meters, the distance equations are: 21  r t = 21  r =  t  ( r − 0.5 )( t + 0.2 ) = 21 

79. Let x = the width of the rectangle. Let y = the length of the rectangle. 2 x + 2 y = 16  xy = 15 

Solve the first equation for r, substitute into the second equation and solve: 783 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities

 21   − 0.5  ( t + 0.2 ) = 21 t   4.2 − 0.5t − 0.1 = 21 21 + t 4.2   − 0.5t − 0.1 = 10t ⋅ ( 21) 10t  21 + t   210t + 42 − 5t 2 − t = 210t 5t 2 + t − 42 = 0

x 2 + (150 − 2 x) 2 = 4500

( 5t − 14 )( t + 3) = 0 5t − 14 = 0 5t = 14

87. Find equations relating area and perimeter: 2 2  x + y = 4500  3x + 3 y + ( x − y ) = 300 Solve the second equation for y, substitute into the first equation and solve: 4 x + 2 y = 300 2 y = 300 − 4 x y = 150 − 2 x x 2 + 22,500 − 600 x + 4 x 2 = 4500

or t + 3 = 0 t = −3

5 x 2 − 600 x + 18, 000 = 0

14 = 2.8 t= 5 t = −3 makes no sense, since time cannot be negative. Solve for r: 21 r= = 7.5 2.8 The average speed of the hare is 7.5 meters per hour, and the average speed for the tortoise is 7 meters per hour.

85. Let x = the width of the cardboard. Let y = the length of the cardboard. The width of the box will be x − 4 , the length of the box will be y − 4 , and the height is 2. The volume is V = ( x − 4)( y − 4)(2) . Solve the system of equations: 216  y=  xy = 216 x  2( x − 4)( y − 4) = 224 Solve the first equation for y, substitute into the second equation and solve. 216 ( 2 x − 8 )  − 4  = 224  x  1728 + 32 = 224 432 − 8 x − x 432 x − 8 x 2 − 1728 + 32 x = 224 x

x 2 − 120 x + 3600 = 0 ( x − 60) 2 = 0 x − 60 = 0 x = 60 y = 150 − 2(60) = 30 The sides of the squares are 30 feet and 60 feet.

89. Solve the system for l and w : 2l + 2 w = P  lw= A 

Solve the first equation for l , substitute into the second equation and solve. 2l = P − 2 w P l = −w 2 P   − w w = A 2   P w − w2 = A 2 P 2 w − w+ A = 0 2 P ± P2 − 16 A P ± P2 − 4 A 2 4 4 2 4 w= = 2 2 =

8 x 2 − 240 x + 1728 = 0

P± 2

x 2 − 30 x + 216 = 0

( x − 12 )( x − 18) = 0 x − 12 = 0

or x − 18 = 0 x = 12 x = 18 The cardboard should be 12 centimeters by 18 centimeters. 784 Copyright © 2016 Pearson Education, Inc.

P 2 − 16 A

2 2

=

P ± P 2 − 16 A 4

Section 12.6: Systems of Nonlinear Equations

If w = l=

95. Solve the system: 2 2 2 x + 3 y = 14  y = mx + b  Solve the system by substitution:

P + P 2 − 16 A then 4

P P + P 2 − 16 A P − P 2 − 16 A − = 2 4 4

2

2 x 2 + 3 ( mx + b ) = 14

P − P 2 − 16 A If w = then 4 2

(3m

2

w=

P − P 2 − 16 A P + P 2 − 16 A and l = 4 4

m 2 − 8m + 16 = 0

( m − 4)

)

+ 2 x 2 + 6mbx + 3b 2 − 14 = 0

(3m 2 + 2) x 2 + (12m − 6m 2 ) x + (3m 2 − 12m − 2) = 0 Find when the discriminant equals 0: (12m − 6m 2 ) 2 − 4(3m 2 + 2)(3m 2 − 12m − 2) = 0

91. Solve the equation: m 2 − 4(2m − 4) = 0 2

2

Note that the tangent line passes through (1, 2). Find the relation between m and b: 2 = m(1) + b  b = 2 − m Substitute into the quadratic to eliminate b: (3m 2 + 2) x 2 + 6m(2 − m) x + 3(2 − m) 2 − 14 = 0

P P − P − 16 A P + P − 16 A − = 2 4 4 If it is required that length be greater than width, then the solution is: l=

2 x 2 + 3m 2 x 2 + 6mbx + 3b 2 = 14

144m 2 + 96m + 16 = 0

=0

9m 2 + 6 m + 1 = 0

m=4 Use the point-slope equation with slope 4 and the point (2, 4) to obtain the equation of the tangent line: y − 4 = 4( x − 2)  y − 4 = 4 x − 8  y = 4 x − 4

(3m + 1) 2 = 0 3m + 1 = 0 m=−

1 3

 1 7 b = 2−m = 2−−  =  3 3

93. Solve the system:  y = x 2 + 2   y = mx + b Solve the system by substitution: x 2 + 2 = mx + b  x 2 − mx + 2 − b = 0 Note that the tangent line passes through (1, 3). Find the relation between m and b: 3 = m(1) + b  b = 3 − m Substitute into the quadratic to eliminate b: x 2 − mx + 2 − (3 − m) = 0  x 2 − mx + (m − 1) = 0 Find when the discriminant equals 0:

1 7 The equation of the tangent line is y = − x + . 3 3 97. Solve the system: 2 2  x − y = 3  y = mx + b  Solve the system by substitution: 2

x 2 − ( mx + b ) = 3 x 2 − m 2 x 2 − 2mbx − b 2 = 3

(1 − m ) x 2

( −m )2 − 4 (1)( m − 1) = 0

2

− 2mbx − b 2 − 3 = 0

Note that the tangent line passes through (2, 1). Find the relation between m and b: 1 = m(2) + b  b = 1 − 2m Substitute into the quadratic to eliminate b: (1 − m 2 ) x 2 − 2m(1 − 2m) x − (1 − 2m) 2 − 3 = 0

m 2 − 4m + 4 = 0

( m − 2 )2 = 0 m−2 = 0 m=2 b = 3− m = 3− 2 =1 The equation of the tangent line is y = 2 x + 1 .

(1 − m 2 ) x 2 + (− 2m + 4m 2 ) x − 1 + 4m − 4m 2 − 3 = 0 (1 − m 2 ) x 2 + (− 2m + 4m 2 ) x + ( − 4m 2 + 4m − 4) = 0 Find when the discriminant equals 0:

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Chapter 12: Systems of Equations and Inequalities

( −2m + 4m )

2 2

(

− 4 1 − m2

)( −4m

2

)

+ 4m − 4 = 0

4m 2 − 16m3 + 16m 4 − 16m 4 + 16m3 − 16m + 16 = 0 4m 2 − 16m + 16 = 0

101. Since the area of the square piece of sheet metal is 100 square feet, the sheet’s dimensions are 10 feet by 10 feet. Let x = the length of the cut. x

m 2 − 4m + 4 = 0

x

( m − 2 )2 = 0

x

m=2 The equation of the tangent line is y = 2 x − 3 .

10 – 2x

99. Solve for r1 and r2 :

The dimensions of the box are: length = 10 − 2 x; width = 10 − 2 x; height = x . Note that each of these expressions must be positive. So we must have x > 0 and 10 − 2 x > 0  x < 5, that is, 0 < x < 5 . So the volume of the box is given by V = ( length ) ⋅ ( width ) ⋅ ( height )

b a

= (10 − 2 x )(10 − 2 x )( x )

b c   −r2 −  r2 = a a  b c −r2 2 − r2 − = 0 a a ar2 2 + br2 + c = 0

= (10 − 2 x )

a.

=

( x)

In order to get a volume equal to 9 cubic feet, 2

( x ) = 9.

(10 − 2 x )2 ( x ) = 9

2

(100 − 40 x + 4 x ) x = 9 2

100 x − 40 x 2 + 4 x3 = 9 So we need to solve the equation 4 x3 − 40 x 2 + 100 x − 9 = 0 .

Graphing y1 = 4 x3 − 40 x 2 + 100 x − 9 on a calculator yields the graph

b  b 2 − 4ac 2b − 2a 2a

80

−b  b 2 − 4ac 2a The solutions are: =

r1 =

2

we solve (10 − 2 x )

−b ± b − 4ac 2a b r1 = − r2 − = a  −b ± b 2 − 4ac  b − = −   a 2a  

r2 =

10

10 – 2x

b  r1 + r2 = − a   rr =c  1 2 a Substitute and solve: r1 = − r2 −

10

−2

−b + b 2 − 4ac −b − b 2 − 4ac . and r2 = 2a 2a

10 −40 80

−2

10 −40

786 Copyright © 2016 Pearson Education, Inc.

Section 12.7: Systems of Inequalities 80

−2

3. x 2 + y 2 = 9 The graph is a circle. Center: (0, 0) ; Radius: 3 10

−40

The graph indicates that there are three real zeros on the interval [0, 6]. Using the ZERO feature of a graphing calculator, we find that the three roots shown occur at x ≈ 0.093 , x ≈ 4.274 and x ≈ 5.632 . But we’ve already noted that we must have 0 1 is false, shade the opposite side of the circle from (0, 0).

21. xy ≥ 4 Graph the hyperbola xy = 4 . Use a solid line since the inequality uses ≥ . Choose a test point not on the hyperbola, such as (0, 0). Since 0 ⋅ 0 ≥ 4 is false, shade the opposite side of the hyperbola from (0, 0).

 x+ y ≤ 2 23.  2 x + y ≥ 4 Graph the line x + y = 2 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≤ 2 is true, shade the side of the line containing (0, 0). Graph the line 2 x + y = 4 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≥ 4 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

19. y ≤ x 2 − 1 Graph the parabola y = x 2 − 1 . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 0). Since 0 ≤ 02 − 1 is false, shade the opposite side of the parabola from (0, 0).

 2x − y ≤ 4 25.  3 x + 2 y ≥ − 6 Graph the line 2 x − y = 4 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) − 0 ≤ 4 is true, shade the side of the line containing (0, 0). Graph the line 3x + 2 y = − 6 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 3(0) + 2(0) ≥ − 6 is true, shade the side of the line containing (0, 0). The overlapping region is the solution.

788 Copyright © 2016 Pearson Education, Inc.

Section 12.7: Systems of Inequalities

The overlapping region is the solution.

2 x − 3 y ≤ 0 27.  3x + 2 y ≤ 6 Graph the line 2 x − 3 y = 0 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 3). Since 2(0) − 3(3) ≤ 0 is true, shade the side of the line containing (0, 3). Graph the line 3 x + 2 y = 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) + 2(0) ≤ 6 is true, shade the side of the line containing (0, 0). The overlapping region is the solution.

2 x + y ≥ − 2 31.  2 x + y ≥ 2 Graph the line 2 x + y = − 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≥ − 2 is true, shade the side of the line containing (0, 0). Graph the line 2 x + y = 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

 x − 2y ≤ 6 29.  2 x − 4 y ≥ 0 Graph the line x − 2 y = 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 − 2(0) ≤ 6 is true, shade the side of the line containing (0, 0). Graph the line 2 x − 4 y = 0 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 2). Since 2(0) − 4(2) ≥ 0 is false, shade the opposite side of the line from (0, 2).

789 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities

2 x + 3 y ≥ 6 33.  2 x + 3 y ≤ 0 Graph the line 2 x + 3 y = 6 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) + 3(0) ≥ 6 is false, shade the opposite side of the line from (0, 0). Graph the line 2 x + 3 y = 0 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 2). Since 2(0) + 3(2) ≤ 0 is false, shade the opposite side of the line from (0, 2). Since the regions do not overlap, the solution is an empty set.

2 2  x + y ≤ 9 35.   x + y ≥ 3 Graph the circle x 2 + y 2 = 9 . Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since 02 + 02 ≤ 9 is true, shade the same side of the circle as (0, 0). Graph the line x + y = 3 . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≥ 3 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

2  y ≥ x − 4 37.   y ≤ x − 2 Graph the parabola y = x 2 − 4 . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 0 ≥ 02 − 4 is true, shade the same side of the parabola as (0, 0). Graph the line y = x − 2 . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 0 ≤ 0 − 2 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution.

39.  x 2 + y 2 ≤ 16  2  y ≥ x − 4

Graph the circle x 2 + y 2 = 16 . Use a sold line since the inequality is not strict. Choose a test point not on the circle, such as (0, 0) . Since 02 + 02 ≤ 16 is true, shade the side of the circle containing (0, 0) . Graph the parabola y = x 2 − 4 . Use a solid line since the inequality is not strict. Choose a test point not on the parabola, such as (0, 0) . Since 0 ≥ 02 − 4 is true, shade the side of the parabola that contains (0, 0) . The overlapping region is the solution.

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Section 12.7: Systems of Inequalities

 xy ≥ 4 41.  2  y ≥ x + 1 Graph the hyperbola xy = 4 . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 0 ⋅ 0 ≥ 4 is false, shade the opposite side of the hyperbola from (0, 0). Graph the parabola y = x 2 + 1 . Use a solid line since the inequality uses ≥ . Choose a test point not on the parabola, such as (0, 0). Since 0 ≥ 02 + 1 is false, shade the opposite side of the parabola from (0, 0).The overlapping region is the solution.

x≥0   y≥0  43.  2 x + y ≤ 6  x + 2 y ≤ 6 Graph x ≥ 0; y ≥ 0 . Shaded region is the first quadrant. Graph the line 2 x + y = 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≤ 6 is true, shade the side of the line containing (0, 0). Graph the line x + 2 y = 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 + 2(0) ≤ 6 is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded. Find the vertices:

2(6 − 2 y ) + y = 6 12 − 4 y + y = 6 12 − 3 y = 6 −3 y = −6 y=2 x = 6 − 2(2) = 2 The point of intersection is (2, 2). The four corner points are (0, 0), (0, 3), (3, 0), and (2, 2).

x≥0   y≥0  45.  x + y≥2  2 x + y ≥ 4 Graph x ≥ 0; y ≥ 0 . Shaded region is the first quadrant. Graph the line x + y = 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0). Graph the line 2 x + y = 4 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≥ 4 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. The graph is unbounded.

The x-axis and y-axis intersect at (0, 0). The intersection of x + 2 y = 6 and the y-axis is (0, 3). The intersection of 2 x + y = 6 and the x-axis is (3, 0). To find the intersection of x + 2 y = 6 and 2 x + y = 6 , solve the system:  x + 2y = 6  2 x + y = 6 Solve the first equation for x: x = 6 − 2 y . Substitute and solve:

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Chapter 12: Systems of Equations and Inequalities

Find the vertices: The intersection of x + y = 2 and the x-axis is (2, 0). The intersection of 2 x + y = 4 and the yaxis is (0, 4). The two corner points are (2, 0), and (0, 4).

x≥0   y≥0  47.  x + y ≥ 2 2 x + 3 y ≤ 12   3x + y ≤ 12 Graph x ≥ 0; y ≥ 0 . Shaded region is the first quadrant. Graph the line x + y = 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0). Graph the line 2 x + 3 y = 12 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) + 3(0) ≤ 12 is true, shade the side of the line containing (0, 0). Graph the line 3x + y = 12 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 3(0) + 0 ≤ 12 is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.

Find the vertices: The intersection of x + y = 2 and the y-axis is (0, 2). The intersection of x + y = 2 and the xaxis is (2, 0). The intersection of 2 x + 3 y = 12 and the y-axis is (0, 4). The intersection of 3x + y = 12 and the x-axis is (4, 0). To find the intersection of 2 x + 3 y = 12 and 3x + y = 12 , solve the system: 2 x + 3 y = 12   3 x + y = 12 Solve the second equation for y: y = 12 − 3x .

Substitute and solve: 2 x + 3(12 − 3x) = 12 2 x + 36 − 9 x = 12 −7 x = −24 24 x= 7 72 12  24  y = 12 − 3   = 12 − = 7 7  7   24 12  The point of intersection is  ,  .  7 7 The five corner points are (0, 2), (0, 4), (2, 0),  24 12  (4, 0), and  ,  .  7 7

x≥ 0   y≥ 0  49.  x + y ≥ 2  x+ y ≤ 8  2 x + y ≤ 10 Graph x ≥ 0; y ≥ 0 . Shaded region is the first quadrant. Graph the line x + y = 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0). Graph the line x + y = 8 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≤ 8 is true, shade the side of the line containing (0, 0). Graph the line 2 x + y = 10 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≤ 10 is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.

Find the vertices: The intersection of x + y = 2 and the y-axis is (0, 2). The intersection of x + y = 2 and the x-axis is (2, 0). The intersection of x + y = 8 and the y-

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Section 12.7: Systems of Inequalities

axis is (0, 8). The intersection of 2 x + y = 10 and the x-axis is (5, 0). To find the intersection of x + y = 8 and 2 x + y = 10 , solve the system:  x+ y =8  2 x + y = 10 Solve the first equation for y: y = 8 − x . Substitute and solve: 2 x + 8 − x = 10 x=2 y = 8−2 = 6 The point of intersection is (2, 6). The five corner points are (0, 2), (0, 8), (2, 0), (5, 0), and (2, 6).

53. The system of linear inequalities is: x≤4  x + y ≤ 6   x≥ 0   y≥ 0 55. The system of linear inequalities is: x ≤ 20   y ≥ 15   x + y ≤ 50 x − y ≤ 0  x≥ 0 

x≥ 0   y≥ 0  51.  x + 2 y ≥ 1  x + 2 y ≤ 10 Graph x ≥ 0; y ≥ 0 . Shaded region is the first quadrant. Graph the line x + 2 y = 1 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 + 2(0) ≥ 1 is false, shade the opposite side of the line from (0, 0). Graph the line x + 2 y = 10 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 + 2(0) ≤ 10 is true, shade the side of the line containing (0, 0). The overlapping region is the solution. The graph is bounded.

Find the vertices: The intersection of x + 2 y = 1 and the y-axis is (0, 0.5). The intersection of x + 2 y = 1 and the x-axis is (1, 0). The intersection of x + 2 y = 10 and the y-axis is (0, 5). The intersection of x + 2 y = 10 and the x-axis is (10, 0). The four corner points are (0, 0.5), (0, 5), (1, 0), and (10, 0).

57. a.

Let x = the amount invested in Treasury bills, and let y = the amount invested in corporate bonds. The constraints are: x + y ≤ 50, 000 because the total investment cannot exceed $50,000. x ≥ 35, 000 because the amount invested in Treasury bills must be at least $35,000. y ≤ 10, 000 because the amount invested in corporate bonds must not exceed $10,000. x ≥ 0, y ≥ 0 because a non-negative amount must be invested. The system is  x + y ≤ 50, 000  x ≥ 35, 000  y ≤ 10, 000   x≥0  y≥0 

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Chapter 12: Systems of Equations and Inequalities b. Graph the system.

b. Graph the system.

The corner points are (35,000, 0), (35,000, 10,000), (40,000, 10,000), (50,000, 0). 59. a.

Let x = the # of packages of the economy blend, and let y = the # of packages of the superior blend. The constraints are: x ≥ 0, y ≥ 0 because a non-negative # of packages must be produced. 4 x + 8 y ≤ 75 ⋅16 because the total amount of “A grade” coffee cannot exceed 75 pounds. (Note: 75 pounds = (75)(16) ounces.) 12 x + 8 y ≤ 120 ⋅16 because the total amount of “B grade” coffee cannot exceed 120 pounds. (Note: 120 pounds = (120)(16) ounces.) Simplifying the inequalities, we obtain: 4 x + 8 y ≤ 75 ⋅16

12 x + 8 y ≤ 120 ⋅16

x + 2 y ≤ 75 ⋅ 4

3x + 2 y ≤ 120 ⋅ 4

x + 2 y ≤ 300

3x + 2 y ≤ 480

The corner points are (0, 0), (0, 150), (90, 105), (160, 0). 61. a.

Let x = the # of microwaves, and let y = the # of printers. The constraints are: x ≥ 0, y ≥ 0 because a non-negative # of items must be shipped. 30 x + 20 y ≤ 1600 because a total cargo weight cannot exceed 1600 pounds. 2 x + 3 y ≤ 150 because the total cargo volume cannot exceed 150 cubic feet. Note that the inequality 30 x + 20 y ≤ 1600 can be simplified: 3x + 2 y ≤ 160 . The system is: 3x + 2 y ≤ 160  2 x + 3 y ≤ 150  x ≥ 0; y ≥ 0

b. Graph the system.

The system is: x ≥ 0 y ≥ 0    x + 2 y ≤ 300 3x + 2 y ≤ 480

The corner points are (0, 0), (0, 50), (36, 26),  160  ,0  .   3 

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Section 12.8: Linear Programming

63.

3. z = x + y

3r = sin θ 3r 2 = r sin θ 3( x 2 + y 2 ) = y 2 3x + 3 y 2 − y = 0 1 x2 + y 2 − y = 0 3 1 1 1 2 2 x + y − y+ = 3 36 36 2 2 1 1  x2 +  y −  =   6 6 

Vertex Value of z = x + y

z = 0+3 = 3 z = 0+6 = 6 (5, 6) z = 5 + 6 = 11 (5, 2) z = 5+2 = 7 (4, 0) z = 4+0 = 4 The maximum value is 11 at (5, 6), and the minimum value is 3 at (0, 3). (0, 3) (0, 6)

1 The graph is a circle with center  0,  and 

radius

6

1 . 6

5. z = x + 10 y Vertex Value of z = x + 10 y

z = 0 + 10(3) = 30 z = 0 + 10(6) = 60 (5, 6) z = 5 + 10(6) = 65 (5, 2) z = 5 + 10(2) = 25 (4, 0) z = 4 + 10(0) = 4 The maximum value is 65 at (5, 6), and the minimum value is 4 at (4, 0). (0, 3)

(0, 6)

7. z = 5 x + 7 y Vertex Value of z = 5 x + 7 y

z = 5(0) + 7(3) = 21 z = 5(0) + 7(6) = 42 (5, 6) z = 5(5) + 7(6) = 67 (5, 2) z = 5(5) + 7(2) = 39 (4, 0) z = 5(4) + 7(0) = 20 The maximum value is 67 at (5, 6), and the minimum value is 20 at (4, 0). (0, 3)

(0, 6)

65.

2 cos 2 θ − cos θ − 1 = 0 (2 cos θ + 1)(cos θ − 1) = 0 2 cos θ + 1 = 0 cos θ = −

or cos θ − 1 = 0 1 2

2π 4π θ= , 3 3

cos θ = 1

θ=0

On 0 ≤ θ < 2π , the solution set is 0,

2π 4π , . 3 3

9. Maximize z = 2 x + y subject to x ≥ 0, y ≥ 0, x + y ≤ 6, x + y ≥ 1 . Graph the constraints. y

(0,6)

x+y=6

Section 12.8 1. objective function

(0,1) (6,0)

x

(1,0) x+y=1

The corner points are (0, 1), (1, 0), (0, 6), (6, 0). Evaluate the objective function: 795 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities

Vertex Value of z = 2 x + y

z = 2(0) + 1 = 1 (0, 6) z = 2(0) + 6 = 6 (1, 0) z = 2(1) + 0 = 2 (6, 0) z = 2(6) + 0 = 12 The maximum value is 12 at (6, 0). (0, 1)

11. Minimize z = 2 x + 5 y subject to x ≥ 0, y ≥ 0, x + y ≥ 2, x ≤ 5, y ≤ 3 . Graph the constraints. y

x=5

(0,3)

(5,3)

(5,0)

3 Solve the second equation for y: y = 6 − x 2 Substitute and solve: 3   2 x + 3  6 − x  = 12 2   9 2 x + 18 − x = 12 2 5 − x = −6 2 12 x= 5 3  12  18 12 y = 6−   = 6− = 2 5  5 5

The point of intersection is ( 2.4, 2.4 ) .

y=3

The corner points are (0, 2), (2, 0), (0, 4), (4, 0), (2.4, 2.4). Evaluate the objective function:

(0,2) (2,0)

2 x + 3 y = 12  3x + 2 y = 12

x

x+y=2

The corner points are (0, 2), (2, 0), (0, 3), (5, 0), (5, 3). Evaluate the objective function: Vertex Value of z = 2 x + 5 y

z = 2(0) + 5(2) = 10 z = 2(0) + 5(3) = 15 (2, 0) z = 2(2) + 5(0) = 4 (5, 0) z = 2(5) + 5(0) = 10 (5, 3) z = 2(5) + 5(3) = 25 The minimum value is 4 at (2, 0). (0, 2) (0, 3)

Vertex Value of z = 3x + 5 y z = 3(0) + 5(2) = 10 (0, 2) z = 3(0) + 5(4) = 20 (0, 4) z = 3(2) + 5(0) = 6 (2, 0) z = 3(4) + 5(0) = 12 (4, 0) (2.4, 2.4) z = 3(2.4) + 5(2.4) = 19.2 The maximum value is 20 at (0, 4).

15. Minimize z = 5 x + 4 y subject to x ≥ 0, y ≥ 0, x + y ≥ 2, 2 x + 3 y ≤ 12, 3x + y ≤ 12 . Graph the constraints. y

13. Maximize z = 3 x + 5 y subject to x ≥ 0, y ≥ 0, x + y ≥ 2, 2 x + 3 y ≤ 12, 3x + 2 y ≤ 12 . Graph the constraints.

3x + y = 12

(0,4)

y

( (0,2) 3x + 2y = 12

7

, 12 ) 7

2x + 3y = 12 (2,0)

(0,4)

24

(4,0)

x

x+y=2 (2.4,2.4)

(0,2) (2,0)

To find the intersection of 2 x + 3 y = 12 and 3x + y = 12 , solve the system:

2x + 3y = 12 (4,0) x

x+y=2

To find the intersection of 2 x + 3 y = 12 and 3x + 2 y = 12 , solve the system:

2 x + 3 y = 12   3 x + y = 12 Solve the second equation for y: y = 12 − 3 x Substitute and solve:

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Section 12.8: Linear Programming

The point of intersection is (10/3 10/3). The corner points are (0, 10), (10, 0), (10/3, 10/3). Evaluate the objective function:

2 x + 3(12 − 3 x) = 12 2 x + 36 − 9 x = 12 −7 x = −24 x = 24 7

y = 12 − 3

( 247 ) = 12 − 727 = 127

The point of intersection is

 24 12   , .  7 7

The corner points are (0, 2), (2, 0), (0, 4), (4, 0),  24 12   , .  7 7

Evaluate the objective function:

Vertex

Value of z = 5 x + 4 y

(0, 2)

z = 5(0) + 4(2) = 8

(0, 4)

z = 5(0) + 4(4) = 16

(2, 0)

z = 5(2) + 4(0) = 10

(4, 0)

z = 5(4) + 4(0) = 20

( 247 , 127 )

z=5

Value of z = 5 x + 2 y z = 5(0) + 2(10) = 20 z = 5(10) + 2(0) = 50

Vertex (0, 10) (10, 0)

( 247 ) + 4 ( 127 ) = 24

The minimum value is 8 at (0, 2). 17. Maximize z = 5 x + 2 y subject to x ≥ 0, y ≥ 0, x + y ≤ 10, 2 x + y ≥ 10, x + 2 y ≥ 10 . Graph the constraints.

 10 10   10   10  70 = 23 13  ,  z = 5  + 2  =  3 3  3  3 3

The maximum value is 50 at (10, 0). 19. Let x = the number of downhill skis produced, and let y = the number of cross-country skis produced. The total profit is: P = 70 x + 50 y . Profit is to be maximized, so this is the objective function. The constraints are: x ≥ 0, y ≥ 0 A positive number of skis must be produced. 2 x + y ≤ 40 Manufacturing time available. x + y ≤ 32 Finishing time available. Graph the constraints. y

2x + y = 40 (0,32)

y

(8,24)

(0,10) x + y = 10

x + y = 32 (20,0)

x

(0,0)

(103 , 103 )

To find the intersection of x + y = 32 and 2 x + y = 40 , solve the system:

(10,0) x x + 2y = 10 2x + y = 10

To find the intersection of 2 x + y = 10 and x + 2 y = 10 , solve the system: 2 x + y = 10   x + 2 y = 10 Solve the first equation for y: y = 10 − 2 x . Substitute and solve: x + 2(10 − 2 x) = 10 x + 20 − 4 x = 10 −3 x = −10 10 x= 3 20 10  10  = y = 10 − 2   = 10 − 3 3  3

 x + y = 32  2 x + y = 40 Solve the first equation for y: y = 32 − x . Substitute and solve: 2 x + (32 − x) = 40

x=8 y = 32 − 8 = 24 The point of intersection is (8, 24). The corner points are (0, 0), (0, 32), (20, 0), (8, 24). Evaluate the objective function: Vertex Value of P = 70 x + 50 y (0, 0)

P = 70(0) + 50(0) = 0

(0, 32)

P = 70(0) + 50(32) = 1600

(20, 0) P = 70(20) + 50(0) = 1400 (8, 24) P = 70(8) + 50(24) = 1760

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Chapter 12: Systems of Equations and Inequalities

The maximum profit is $1760, when 8 downhill skis and 24 cross-country skis are produced. With the increase of the manufacturing time to 48 hours, we do the following: The constraints are: x ≥ 0, y ≥ 0

A positive number of skis must be produced. 2 x + y ≤ 48 Manufacturing time available. x + y ≤ 32 Finishing time available. Graph the constraints.

x + y ≤ 35 Maximum number of tables. 6 x + 10 y ≥ 250 Number of guests. x ≤ 15 Rectangular tables available. Graph the constraints. y x = 15 x + y = 35 (0, 35) 6x + 10y = 250 (0, 25)

(15, 20) (15, 16)

10

y

25

2x + y = 48

x

(0,32)

(16,16) x + y = 32 (24,0)

x

(0,0)

To find the intersection of x + y = 32 and 2 x + y = 48 , solve the system:  x + y = 32  2 x + y = 48 Solve the first equation for y: y = 32 − x . Substitute and solve: 2 x + (32 − x) = 48 x = 16 y = 32 − 16 = 16 The point of intersection is (16, 16). The corner points are (0, 0), (0, 32), (24, 0), (16, 16). Evaluate the objective function: Vertex Value of P = 70 x + 50 y P = 70(0) + 50(0) = 0 (0, 0) (0, 32) P = 70(0) + 50(32) = 1600 (24, 0) P = 70(24) + 50(0) = 1680 (16, 16) P = 70(16) + 50(16) = 1920 The maximum profit is $1920, when 16 downhill skis and 16 cross-country skis are produced.

21. Let x = the number of rectangular tables rented, and let y = the number of round tables rented. The cost for the tables is: C = 28 x + 52 y . Cost is to be minimized, so this is the objective function. The constraints are: x ≥ 0, y ≥ 0 A non-negative number of tables must be used.

The corner points are (0, 25), (0, 35), (15, 20), (15, 16). Evaluate the objective function: Vertex Value of C = 28 x + 52 y (0, 25) C = 28(0) + 52(25) = 1300 (0, 35) C = 28(0) + 52(35) = 1820 (15, 20) C = 28(15) + 52(20) = 1460 (15, 16) C = 28(15) + 52(16) = 1252 Kathleen should rent 15 rectangular tables and 16 round tables in order to minimize the cost. The minimum cost is $1252.00. 23. Let x = the amount invested in junk bonds, and let y = the amount invested in Treasury bills. The total income is: I = 0.09 x + 0.07 y . Income is to be maximized, so this is the objective function. The constraints are: x ≥ 0, y ≥ 0 A non-negative amount must be invested. x + y ≤ 20, 000 Total investment cannot exceed $20,000. Amount invested in junk bonds x ≤ 12, 000 must not exceed $12,000. y ≥ 8000 Amount invested in Treasury bills must be at least $8000. a. y ≥ x Amount invested in Treasury bills must be equal to or greater than the amount invested in junk bonds. Graph the constraints.

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Section 12.8: Linear Programming

(0,20000) x + y = 20000

Vertex (12000, 8000)

Value of I = 0.09 x + 0.07 y I = 0.09(12000) + 0.07(8000)

(8000, 8000)

I = 0.09(8000) + 0.07(8000) = 1280

= 1640

y=x (10000,10000) (0,8000)

(8000,8000)

(10000, 10000) I = 0.09(10000) + 0.07(10000)

y = 8000

= 1600

The maximum income is $1640, when $12,000 is invested in junk bonds and $8000 is invested in Treasury bills.

x = 12000

The corner points are (0, 20,000), (0, 8000), (8000, 8000), (10,000, 10,000). Evaluate the objective function: Value of I = 0.09 x + 0.07 y I = 0.09(0) + 0.07(20000) = 1400 (0, 8000) I = 0.09(0) + 0.07(8000) = 560 (8000, 8000) I = 0.09(8000) + 0.07(8000) = 1280 (10000, 10000) I = 0.09(10000) + 0.07(10000) = 1600

Vertex (0, 20000)

The maximum income is $1600, when $10,000 is invested in junk bonds and $10,000 is invested in Treasury bills. b.

y≤x

Amount invested in Treasury bills must not exceed the amount invested in junk bonds. Graph the constraints.

25. Let x = the number of pounds of ground beef, and let y = the number of pounds of ground pork. The total cost is: C = 0.75 x + 0.45 y . Cost is to be minimized, so this is the objective function. The constraints are: x ≥ 0, y ≥ 0 A positive number of pounds must be used. x ≤ 200 Only 200 pounds of ground beef are available. y ≥ 50 At least 50 pounds of ground pork must be used. 0.75 x + 0.60 y ≥ 0.70( x + y ) Leanness condition (Note that the last equation will simplify to 1 y ≤ x .) Graph the constraints. 2 y (200,100)

x + y = 20000

(100,50) (200,50) y=x

(8000,8000)

(10000,10000) y = 8000 (12000,8000)

x = 12000

The corner points are (12,000, 8000), (8000, 8000), (10,000, 10,000). Evaluate the objective function:

x

The corner points are (100, 50), (200, 50), (200, 100). Evaluate the objective function: Vertex Value of C = 0.75 x + 0.45 y (100, 50) C = 0.75(100) + 0.45(50) = 97.50 (200, 50) C = 0.75(200) + 0.45(50) = 172.50 (200, 100) C = 0.75(200) + 0.45(100) = 195

The minimum cost is $97.50, when 100 pounds of ground beef and 50 pounds of ground pork are used.

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Chapter 12: Systems of Equations and Inequalities 27. Let x = the number of racing skates manufactured, and let y = the number of figure skates manufactured. The total profit is: P = 10 x + 12 y . Profit is to be maximized, so this is the objective function. The constraints are: x ≥ 0, y ≥ 0

A positive number of skates must be manufactured. 6 x + 4 y ≤ 120 Only 120 hours are available for fabrication. x + 2 y ≤ 40 Only 40 hours are available for finishing. Graph the constraints.

29. Let x = the number of metal fasteners, and let y = the number of plastic fasteners. The total cost is: C = 9 x + 4 y . Cost is to be minimized, so this is the objective function. The constraints are: x ≥ 2, y ≥ 2 At least 2 of each fastener must be made. x+ y ≥6 At least 6 fasteners are needed. 4 x + 2 y ≤ 24 Only 24 hours are available. Graph the constraints. y (2,8)

y (2,4) (5,2) (4,2) x

(0,20)

(10,15)

(20,0)

x

(0,0)

To find the intersection of 6 x + 4 y = 120 and x +2y = 40 , solve the system: 6 x + 4 y = 120   x + 2 y = 40 Solve the second equation for x: x = 40 − 2 y Substitute and solve: 6(40 − 2 y ) + 4 y = 120 240 − 12 y + 4 y = 120 − 8 y = −120 y = 15 x = 40 − 2(15) = 10 The point of intersection is (10, 15). The corner points are (0, 0), (0, 20), (20, 0), (10, 15). Evaluate the objective function: Vertex Value of P = 10 x + 12 y P = 10(0) + 12(0) = 0 (0, 0) (0, 20) P = 10(0) + 12(20) = 240 (20, 0) P = 10(20) + 12(0) = 200 (10, 15) P = 10(10) + 12(15) = 280 The maximum profit is $280, when 10 racing skates and 15 figure skates are produced.

The corner points are (2, 4), (2, 8), (4, 2), (5, 2). Evaluate the objective function: Vertex Value of C = 9 x + 4 y (2, 4) C = 9(2) + 4(4) = 34 (2, 8) C = 9(2) + 4(8) = 50 (4, 2) C = 9(4) + 4(2) = 44 (5, 2) C = 9(5) + 4(2) = 53 The minimum cost is $34, when 2 metal fasteners and 4 plastic fasteners are ordered. 31. Let x = the number of first class seats, and let y = the number of coach seats. Using the hint, the revenue from x first class seats and y coach seats is Fx + Cy, where F > C > 0. Thus, R = Fx + Cy is the objective function to be maximized. The constraints are: 8 ≤ x ≤ 16 Restriction on first class seats. 80 ≤ y ≤ 120 Restriction on coach seats. a.

x 1 ≤ Ratio of seats. y 12 The constraints are: 8 ≤ x ≤ 16 80 ≤ y ≤ 120 12x ≤ y

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Section 12.8: Linear Programming

Graph the constraints.

Vertex Value of R = Fx + Cy (8, 80) R = 8 F + 80C (8, 120) R = 8F + 120C (15, 120) R = 15 F + 120C (10, 80) R = 10 F + 80C

Since F > 0 and C > 0, 120C > 96C , the maximum value of R occurs at (15, 120). The maximum revenue occurs when the aircraft is configured with 15 first class seats and 120 coach seats. The corner points are (8, 96), (8, 120), and (10, 120). Evaluate the objective function:

c. 33.

Vertex Value of R = Fx + Cy (8, 96) R = 8F + 96C (8, 120) R = 8 F + 120C (10, 120) R = 10 F + 120C

2m 2/5 − m1/5 = 1 2m 2/5 − m1/5 − 1 = 0 (2m1/5 + 1)(m1/5 − 1) = 0 (2m1/5 + 1) = 0 or (m1/5 − 1) = 0

Since C > 0, 120C > 96C , so 8F + 120C > 8 F + 96C. Since F > 0, 10 F > 8F , so 10 F + 120C > 8 F + 120C. Thus, the maximum revenue occurs when the aircraft is configured with 10 first class seats and 120 coach seats. x 1 ≤ b. y 8 The constraints are: 8 ≤ x ≤ 16 80 ≤ y ≤ 120 8x ≤ y Graph the constraints.

Answers will vary.

2m1/5 = −1

or m1/5 = 1

1 2

or m1/5 = 1

m1/5 = −

5

1 2 1 m=− 32 m= −

or m = 15 or m = 1

The solution set is − 35.

1 ,1 . 32

A(t ) = A0 e rt 1 = 2e63t ln 0.5 = 63t

r = −0.011 Find t when A = 75 and A0 = 200 : 75 = 200e −0.011t 0.375 = e −0.011t ln 0.375 = −0.011t

t=

The corner points are (8, 80), (8, 120), (15, 120), and (10, 80). Evaluate the objective function:

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ln 0.375 ≈ 89.1 years −0.011

Chapter 12: Systems of Equations and Inequalities

Chapter 12 Review Exercises  2x − y = 5 1.  5 x + 2 y = 8 Solve the first equation for y: y = 2 x − 5 . Substitute and solve: 5 x + 2(2 x − 5) = 8 5 x + 4 x − 10 = 8 9 x = 18 x=2 y = 2(2) − 5 = 4 − 5 = −1 The solution is x = 2, y = −1 or (2, −1) . 3 x − 4 y = 4  2.  1  x − 3 y = 2 1 2 Substitute into the first equation and solve: 1  3 3 y +  − 4 y = 4 2  3 9y + − 4y = 4 2 5 5y = 2 1 y= 2 1 1 x = 3  + = 2 2 2 1  1 The solution is x = 2, y = or  2,  . 2  2

Solve the second equation for x: x = 3 y +

 x − 2y − 4 = 0 3.  3 x + 2 y − 4 = 0 Solve the first equation for x: x = 2 y + 4 Substitute into the second equation and solve: 3(2 y + 4) + 2 y − 4 = 0 6 y + 12 + 2 y − 4 = 0 8 y = −8 y = −1 x = 2(−1) + 4 = 2 The solution is x = 2, y = −1 or (2, −1) .

 y = 2x − 5 4.   x = 3y + 4 Substitute the first equation into the second equation and solve: x = 3(2 x − 5) + 4 x = 6 x − 15 + 4 −5 x = −11 11 x= 5 3  11  y = 2  − 5 = − 5 5   11 3  11 3  The solution is x = , y = − or  , −  . 5 5  5 5  x − 3y + 4 = 0  5.  1 3 4  2 x − 2 y + 3 = 0 Multiply each side of the first equation by 3 and each side of the second equation by −6 and add:

 3 x − 9 y + 12 = 0  −3x + 9 y − 8 = 0 4=0 There is no solution to the system. The system is inconsistent. 2 x + 3 y − 13 = 0 6.  =0  3x − 2 y Multiply each side of the first equation by 2 and each side of the second equation by 3, and add to eliminate y: 4 x + 6 y − 26 = 0  =0  9 x − 6 y 13 x − 26 = 0 13 x = 26 x=2 Substitute and solve for y: 3(2) − 2 y = 0 − 2y = −6 y=3 The solution is x = 2, y = 3 or (2, 3).  2 x + 5 y = 10 7.  4 x + 10 y = 20 Multiply each side of the first equation by –2 and add to eliminate x:

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Chapter 12 Review Exercises

−4 x − 10 y = − 20   4 x + 10 y = 20 0= 0 The system is dependent. 2 x + 5 y = 10 5 y = −2 x + 10 2 y = − x+2 5 2 The solution is y = − x + 2 , x is any real number 5  2  or ( x, y ) y = − x + 2, x is any real number  . 5    x + 2y − z = 6  8. 2 x − y + 3z = −13 3x − 2 y + 3 z = −16 

Multiply each side of the first equation by –2 and add to the second equation to eliminate x;  −2 x − 4 y + 2 z = −12   2 x − y + 3 z = −13 − 5 y + 5 z = − 25 y−z =5 Multiply each side of the first equation by –3 and add to the third equation to eliminate x: −3x − 6 y + 3 z = −18 3x − 2 y + 3z = −16 − 8 y + 6 z = −34 Multiply each side of the first result by 8 and add to the second result to eliminate y: 8 y − 8 z = 40 − 8 y + 6 z = −34 − 2z = 6 z = −3 Substituting and solving for the other variables: y − (−3) = 5 x + 2(2) − (−3) = 6 y=2 x+4+3= 6 x = −1 The solution is x = −1, y = 2, z = −3 or (−1, 2, −3) .

 2 x − 4 y + z = − 15  9.  x + 2 y − 4 z = 27 5 x − 6 y − 2 z = −3 

Multiply the first equation by −1 and the second equation by 2, and then add to eliminate x: −2 x + 4 y − z = 15  2 x + 4 y − 8 z = 54  8 y − 9 z = 69 Multiply the second equation by −5 and add to the third equation to eliminate x: − 5 x − 10 y + 20 z = − 135 5 x − 6 y − 2 z = −3 − 16 y + 18 z = −138 Multiply both sides of the first result by 2 and add to the second result to eliminate y: 16 y − 18 z = 138 −16 y + 18 z = −138 0=0 The system is dependent. −16 y + 18 z = −138 18 z + 138 = 16 y 9 69 y = z+ 8 8 Substituting into the second equation and solving for x: 69  9 x + 2  z +  − 4 z = 27 8  8 9 69 x + z + − 4 z = 27 4 4 7 39 x= z+ 4 4 9 69 7 39 , z is The solution is x = z + , y = z + 8 8 4 4  7 39 any real number or ( x, y, z ) x = z + , 4 4  y=

9 69  z + , z is any real number  . 8 8 

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Chapter 12: Systems of Equations and Inequalities

 x − 4 y + 3 z = 15  10. −3x + y − 5 z = − 5  −7 x − 5 y − 9 z = 10  Multiply the first equation by 3 and then add the second equation to eliminate x: 3x − 12 y + 9 z = 45 − 3x + y − 5 z = − 5  − 11y + 4 z = 40 Multiply the first equation by 7 and add to the third equation to eliminate x: 7 x − 28 y + 21z = 105

− 7 x − 5 y − 9 z = 10

 1 0

0  4 −3  1 2 −    −1 2  1(−3) + 0(1) 1(0) + 0( −2)   1(4) + 0(1) =  2(4) + 4(1) 2( −3) + 4(1) 2(0) + 4( −2)   −1(4) + 2(1) −1( −3) + 2(1) −1(0) + 2(−2) 

15. AB =  2 4  ⋅  1

0  4 −3 =  12 − 2 − 8  − 2 5 − 4 

3 − 4 0   4 −3 ⋅ 5  1 − 1 1 2   5 2    4(3) − 3(1) + 0(5) 4(−4) − 3(5) + 0(2)  =   1(3) + 1(1) − 2(5) 1( −4) + 1(5) − 2(2) 

16. BC = 

− 33 y + 12 z = 115 115 3 Multiply the first result by −1 and adding it to the second result:  11y − 4 z = −40  115  − 11y + 4 z = 3  − 11 y + 4 z =

5 3 The system has no solution. The system is inconsistent. 0=−

3x + 2 y = 8 11.   x + 4 y = −1

 9 −31 =   − 6 −3

 4 6 17. A =    1 3 Augment the matrix with the identity and use row operations to find the inverse: 4 6 1 0  1 3 0 1    1 3 0 1  Interchange  →     4 6 1 0   r1 and r2  3 0 1 1 →  ( R2 = − 4r1 + r2 ) 0 − 6 1 − 4

 x + 2 y + 5z = − 2  12. 5 x − 3z = 8 2 x − y =0   1 0  3 − 4  13. A + C =  2 4  +  1 5 2   −1 2  5  1 + 3 0 + (− 4)   4 − 4  =  2 + 1 4 + 5 =  3 9   −1 + 5 2 + 2   4 4 

1 → 0 1 → 0

1

3 0 1 − 16

2 3

1 0 2 1 − 16

−1 2  3

 1 Thus, A−1 =  12  − 6

 1 0   6 ⋅1 6 ⋅ 0   6 0  14. 6 A = 6 ⋅  2 4  =  6 ⋅ 2 6 ⋅ 4  =  12 24   −1 2  6(−1) 6 ⋅ 2   − 6 12 

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( R2 = − 16 r2 ) ( R1 = −3r2 + r1 )

−1 . 2 3 

Chapter 12 Review Exercises

1 3 3 18. A = 1 2 1 1 −1 2  Augment the matrix with the identity and use row operations to find the inverse: 1 3 3 1 0 0  1 2 1 0 1 0    1 −1 2 0 0 1 3 3 1 1  → 0 −1 − 2 −1 0 − 4 −1 −1 3 3 1 1  1 2 1 → 0 0 − 4 −1 −1 1 → 0 0 1  → 0  0

0 0 1 0  0 1 0 0 −1 0  0 1

0 −3 − 2 3 0 1 2 1 −1 0  0 7 3 − 4 1 0 −3 − 2 3 0  1 2 1 −1 0   3 0 1 − 74 71  7

1 0 0 − 5 7  1 → 0 1 0 7  3 0 0 1 7 

9 7 1 7 − 74

 − 75  =  17  3  7

9 7 1 7 4 7

Thus, A−1

−

3 7 − 72   1 7

 R2 = − r1 + r2     R3 = −r1 + r3 

( R2 = − r2 )  R1 = −3 r2 + r1     R3 = 4 r2 + r3 

( R3 = ) 1r 7 3

obtain the identity on the left side. The matrix is singular.  3x − 2 y = 1 20.  10 x + 10 y = 5 Write the augmented matrix:  3 − 2 1 10 10 5  

3 − 2 1 →   1 16 2 

( R2 = −3r1 + r2 )

 16 2  → 1  3 − 2 1

 Interchange     r1 and r2 

 16 2  → 1  0 −50 −5  1 16 2  →  1 0 1 10 

( R2 = −3r1 + r2 )

 0 → 1 0 1

2 5



1  10 

The solution is x =  R1 = 3 r3 + r1     R2 = − 2 r3 + r2 

3 7  − 72  . 1  7

 4 − 8 19. A =    −1 2  Augment the matrix with the identity and use row operations to find the inverse:  4 − 8 1 0  −1 2 0 1    −1 2 0 1  Interchange  →     4 − 8 1 0   r1 and r2   −1 2 0 1 → ( R2 = 4r1 + r2 )   0 0 1 4  1 − 2 0 −1 → ( R1 = −r1 ) 0 1 4  0 There is no inverse because there is no way to

( R2 = − 501 r2 ) ( R1 = −16r2 + r1 ) 2 1 2 1  ,y= or  ,  . 5 10  5 10 

5 x − 6 y − 3 z = 6  21. 4 x − 7 y − 2 z = −3 3 x + y − 7 z = 1 

Write the augmented matrix:  5 −6 −3 6     4 −7 − 2 −3 3 1 −7 1   1 1 −1 9    →  4 −7 − 2 −3 ( R1 = −r2 + r1 ) 3 1 −7 1   1 9 1 −1    R2 = − 4r1 + r2  → 0 −11 2 −39    0 −2 − 4 − 26   R3 = − 3r1 + r3     1 1 −1 9  1   R2 = − 11 r2  39  2   → 0 1 − 11 11  1  R3 = − r3    2   2 13 0 1 

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Chapter 12: Systems of Equations and Inequalities

9  1 0 − 11  2 → 0 1 − 11 24 0 0 11 

60  11 39  11  104  11 

9  1 0 − 11  2 → 0 1 − 11 0 0 1 

60  11 39  11  13  3

1 0 0  → 0 1 0  0 0 1

 R1 = − r2 + r1     R3 = − r2 + r3 

( R3 = 1124 r3 )

9 

9  R1 = 11 r3 + r1    2  R2 = r3 + r2    11

13 3 13   3

The solution is x = 9, y =

13 13 ,z= or 3 3

 13 13   9, ,  .  3 3 2 x + y + z = 5  22.  4 x − y − 3z = 1 8x + y − z = 5 

Write the augmented matrix:  2 1 1 5    4 −1 −3 1  8 1 −1 5   2 5 1 1   →  0 −3 −5 −9   0 −3 −5 −15   1  → 0 0  1  → 0  0

1 2

1 2 5 3

5 2

 1 3 −3 −5 −15 1  5 3 3  0 − 6

0 − 13 1 0

 R2 = − 2r1 + r2     R3 = − 4r1 + r3   R1 = 12 r1     R2 = − 1 r2  3  

− 2z = 1  x  23. 2 x + 3 y = −3  4x − 3 y − 4z = 3  Write the augmented matrix: 1 1 0 −2 2 3 0 −3   4 −3 − 4 3 1  → 0  0 1  → 0 0  1  → 0 0 

0 −2 1 3 4 −5 4 −1 −3 0 −2 1 5 4 1 − 3 3 −3 4 −1 0 −2 1

4 3

 R2 = − 2r1 + r2     R3 = − 4r1 + r3 

( R2 = 13 r2 )

1

 − 53 

0 8 − 6  1 0 −2 1   4 → 0 1 − 53  3   1 − 34  0 0 1 0 0 − 1  2  → 0 1 0 − 23   3 0 0 1 − 4 

( R3 = 3r2 + r3 )

( R3 = 18 r3 )  R1 = 2r3 + r1    4  R2 = − 3 r3 + r2 

1 2 3 The solution is x = − , y = − , z = − or 2 3 4  1 2 3 − ,− ,−  .  2 3 4

 R1 = − 12 r2 + r1     R3 = 3 r2 + r3 

There is no solution; the system is inconsistent.

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Chapter 12 Review Exercises

 x− y+ z= 0  24.  x − y − 5 z = 6 2 x − 2 y + z = 1  Write the augmented matrix:  1 −1 1 0   1 −1 −5 6     2 − 2 1 1 1 0  1 −1  →  0 0 − 6 6   0 0 −1 1  1 −1 1 →  0 0 1  0 0 −1  1 −1 0  → 0 0 1  0 0 0

1 0 → 0  0 1 0 → 0  0 1 0 → 0  0

0 −1 R2 = − 16 r2 1 1  R1 = −r2 + r1  −1   R3 = r2 + r3    0 The system is dependent. x = y +1   z = −1 The solution is x = y + 1 , z = −1 , y is any real

)

number or {( x, y, z ) x = y + 1, z = −1, y is any real number} .

 x− y− z− t = 1 2 x + y − z + 2t = 3  25.   x − 2 y − 2 z − 3t = 0  3x − 4 y + z + 5t = −3 Write the augmented matrix:  1 −1 −1 −1 1 2 1 −1 2 3   1 − 2 − 2 −3 0    1 5 −3 3 − 4

−1 4

1  R2 = − 2r1 + r2  1   R3 = − r1 + r3  −1 −1 − 2 −1  R = − 3 r1 + r4   8 − 6  4 −1 4 1 −1 −1 −1 −1 −1 − 2 −1  Interchange    3 1 4 1  r2 and r3   −1 4 8 − 6 −1 −1 −1 1 1 1 2 1 ( R2 = −r2 ) 3 1 4 1  −1 4 8 − 6  1 2 1 0 0  R1 = r2 + r1  0 1 1 2 1    R = −3 r2 + r3  →  0 0 −2 − 2 − 2   3    R4 = r2 + r4   0 0 5 10 −5  1 0 0 1 2  0 1 1 2 1  R3 = − 12 r3     →  R4 = 1 r4   0 0 1 1 1 5      0 0 1 2 −1  1 0 0 1 2 0 1 0 1 0   R2 = − r3 + r2   →    0 0 1 1 1  R4 = − r3 + r4     0 0 0 1 −2   1 0 0 0 4  R1 = − r4 + r1  0 1 0 0 2     → R2 = −r4 + r2   0 0 1 0  3  R = −r + r    4 3   3 0 0 0 1 − 2   The solution is x = 4, y = 2, z = 3, t = −2 or (4, 2, 3, −2) .

 R2 = − r1 + r2     R3 = − 2r1 + r3 

(

−1 −1 3 1

26.

27.

3 4 1 3

= 3(3) − 4(1) = 9 − 4 = 5

1 4 0 6 6 −1 2 6 = 1 2 − 4 −1 + 0 −1 2 1 3 4 3 4 1 4 1 3 = 1(6 − 6) − 4(−3 − 24) + 0(−1 − 8) = 1(0) − 4(− 27) + 0(−9) = 0 + 108 + 0 = 108

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Chapter 12: Systems of Equations and Inequalities

28.

2 1 −3 0 1 5 1 5 0 5 0 −1 + (−3) 1 =2 6 0 2 0 2 6 2 6 0 = 2(0 − 6) − 1(0 − 2) − 3(30 − 0) = 2(−6) − 1(−2) − 3(30) = −12 + 2 − 90 = −100

Set up and evaluate the determinants to use Cramer’s Rule: 1 −2 = 1(2) − 3(−2) = 2 + 6 = 8 3 2

Dx =

4 −2 = 4(2) − 4(−2) = 8 + 8 = 16 4 2

Dy =

1 4 = 1(4) − 3(4) = 4 − 12 = −8 3 4

The solution is x =

D Dx 16 −8 = −1 = =2, y= y = D 8 8 D

or (2, −1) . 2 x + 3 y − 13 = 0 30.  3x − 2 y = 0  Write the system is standard form: 2 x + 3 y = 13  3 x − 2 y = 0 Set up and evaluate the determinants to use Cramer’s Rule: 2 3 D= = − 4 − 9 = −13 3 −2

Dx =

13 3 = −26 − 0 = −26 0 −2

y=

Dy D

=

Set up and evaluate the determinants to use Cramer’s Rule: 1 2 −1 D = 2 −1 3 3 −2 3

Dx − 26 = =2, −13 D

−39 = 3 or (2, 3). −13

−1 3 −1 3 2 −1 −2 + (−1) 3 −2 −2 3 −2 3

= 1( −3 + 6 ) − 2(−3 + 6) + ( −1)(−4 + 3) = 3 + 6 + 1 = 10 6 2 −1 Dx = −13 −1 3 −16 −2 3 =6

−1 3 −13 3 −13 −1 −2 + (−1) −2 3 −16 3 −16 −2

= 6 ( −3 + 6 ) − 2(−39 + 48) + (−1)(26 − 16) = 18 − 18 − 10 = −10 1 6 −1 Dy = 2 −13 3 3 −16 3 =1

2 3 2 −13 −13 3 −6 + (−1) 3 3 3 −16 −16 3

= 1( −39 + 48 ) − 6(6 − 9) + (−1)( −32 + 39) = 9 + 18 − 7 = 20 1 2 6 Dz = 2 −1 −13 3 −2 −16 =1

−1 −13 2 −13 2 −1 −2 +6 −2 −16 3 −16 3 −2

= 1(16 − 26 ) − 2(−32 + 39) + 6(−4 + 3)

Dy = 2 13 = 0 − 39 = −39 3 0 The solution is x =

6

3 x − 2 y + 3 z = −16 

=1

 x − 2y = 4 29.  3x + 2 y = 4

D=

 x + 2y − z = 

31.  2 x − y + 3z = −13

= −10 − 14 − 6 = −30 D −10 The solution is x = x = = −1 , D 10 Dy 20 D −30 y= = = 2, z = z = = − 3 or D 10 D 10

(−1, 2, −3) .

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Chapter 12 Review Exercises

32. Let

x y = 8. a b

2x

y

= 2 ( 8 ) = 16 by Theorem (14). 2a b The value of the determinant is multiplied by k when the elements of a column are multiplied by k.

Then

33. Let

x y = 8. a b

y

x

= −8 by Theorem (11). The b a value of the determinant changes sign when any 2 columns are interchanged. Then

34. Find the partial fraction decomposition:  6  B  A x( x − 4)   = x( x − 4)  +   x x−4  x( x − 4)  6 = A( x − 4) + Bx Let x =4, then 6 = A(4 − 4) + B (4) 4B = 6 3 B= 2 Let x = 0, then 6 = A(0 − 4) + B (0) −4 A = 6 3 A=− 2 3 3 − 6 2 = + 2 x( x − 4) x x−4

2 − 4 = A(2)(2 − 1) + B (2 − 1) + C (2) 2 −2 = 2 A + B + 4C 2 A = − 2 − 4 − 4(−3) 2A = 6 A=3 x−4 3 4 −3 = + 2+ 2 x −1 x ( x − 1) x x

36. Find the partial fraction decomposition: x A Bx + C = + ( x 2 + 9)( x + 1) x + 1 x 2 + 9

Multiply both sides by ( x + 1)( x 2 + 9) . x = A( x 2 + 9) + ( Bx + C )( x + 1) Let x = −1 , then

(

)

−1 = A ( −1) + 9 + ( B ( −1) + C ) ( −1 + 1) 2

−1 = A(10) + (− B + C )(0) −1 = 10 A 1 A=− 10 Let x = 0 , then

(

)

0 = A 02 + 9 + ( B ( 0 ) + C ) ( 0 + 1) 0 = 9A + C

 1 0 = 9 −  + C  10  9 C= 10

(

)

Let x = 1 , then 1 = A 12 + 9 + ( B (1) + C ) (1 + 1)

35. Find the partial fraction decomposition: x−4 A B C = + 2+ 2 x −1 x ( x − 1) x x

Multiply both sides by x 2 ( x − 1) x − 4 = Ax( x − 1) + B ( x − 1) + Cx 2 Let x = 1 , then 1 − 4 = A(1)(1 − 1) + B(1 − 1) + C (1) 2 −3 = C C = −3 Let x = 0 , then 0 − 4 = A(0)(0 − 1) + B(0 − 1) + C (0) 2 −4 = − B B=4 Let x = 2 , then

1 = A(10) + ( B + C )(2) 1 = 10 A + 2 B + 2C

 1 9 1 = 10  −  + 2 B + 2    10   10  9 1 = −1 + 2 B + 5 1 2B = 5 1 B= 10 1 1 9 − x+ x 10 10 10 = + 2 ( x 2 + 9)( x + 1) x + 1 x +9

809 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities 37. Find the partial fraction decomposition: x3 Ax + B Cx + D = 2 + 2 2 x + 4 ( x 2 + 4) 2 ( x + 4)

Multiply both sides by ( x 2 + 4) 2 . 3

2

x = ( Ax + B )( x + 4) + Cx + D x3 = Ax3 + Bx 2 + 4 Ax + 4 B + Cx + D x3 = Ax3 + Bx 2 + (4 A + C ) x + 4 B + D A = 1; B = 0 4A + C = 0 4(1) + C = 0 C = −4 4B + D = 0 4(0) + D = 0 D=0 x3 2

( x + 4)

2

=

x 2

x +4

+

− 4x 2

( x + 4) 2

38. Find the partial fraction decomposition: x2 ( x 2 + 1)( x 2 − 1)

=

x2

( x 2 + 1)( x − 1)( x + 1) A B Cx + D = + + 2 x −1 x +1 x +1

Multiply both sides by ( x − 1)( x + 1)( x 2 + 1) . x 2 = A( x + 1)( x 2 + 1) + B ( x − 1)( x 2 + 1) + (Cx + D)( x − 1)( x + 1) Let x = 1 , then 2

2

2

1 = A(1 + 1)(1 + 1) + B (1 − 1)(1 + 1) + (C (1) + D)(1 − 1)(1 + 1) 1 = 4A 1 4 Let x = −1 , then A=

(−1)2 = A(−1 + 1)((−1) 2 + 1) + B (−1 − 1)((−1) 2 + 1) + (C (−1) + D)(−1 − 1)(−1 + 1) 1 = − 4B 1 4 Let x = 0 , then

02 = A(0 + 1)(02 + 1) + B (0 − 1)(02 + 1) + (C (0) + D)(0 − 1)(0 + 1) = A − B − D 0 1  1 −− − D 4  4 1 D= 2 Let x = 2 , then 22 = A(2 + 1)(22 + 1) + B (2 − 1)(22 + 1) + (C (2) + D)(2 − 1)(2 + 1) 4 = 15 A + 5B + 6C + 3D 0=

1  1 1 4 = 15   + 5  −  + 6C + 3   4  4 2 15 5 3 6C = 4 − + − 4 4 2 6C = 0 C=0 1 1 1 − x2 = 4 + 4 + 22 x −1 x + 1 x +1 x2 + 1 x2 − 1

(

)(

)

39. Solve the first equation for y, substitute into the second equation and solve: 2 x + y + 3 = 0 → y = − 2 x − 3  2 2  x + y = 5 x 2 + (− 2 x − 3) 2 = 5 → x 2 + 4 x 2 + 12 x + 9 = 5 5 x 2 + 12 x + 4 = 0 → (5 x + 2)( x + 2) = 0 2 or x = − 2 5 11 y=− y =1 5  2 11  Solutions:  − , −  , (−2, 1) . 5  5 x=−

40. Multiply each side of the second equation by 2 and add the equations to eliminate xy:  2 xy + y 2 = 10 ⎯⎯ → 2 xy + y 2 = 10  2 2 → − 2 xy + 6 y 2 = 4 − xy + 3 y = 2 ⎯⎯

B=−

7 y 2 = 14 y2 = 2 y=± 2

810 Copyright © 2016 Pearson Education, Inc.

Chapter 12 Review Exercises

If y = 2 : 2x

( 2)+( 2)

2

) (

2x − 2 + − 2

Solutions:

(2

)

2

= 10 → − 2 2 x = 8 → x = −2 2

)(

2, 2 , −2 2, − 2

)

41. Substitute into the second equation into the first equation and solve:  x 2 + y 2 = 6 y  x2 = 3 y  3y + y2 = 6 y y2 − 3 y = 0 y ( y − 3) = 0 → y = 0 or y = 3 If y = 0 :

2 2 : 3

If y =

−2 2 : 3

= 10 → 2 2 x = 8 → x = 2 2

If y = − 2 :

(

If y =

x 2 = 3(0) → x 2 = 0 → x = 0

If y = 3 : x 2 = 3(3) → x 2 = 9 → x = ±3 Solutions: (0, 0), (–3, 3), (3, 3).

42. Factor the second equation, solve for x, substitute into the first equation and solve: 3x 2 + 4 xy + 5 y 2 = 8  2 2  x + 3 xy + 2 y = 0 x 2 + 3 xy + 2 y 2 = 0 ( x + 2 y )( x + y ) = 0 → x = − 2 y or x = − y Substitute x = − 2 y and solve: 3 x 2 + 4 xy + 5 y 2 = 8 3(− 2 y ) 2 + 4(− 2 y ) y + 5 y 2 = 8 12 y 2 − 8 y 2 + 5 y 2 = 8 9 y2 = 8 y2 =

8 2 2  y=± 9 3

If y = 2 :

 2 2  −4 2 x = − 2   = 3  3   −2 2  4 2 x = − 2   = 3  3  x=− 2

If y = − 2 : x = 2 Solutions:  −4 2 2 2   4 2 −2 2  , ,  ,   , − 2, 2 , 3   3 3   3

(

(

2, − 2

)

 x 2 − 3x + y 2 + y = − 2  43.  x 2 − x + y +1 = 0  y  Multiply each side of the second equation by –y and add the equations to eliminate y: x 2 − 3x + y 2 + y = − 2 − x2 + x − y2 − y = 0 − 2 x = −2 x =1 If x = 1: 12 − 3(1) + y 2 + y = − 2 y2 + y = 0 y ( y + 1) = 0 y = 0 or y = −1 Note that y ≠ 0 because that would cause division by zero in the original system. Solution: (1, –1)

44. 3x + 4 y ≤ 12 Graph the line 3 x + 4 y = 12 . Use a solid line since the inequality uses ≤ . Choose a test point not on the line, such as (0, 0). Since 3 ( 0 ) + 4 ( 0 ) ≤ 12 is true, shade the side of the line

containing (0, 0). y

Substitute x = − y and solve:

5

3x 2 + 4 xy + 5 y 2 = 8

3x + 4y ≤ 12

3(− y ) 2 + 4(− y ) y + 5 y 2 = 8 3y2 − 4 y2 + 5 y2 = 8

−5

4 y2 = 8 y2 = 2  y = ± 2

)

5 −5

811 Copyright © 2016 Pearson Education, Inc.

x

Chapter 12: Systems of Equations and Inequalities

45. y ≤ x 2

Graph the parabola y = x 2 . Use a solid curve since the inequality uses ≤ . Choose a test point not on the parabola, such as (0, 1). Since 0 ≤ 12 is false, shade the opposite side of the parabola from (0, 1). y

5 y ≤ x2

−5

5

x

−5

− 2 x + y ≤ 2 46.   x+ y ≥ 2 Graph the line − 2 x + y = 2 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since − 2(0) + 0 ≤ 2 is true, shade the side of the line containing (0, 0). Graph the line x + y = 2 . Use a solid line since the inequality uses ≥. Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. y 5

x≥0   y≥0  47.   x+ y ≤ 4 2 x + 3 y ≤ 6 Graph x ≥ 0; y ≥ 0 . Shaded region is the first quadrant. Graph the line x + y = 4 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≤ 4 is true, shade the side of the line containing (0, 0). Graph the line 2 x + 3 y = 6 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) + 3(0) ≤ 6 is true, shade the side of the line containing (0, 0). y 8

x+ y=4 (0, 2) x (0, 0) (3, 0) –2 2x + 3y = 6

8

The overlapping region is the solution. The graph is bounded. Find the vertices: The x-axis and yaxis intersect at (0, 0). The intersection of 2 x + 3 y = 6 and the y-axis is (0, 2). The intersection of 2 x + 3 y = 6 and the x-axis is (3, 0). The three corner points are (0, 0), (0, 2), and (3, 0).

x+y=2 x –5

The point of intersection is (0, 2). The corner point is (0, 2).

5

–2x + y = 2 –5

The graph is unbounded. Find the vertices: To find the intersection of x + y = 2 and −2 x + y = 2 , solve the system:  x+ y = 2  − 2 x + y = 2 Solve the first equation for x: x = 2 − y . Substitute and solve: − 2(2 − y ) + y = 2 − 4 + 2y + y = 2 3y = 6 y=2 x = 2−2 = 0

x≥0   y≥0  48.  2 x + y ≤ 8  x + 2 y ≥ 2 Graph x ≥ 0; y ≥ 0 . Shaded region is the first quadrant. Graph the line 2 x + y = 8 . Use a solid line since the inequality uses ≤. Choose a test point not on the line, such as (0, 0). Since 2(0) + 0 ≤ 8 is true, shade the side of the line containing (0, 0). Graph the line x + 2 y = 2 . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 + 2(0) ≥ 2 is false, shade the opposite side of the line from (0, 0).

812 Copyright © 2016 Pearson Education, Inc.

Chapter 12 Review Exercises y

since the inequality uses ≤ . Choose a test point not on the hyperbola, such as (1, 2). Since 1 ⋅ 2 ≤ 4 is true, shade the same side of the hyperbola as (1, 2). The overlapping region is the solution.

9

(0, 8)

2x + y = 8

(0, 1) x + 2y = 2

(4, 0)

–1–1

y 5

x 9

(2, 0)

The overlapping region is the solution. The graph is bounded. Find the vertices: The intersection of x + 2 y = 2 and the y-axis is (0, 1). The intersection of x + 2 y = 2 and the x-axis is (2, 0). The intersection of 2 x + y = 8 and the y-axis is (0, 8). The intersection of 2 x + y = 8 and the xaxis is (4, 0). The four corner points are (0, 1), (0, 8), (2, 0), and (4, 0).

–5 5 x xy = 4 –5

51. Maximize z = 3x + 4 y subject to x ≥ 0 , y ≥ 0 , 3x + 2 y ≥ 6 , x + y ≤ 8 . Graph the constraints. y

49. Graph the system of inequalities: 2 2  x + y ≤ 16   x + y ≥ 2

Graph the circle x 2 + y 2 = 16 .Use a solid line since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since 02 + 02 ≤ 16 is true, shade the side of the circle containing (0, 0). Graph the line x + y = 2 . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 0). Since 0 + 0 ≥ 2 is false, shade the opposite side of the line from (0, 0). The overlapping region is the solution. y 5 x2+ y2 = 16 x –5

5

–5

y = x2

(0,8)

(0,3)

(2,0)

(8,0) x

The corner points are (0, 3), (2, 0), (0, 8), (8, 0). Evaluate the objective function: Vertex Value of z = 3 x + 4 y (0, 3) z = 3(0) + 4(3) = 12 (0, 8) z = 3(0) + 4(8) = 32 (2, 0) z = 3(2) + 4(0) = 6 (8, 0) z = 3(8) + 4(0) = 24 The maximum value is 32 at (0, 8).

x+y=2

50. Graph the system of inequalities:  y ≤ x 2   xy ≤ 4

Graph the parabola y = x 2 . Use a solid line since the inequality uses ≤ . Choose a test point not on the parabola, such as (1, 2). Since 2 ≤ 12 is false, shade the opposite side of the parabola from (1, 2). Graph the hyperbola xy = 4 . Use a solid line 813 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities 52. Minimize z = 3 x + 5 y subject to x ≥ 0 , y ≥ 0 , x + y ≥ 1 , 3 x + 2 y ≤ 12 , x + 3 y ≤ 12 . Graph the constraints. y

(0,4)

(127 , 247 )

(0,1)

(4,0)

x

(1,0)

To find the intersection of 3x + 2 y = 12 and x + 3 y = 12 , solve the system: 3x + 2 y = 12   x + 3 y = 12 Solve the second equation for x: x = 12 − 3 y Substitute and solve: 3(12 − 3 y ) + 2 y = 12 36 − 9 y + 2 y = 12 −7 y = − 24 24 y= 7 72 12  24  x = 12 − 3   = 12 − = 7 7  7   12 24  The point of intersection is  ,  . 7 7  The corner points are (0, 1), (1, 0), (0, 4), (4, 0),  12 24   , . 7 7 

Evaluate the objective function: Vertex Value of z = 3x + 5 y (0, 1) z = 3(0) + 5(1) = 5 (0, 4) z = 3(0) + 5(4) = 20 (1, 0) z = 3(1) + 5(0) = 3 (4, 0) z = 3(4) + 5(0) = 12  12 24   12   24  156  ,  z = 3  + 5  = 7 7 7  7  7  The minimum value is 3 at (1, 0). 2 x + 5 y = 5 53.  4 x + 10 y = A Multiply each side of the first equation by –2 and eliminate x:

−4 x − 10 y = −10  4 x + 10 y = A  0 = A − 10 If there are to be infinitely many solutions, the result of elimination should be 0 = 0. Therefore, A − 10 = 0 or A = 10 .

2 x + 5 y = 5 54.  4 x + 10 y = A Multiply each side of the first equation by –2 and eliminate x: −4 x − 10 y = −10  4 x + 10 y = A  0 = A − 10 If the system is to be inconsistent, the result of elimination should be 0 = any number except 0. Therefore, A − 10 ≠ 0 or A ≠ 10 . 55. y = ax 2 + bx + c At (0, 1) the equation becomes: 1 = a (0) 2 + b(0) + c c =1

At (1, 0) the equation becomes: 0 = a(1) 2 + b(1) + c 0 = a+b+c a+b+c = 0 At (–2, 1) the equation becomes: 1 = a(− 2) 2 + b(− 2) + c 1 = 4a − 2b + c 4a − 2b + c = 1 The system of equations is:  a+ b+c = 0  4a − 2b + c = 1  c=1  Substitute c = 1 into the first and second equations and simplify: 4a − 2b + 1 = 1 a + b +1 = 0 a + b = −1 4a − 2b = 0 a = −b − 1 Solve the first equation for a, substitute into the second equation and solve:

814 Copyright © 2016 Pearson Education, Inc.

Chapter 12 Review Exercises 4(−b − 1) − 2b = 0

Substituting and solving for the other variables: 5 + 3z = 11 x + 5 + 2(2) = 10 3z = 6 x + 9 = 10 z=2 x =1 Thus, 1 small box, 5 medium boxes, and 2 large boxes of cookies should be purchased.

− 4b − 4 − 2b = 0 − 6b = 4 b=− a=

2 3

2 1 −1 = − 3 3

58. a.

1 2 The quadratic function is y = − x 2 − x + 1 . 3 3

56. Let x = the number of pounds of coffee that costs $6.00 per pound, and let y = the number of pounds of coffee that costs $9.00 per pound. Then x + y = 100 represents the total amount of coffee in the blend. The value of the blend will be represented by the equation: 6 x + 9 y = 6.90(100) . Solve the system of equations:  x + y = 100  6 x + 9 y = 690 Solve the first equation for y: y = 100 − x . Solve by substitution: 6 x + 9(100 − x) = 690 6 x + 900 − 9 x = 690 −3x = − 210 x = 70 y = 100 − 70 = 30 The blend is made up of 70 pounds of the $6.00per-pound coffee and 30 pounds of the $9.00per-pound coffee. 57. Let x = the number of small boxes, let y = the number of medium boxes, and let z = the number of large boxes. Oatmeal raisin equation: x + 2 y + 2 z = 15 Chocolate chip equation: x + y + 2 z = 10 Shortbread equation: y + 3 z = 11  x + 2 y + 2 z = 15   x + y + 2 z = 10  y + 3 z = 11  Multiply each side of the second equation by –1 and add to the first equation to eliminate x:  x + 2 y + 2 z = 15 − x − y − 2 z = −10 

y + 3 z = 11 y =5

Let x = the number of lower-priced packages, and let y = the number of quality packages. 8 x + 6 y ≤ 120(16) Peanut inequality: 4 x + 3 y ≤ 960 Cashew inequality: 4 x + 6 y ≤ 72(16) 2 x + 3 y ≤ 576 The system of inequalities is: x≥0   y≥0   4 x + 3 y ≤ 960 2 x + 3 y ≤ 576

b. Graphing:

To find the intersection of 2 x + 3 y = 576 and 4 x + 3 y = 960 , solve the system: 4 x + 3 y = 960  2 x + 3 y = 576 Subtract the second equation from the first: 4 x + 3 y = 960 −2 x − 3 y = 576 2 x = 384 x = 192 Substitute and solve: 2(192) + 3 y = 576 3 y = 192 y = 64 The corner points are (0, 0), (0, 192), (240, 0), and (192, 64).

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Chapter 12: Systems of Equations and Inequalities 59. Let x = the speed of the boat in still water, and let y = the speed of the river current. The distance from Chiritza to the Flotel Orellana is 100 kilometers. Rate trip downstream x + y trip downstream x − y

Time Distance 5/ 2 3

100 100

The system of equations is: 5  ( x + y ) = 100 2 3( x − y ) = 100 Multiply both sides of the first equation by 6, multiply both sides of the second equation by 5, and add the results. 15 x + 15 y = 600 15 x − 15 y = 500 30 x = 1100 1100 110 x= = 30 3  110  3  − 3 y = 100  3  110 − 3 y = 100 10 = 3 y 10 3 The speed of the boat is 110 / 3 ≈ 36.67 km/hr ; the speed of the current is 10 / 3 ≈ 3.33 km/hr . y=

60. Let x = the number of hours for Bruce to do the job alone, let y = the number of hours for Bryce to do the job alone, and let z = the number of hours for Marty to do the job alone. Then 1/x represents the fraction of the job that Bruce does in one hour. 1/y represents the fraction of the job that Bryce does in one hour. 1/z represents the fraction of the job that Marty does in one hour. The equation representing Bruce and Bryce working together is: 1 1 1 3 + = = = 0.75 x y ( 4 / 3) 4

1 1 1 5 + = = = 0.625 y z (8 / 5) 6

The equation representing Bruce and Marty working together is: 1 1 1 3 + = = = 0.375 x z ( 8 / 3) 8 Solve the system of equations:  x −1 + y −1 = 0.75  −1 −1  y + z = 0.625  −1 −1  x + z = 0.375 Let u = x −1 , v = y −1 , w = z −1 u + v = 0.75  v + w = 0.625 u + w = 0.375 

Solve the first equation for u: u = 0.75 − v . Solve the second equation for w: w = 0.625 − v . Substitute into the third equation and solve: (0.75 − v) + (0.625 − v) = 0.375 −2v = −1 v = 0.5 u = 0.75 − 0.5 = 0.25 w = 0.625 − 0.5 = 0.125 Solve for x, y, and z : x = 4, y = 2, z = 8 (reciprocals) Bruce can do the job in 4 hours, Bryce in 2 hours, and Marty in 8 hours. 61. Let x = the number of gasoline engines produced each week, and let y = the number of diesel engines produced each week. The total cost is: C = 450 x + 550 y . Cost is to be minimized; thus, this is the objective function. The constraints are: 20 ≤ x ≤ 60 number of gasoline engines needed and capacity each week. 15 ≤ y ≤ 40 number of diesel engines needed and capacity each week. x + y ≥ 50 number of engines produced to prevent layoffs. Graph the constraints.

The equation representing Bryce and Marty working together is:

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Chapter 12 Test y = 2x − 7

y

y = 2 ( 3) − 7 = 6 − 7 = −1

The solution of the system is x = 3 , y = −1 or (20,40)

(60,40)

(20,30) (60,15) (35,15) x

The corner points are (20, 30), (20, 40), (35, 15), (60, 15), (60, 40) Evaluate the objective function: Vertex Value of C = 450 x + 550 y (20, 30) C = 450(20) + 550(30) = 25,500 (20, 40) C = 450(35) + 550(40) = 31, 000 (35, 15) C = 450(35) + 550(15) = 24, 000 (60, 15) C = 450(60) + 550(15) = 35, 250 ( 60, 40 ) C = 450 ( 60 ) + 550 ( 40 ) = 49, 000 The minimum cost is $24,000, when 35 gasoline engines and 15 diesel engines are produced. The excess capacity is 15 gasoline engines, since only 20 gasoline engines had to be delivered.

(3, −1) .

Elimination: Multiply each side of the first equation by 2 so that the coefficients of x in the two equations are negatives of each other. The result is the equivalent system −4 x + 2 y = −14   4x + 3y = 9 We can replace the second equation of this system by the sum of the two equations. The result is the equivalent system −4 x + 2 y = −14  5 y = −5  Now we solve the second equation for y. 5 y = −5 −5 y= = −1 5 We back-substitute this value for y into the original first equation and solve for x. −2 x + y = −7 −2 x + ( −1) = −7 −2 x = −6 −6 x= =3 −2 The solution of the system is x = 3 , y = −1 or

62. Answers will vary.

(3, −1) .

Chapter 12 Test 1. −2 x + y = −7   4x + 3y = 9

Substitution: We solve the first equation for y, obtaining y = 2x − 7 Next we substitute this result for y in the second equation and solve for x. 4x + 3 y = 9 4x + 3( 2x − 7) = 9 4 x + 6 x − 21 = 9 10 x = 30 30 x= =3 10 We can now obtain the value for y by letting x = 3 in our substitution for y.

1 2.  x − 2 y = 1 3 5 x − 30 y = 18 We choose to use the method of elimination and multiply the first equation by −15 to obtain the equivalent system −5 x + 30 y = −15   5 x − 30 y = 18

We replace the second equation by the sum of the two equations to obtain the equivalent system −5 x + 30 y = −15  0=3  The second equation is a contradiction and has no solution. This means that the system itself has no solution and is therefore inconsistent.

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Chapter 12: Systems of Equations and Inequalities x − y + 2z = 5

 x − y + 2 z = 5 (1)  3.  3 x + 4 y − z = −2 (2) 5 x + 2 y + 3z = 8 (3) 

We use the method of elimination and begin by eliminating the variable y from equation (2). Multiply each side of equation (1) by 4 and add the result to equation (2). This result becomes our new equation (2). x − y + 2z = 5 4 x − 4 y + 8 z = 20 3x + 4 y − z = −2

3 x + 4 y − z = −2 7x

+ 7 z = 18 (2)

We now eliminate the variable y from equation (3) by multiplying each side of equation (1) by 2 and adding the result to equation (3). The result becomes our new equation (3). x − y + 2z = 5 2x − 2 y + 4 z = 10 5 x + 2 y + 3z = 8

5x + 2 y + 3 z = 8 7x

+ 7 z = 18 (3)

Our (equivalent) system now looks like  x − y + 2 z = 5 (1)  + 7 z = 18 (2) 7 x 7 x + 7 z = 18 (3)  Treat equations (2) and (3) as a system of two equations containing two variables, and eliminate the x variable by multiplying each side of equation (2) by −1 and adding the result to equation (3). The result becomes our new equation (3). 7 x + 7 z = 18 − 7 x − 7 z = −18 7 x + 7 z = 18 7 x + 7 z = 18 0 = 0 (3) We now have the equivalent system  x − y + 2 z = 5 (1)  + 7 z = 18 (2) 7 x  0 = 0 (3)  This is equivalent to a system of two equations with three variables. Since one of the equations contains three variables and one contains only two variables, the system will be dependent. There are infinitely many solutions. We solve equation (2) for x and determine that 18 x = − z + . Substitute this expression into 7 equation (1) to obtain y in terms of z.

18    −z +  − y + 2z = 5 7  18 −z + − y + 2z = 5 7 17 −y + z = 7 17 7 18 17 The solution is x = − z + , y = z − , 7 7  18 z is any real number or ( x, y, z ) x = − z + , 7  17  y = z − , z is any real number  . 7  y = z−

 3x + 2 y − 8 z = −3 (1)  4.  − x − 23 y + z = 1 (2)  6 x − 3 y + 15 z = 8 (3) We start by clearing the fraction in equation (2) by multiplying both sides of the equation by 3.  3x + 2 y − 8 z = −3 (1)  −3x − 2 y + 3z = 3 (2)  6 x − 3 y + 15 z = 8 (3) 

We use the method of elimination and begin by eliminating the variable x from equation (2). The coefficients on x in equations (1) and (2) are negatives of each other so we simply add the two equations together. This result becomes our new equation (2). 3x + 2 y − 8 z = −3 −3 x − 2 y + 3z = 3 − 5 z = 0 (2)

We now eliminate the variable x from equation (3) by multiplying each side of equation (1) by −2 and adding the result to equation (3). The result becomes our new equation (3). 3x + 2 y − 8 z = −3 − 6 x − 4 y + 16 z = 6 6 x − 3 y + 15 z = 8

6 x − 3 y + 15 z = 8 − 7 y + 31z = 14 (3)

Our (equivalent) system now looks like 3x + 2 y − 8 z = −3 (1)  −5 z = 0 (2)   −7 y + 31z = 14 (3) We solve equation (2) for z by dividing both

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Chapter 12 Test

sides of the equation by −5 . −5 z = 0 z=0 Back-substitute z = 0 into equation (3) and solve for y. −7 y + 31z = 14 −7 y + 31(0) = 14 −7 y = 14 y = −2 Finally, back-substitute y = −2 and z = 0 into equation (1) and solve for x. 3x + 2 y − 8 z = −3 3x + 2(−2) − 8(0) = −3 3x − 4 = −3 3x = 1 1 x= 3 The solution of the original system is 1 1  x = , y = −2 , z = 0 or  , −2, 0  . 3 3  5.  4 x − 5 y + z = 0  −2 x − y + 6 = −19  x + 5 y − 5 z = 10 

We first check the equations to make sure that all variable terms are on the left side of the equation and the constants are on the right side. If a variable is missing, we put it in with a coefficient of 0. Our system can be rewritten as  4x − 5 y + z = 0  −2 x − y + 0 z = −25  x + 5 y − 5 z = 10  The augmented matrix is 0   4 −5 1  −2 −1 0 −25    1 5 −5 10  6. The matrix has three rows and represents a system with three equations. The three columns to the left of the vertical bar indicate that the system has three variables. We can let x, y, and z denote these variables. The column to the right of the vertical bar represents the constants on the right side of the equations. The system is  3 x + 2 y + 4 z = −6 3x + 2 y + 4 z = −6   + + = or 1 0 8 2 x y z x + 8z = 2   −2 x + 1y + 3 z = −11 −2 x + y + 3 z = −11  

1 −1  4

6 −3 2   −1 8  −2   4 6  6 4  −8  +  1 −3 = 1 −11 4   −1 8  5 12 

7. 2 A + C = 2 0 −4  +  1  3 2 =  0  6

1 8. A − 3C = 0  3 1 = 0  3

−1  4 6  −4  − 3  1 −3 2   −1 8  −1 12 18   −11 −19  −4  −  3 −9  =  −3 5  2   −3 24   6 −22 

9. Here we are taking the product of a 3 × 2 matrix and a 2 × 3 matrix. Since the number of columns in the first matrix is the same as the number of rows in the second matrix (2 in both cases), the operation can be performed and will result in a 3 × 3 matrix. 4 6 1 −2 5 CB =  1 −3  0 3 1   −1 8   4( −2) + 6 ⋅ 3 4 ⋅ 5 + 6 ⋅1   4 ⋅1 + 6 ⋅ 0  = 1 ⋅ 1 + ( −3)0 1( −2) + ( −3)3 1 ⋅ 5 + ( −3)1    ( −1)1 + 8 ⋅ 0 ( −1)( −2) + 8 ⋅ 3 ( −1)5 + 8 ⋅ 1

 4 10 26  =  1 −11 2     −1 26 3 

10. Here we are taking the product of a 2 × 3 matrix and a 3 × 2 matrix. Since the number of columns in the first matrix is the same as the number of rows in the second matrix (3 in both cases), the operation can be performed and will result in a 2 × 2 matrix. 1 −1  1 −2 5    BA =   0 −4  0 3 1   3 2    = 1 ⋅ 1 + ( −2 ) ⋅ 0 + 5 ⋅ 3  0 ⋅1 + 3 ⋅ 0 + 1⋅ 3  16 17  =   3 −10 

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1 ⋅ ( −1) + ( −2 ) ⋅ ( −4 ) + 5 ⋅ 2  0 ⋅ ( −1) + 3 ( −4 ) + 1 ⋅ 2

 

Chapter 12: Systems of Equations and Inequalities 11. We first form the matrix 3 2 1 0  [ A | I2 ] =   5 4 0 1 

Next we use row operations to transform [ A | I 2 ]

into reduced row echelon form. 1 23 13 0  3 2 1 0   5 4 0 1  →    5 4 0 1  1 →  0

2 3 2 3

1 3 − 53

0  1 

1 23 13 0  →  5 3  0 1 − 2 2  1 0 2 −1 → 5 3  0 1 − 2 2   2 Therefore, A−1 =  5 − 2

(R

1

= 13 r1

)

(R

2

=

(R

1

)

= − 23 r2 + r1

)

−1 3 . 2 

12. We first form the matrix  1 −1 1 1 0 0  B | I = [ 3 ]  2 5 −1 0 1 0  2 3 0 0 0 1  Next we use row operations to transform [ B | I 3 ]

into reduced row echelon form.

1 −1 1 1 0 0   R2 = −2r1 + r2  → 0 7 −3 −2 1 0     R3 = −2r1 + r3  0 5 −2 −2 0 1  1 0 0 1 −1 1   3 → 0 1 − 7 − 72 17 0  R2 = 17 r2 0 5 −2 −2 0 1   

(

( R2 = −5r1 + r2 ) 3r 2 2

1 −1 1 1 0 0   2 5 −1 0 1 0     2 3 0 0 0 1 

1 0 74  → 0 1 − 73  1 0 0 7

5 7 − 72 − 74

1 7 1 7 − 75

1

4 7 − 73

5 7 − 72

1 7 1 7

0

1

1  → 0  0 1 → 0 0

0

0  0  −4 −5 7   3 3 −4  −2 −2 3  −4 −5 7 

0 0 1 0 0 1

Thus, B

−1

0  0  1 

)

 R1 = r2 + r1     R3 = −5r2 + r3 

( R3 = 7r3 )  R1 = − 74 r3 + r1    R = 3r +r   2 7 3 2 

 3 3 −4  =  −2 −2 3   −4 −5 7 

13. 6 x + 3 y = 12   2 x − y = −2

We start by writing the augmented matrix for the system.  6 3 12   2 −1 −2    Next we use row operations to transform the augmented matrix into row echelon form.  6 3 12   2 −1 −2   R1 = r2   2 −1 −2  →  6 3 12   R = r       2 1 1 − 12 −1 →   6 3 12 

( R1 = 12 r1 )

1 − 12 −1 →   0 6 18 

( R2 = −6r1 + r2 )

1 − 12 −1 →  ( R2 = 16 r2 ) 3  0 1 1 0 12  →  ( R2 = 12 r2 + r1 ) 0 1 3 

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Chapter 12 Test

The solution of the system is x =

1 2

, y = 3 or

( 12 , 3) 1  14.  x + y = 7 4  8 x + 2 y = 56 We start by writing the augmented matrix for the system. 1 14 7    8 2 56 

Next we use row operations to transform the augmented matrix into row echelon form. 1 14 7    8 2 56  R2 = −8R1 + r2 1 14 7    0 0 0  The augmented matrix is now in row echelon form. Because the bottom row consists entirely of 0’s, the system actually consists of one equation in two variables. The system is dependent and therefore has an infinite number of solutions. Any ordered pair satisfying the 1 equation x + y = 7 , or y = −4 x + 28 , is a 4 solution to the system.

15.  x + 2 y + 4 z = −3  2 x + 7 y + 15 z = −12  4 x + 7 y + 13z = −10 

We start by writing the augmented matrix for the system.  1 2 4 −3   2 7 15 −12     4 7 13 −10  Next we use row operations to transform the augmented matrix into row echelon form.

 1 2 4 −3   2 7 15 −12     4 7 13 −10  1 2 4 −3 →  0 3 7 −6   0 −1 −3 2  1 2 4 −3 →  0 1 3 −2   0 3 7 −6 

 R2 = −2r1 + r2     R3 = −4r1 + r3   R2 = − r3     R3 = r2 

1 2 4 −3 →  0 1 3 −2  ( R3 = −3r2 + r3 )  0 0 −2 0  1 2 4 −3 →  0 1 3 −2  ( R3 = − 12 r3 )  0 0 1 0  The matrix is now in row echelon form. The last row represents the equation z = 0 . Using z = 0 we back-substitute into the equation y + 3z = −2 (from the second row) and obtain y + 3 z = −2 y + 3 ( 0 ) = −2 y = −2 Using y = −2 and z = 0 , we back-substitute into the equation x + 2 y + 4 z = −3 (from the first row) and obtain x + 2 y + 4 z = −3

x + 2 ( −2 ) + 4 ( 0 ) = −3 x =1 The solution is x = 1 , y = −2 , z = 0 or (1, −2, 0) .

16. 2 x + 2 y − 3 z = 5   x − y + 2z = 8  3x + 5 y − 8 z = −2 

We start by writing the augmented matrix for the system.  2 2 −3 5   1 −1 2 8     3 5 −8 −2  Next we use row operations to transform the augmented matrix into row echelon form.

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Chapter 12: Systems of Equations and Inequalities

 2 2 −3 5   1 −1 2 8     3 5 −8 −2  1 −1 2 8   =  2 2 −3 5   3 5 −8 −2  8  1 −1 2  = 0 4 −7 −11 0 8 −14 −26 

 R1 = r2     R2 = r1 

1 −1 2 8    = 0 1 − 7 − 11  4 4   0 8 14 26 − −  

( R2 = 14 r2 )

Dy =

3

= ( 4 )(19 ) − ( −23)( 3) = 145

19

Dx 58 = = −2 D −29 Dy 145 = = −5 y= D −29 The solution of the system is x = −2 , y = −5 or (−2, −5) . x=

 R2 = −2r1 + r2     R3 = −3r1 + r3 

1 −1 2 8    = 0 1 − 7 − 11  ( R3 = −8r2 + r3 ) 4 4   −4  0 0 0 The last row represents the equation 0 = −4 which is a contradiction. Therefore, the system has no solution and is be inconsistent.

17.

−2 5 = ( −2 )( 7 ) − ( 5 )( 3) = −14 − 15 = −29 3 7

18.

2 −4 6 1 4 0 −1 2 −4 =2

4 −23

4 0 1 0 1 4 − (−4) +6 −1 −4 −1 2 2 −4

= 2 [ 4(−4) − 2(0) ] + 4 [1(−4) − (−1)(0) ] + 6 [1(2) − (−1)4]

20.  4 x − 3 y + 2 z = 15  −2 x + y − 3 z = −15  5 x − 5 y + 2 z = 18 

The determinant D of the coefficients of the variables is 4 −3 2 D = −2 1 −3 5 −5 2 =4

1

−3

−5

2

− ( −3)

−2 −3 5

2

+2

−2

1

5

−5

= 4 ( 2 − 15 ) + 3 ( −4 + 15 ) + 2 (10 − 5 ) = 4 ( −13) + 3 (11) + 2 ( 5 ) = −52 + 33 + 10 = −9 Since D ≠ 0 , Cramer’s Rule can be applied. 15 −3 2 Dx = −15 1 −3 18 −5 2 = 15

1

−3

−5

2

− ( −3)

−15 −3 18

2

+2

−15

1

18

−5

= 15 ( 2 − 15 ) + 3 ( −30 + 54 ) + 2 ( 75 − 18 ) = 15 ( −13) + 3 ( 24 ) + 2 ( 57 )

= 2(−16) + 4(−4) + 6(6) = −32 − 16 + 36 = −12

= −9

19. 4 x + 3 y = −23   3 x − 5 y = 19

The determinant D of the coefficients of the variables is 4 3 D= = ( 4 )( −5 ) − ( 3)( 3) = −20 − 9 = −29 3 −5 Since D ≠ 0 , Cramer’s Rule can be applied. −23 3 Dx = = ( −23)( −5 ) − ( 3)(19 ) = 58 19 −5 822 Copyright © 2016 Pearson Education, Inc.

Chapter 12 Test

4

15

2 y 2 − 3x 2 = 5 22.  y − x = 1  y = x +1  Substitute x + 1 for y into the first equation and solve for x:

2

Dy = −2 −15 −3 5 =4

18

2

−15 −3 18

2

− 15

−2 −3 5

2

+2

−2 −15 5

= 4 ( −30 + 54 ) − 15 ( −4 + 15 ) + 2 ( −36 + 75 ) = 4 ( 24 ) − 15 (11) + 2 ( 39 ) = −9 4

−3

15

Dz = −2

1

−15

5

−5

18

=4

1

−15

−5

18

− ( −3)

2 ( x + 1) − 3x 2 = 5 2

18

(

)

2 x 2 + 2 x + 1 − 3x 2 = 5 2 x 2 + 4 x + 2 − 3x 2 = 5 − x2 + 4 x − 3 = 0

−2 −15 5

18

+ 15

−2

1

5

−5

= 4 (18 − 75 ) + 3 ( −36 + 75 ) + 15 (10 − 5 ) = 4 ( −57 ) + 3 ( 39 ) + 15 ( 5 )

= −36 Dy D −9 9 x= x = = 1, y = = = −1 , D −9 D −9 D −36 z= z = =4 D −9 The solution of the system is x = 1 , y = −1 , z = 4 or (1, −1, 4) . 2 2 3x + y = 12 21.  y2 = 9x  2

Substitute 9x for y into the first equation and solve for x: 3x 2 + ( 9 x ) = 12 3x 2 + 9 x − 12 = 0

x2 − 4x + 3 = 0 ( x − 1)( x − 3) = 0 x = 1 or x = 3 Back substitute these values into the second equation to determine y: x = 1 : y = 1+1 = 2 x = 3 : y = 3 +1 = 4

The solutions of the system are (1, 2) and (3, 4) .  x 2 + y 2 ≤ 100 23.  4 x − 3 y ≥ 0

Graph the circle x 2 + y 2 = 100 . Use a solid curve since the inequality uses ≤ . Choose a test point not on the circle, such as (0, 0). Since 02 + 02 ≤ 100 is true, shade the same side of the circle as (0, 0); that is, inside the circle. Graph the line 4 x − 3 y = 0 . Use a solid line since the inequality uses ≥ . Choose a test point not on the line, such as (0, 1). Since 4(0) − 3(1) ≥ 0 is false, shade the opposite side of the line from (0, 1). The overlapping region is the solution.

x 2 + 3x − 4 = 0 ( x − 1)( x + 4) = 0 x = 1 or x = −4 Back substitute these values into the second equation to determine y: x = 1 : y 2 = 9(1) = 9 y = ±3 x = −4 : y 2 = 9(−4) = −36

y = ± −36 (not real) The solutions of the system are (1, −3) and (1, 3) .

823 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities

24.

3x + 7

( x + 3) The denominator contains the repeated linear factor x + 3 . Thus, the partial fraction decomposition takes on the form 3x + 7 A B = + 2 2 ( x + 3) x + 3 ( x + 3) 2

Clear the fractions by multiplying both sides by

( x + 3)

2

. The result is the identity

3x + 7 = A ( x + 3) + B

or 3x + 7 = Ax + ( 3 A + B ) We equate coefficients of like powers of x to obtain the system 3 = A  7 = 3 A + B Therefore, we have A = 3 . Substituting this result into the second equation gives 7 = 3A + B 7 = 3 ( 3) + B −2 = B Thus, the partial fraction decomposition is 3x + 7 3 −2 . = + 2 2 3 x + 3 x x + ( ) ( + 3)

25.

4x2 − 3

(

x x2 + 3

)

2

The denominator contains the linear factor x and the repeated irreducible quadratic factor x 2 + 3 . The partial fraction decomposition takes on the form 4 x2 − 3 A Bx + C Dx + E = + 2 + 2 2 2 x x +3 x x +3 x2 + 3

(

)

(

)

We clear the fractions by multiplying both sides

(

by x x 2 + 3 2

)

(

2

2

to obtain the identity

4x − 3 = A x + 3

)

2

(

)

+ x x + 3 ( Bx + C ) + x ( Dx + E ) 2

Collecting like terms yields 4 x 2 − 3 = ( A + B ) x 4 + Cx3 + ( 6 A + 3B + D ) x 2 + ( 3C + E ) x + ( 9 A )

Equating coefficients, we obtain the system

A+ B = 0   C=0  6 A + 3 B + D = 4  3C + E = 0  9 A = −3 

1 From the last equation we get A = − . 3 Substituting this value into the first equation 1 gives B = . From the second equation, we 3 know C = 0 . Substituting this value into the fourth equation yields E = 0 . 1 1 Substituting A = − and B = into the third 3 3 equation gives us 6 ( − 13 ) + 3 ( 13 ) + D = 4 −2 + 1 + D = 4 D=5 Therefore, the partial fraction decomposition is 4x2 − 3 x ( x + 3) 2

2

=

−

1 3+

x

1 x 3

+

5x

( x + 3) ( x + 3) 2 2

2

26.  x ≥ 0 y ≥ 0   x + 2 y ≥ 8 2 x − 3 y ≥ 2

The inequalities x ≥ 0 and y ≥ 0 require that the graph be in quadrant I. x + 2y ≥ 8 1 y ≥ − x+4 2 Test the point ( 0, 0 ) . x + 2y ≥ 8 0 + 2 ( 0) ≥ 8 ? 0 ≥ 8 false The point ( 0, 0 ) is not a solution. Thus, the

graph of the inequality x + 2 y ≥ 8 includes the 1 half-plane above the line y = − x + 4 . Because 2 the inequality is non-strict, the line is also part of the graph of the solution.

824 Copyright © 2016 Pearson Education, Inc.

Chapter 12 Test y

2x − 3y ≥ 2 2 2 x− 3 3 Test the point ( 0, 0 ) . y≤

8

2x − 3 y ≥ 2

(0, 8)

4

2 ( 0) − 3( 0) ≥ 2 ? 0 ≥ 2 false The point ( 0, 0 ) is not a solution. Thus, the

(0, 1)

2 2 x− . 3 3 Because the inequality is non-strict, the line is also part of the graph of the solution. The overlapping shaded region (that is, the shaded region in the graph below) is the solution to the system of linear inequalities.

x

8

4

graph of the inequality 2 x − 3 y ≥ 2 includes the half-plane below the line y =

x − 3 y = −3

(3, 2)

2x + y = 8

Corner point, ( x, y ) Value of obj. function, z

( 0,1) ( 3, 2 ) ( 0,8 )

z = 5 ( 0 ) + 8 (1) = 8

z = 5 ( 3) + 8 ( 2 ) = 31

z = 5 ( 0 ) + 8 ( 8 ) = 64

From the table, we can see that the maximum value of z is 64, and it occurs at the point ( 0,8 ) . 28. Let j = unit price for flare jeans, c = unit price for camisoles, and t = unit price for t-shirts. The given information yields a system of equations with each of the three women yielding an equation. 2 j + 2c + 4t = 90 (Megan)  + 3t = 42.5 (Paige)  j  j + 3c + 2t = 62 (Kara) 

The graph is unbounded. The corner points are ( 4, 2 ) and ( 8, 0 ) . 27. The objective function is z = 5 x + 8 y . We seek the largest value of z that can occur if x and y are solutions of the system of linear inequalities x ≥ 0  2 x + y ≤ 8  x − 3 y ≤ −3  2x + y = 8 x − 3 y = −3 y = −2 x + 8 −3 y = − x − 3 1 y = x +1 3 The graph of this system (the feasible points) is shown as the shaded region in the figure below. The corner points of the feasible region are ( 0,1) , ( 3, 2 ) , and ( 0,8 ) .

We can solve this system by using matrices.  2 2 4 90  1 1 2 45   1 0 3 42.5 = 1 0 3 42.5 ( R1 = 12 r1 )      1 3 2 62  1 3 2 62  1 = 0  0 1 = 0  0

1 2 45  −1 1 −2.5  2 0 17 

 R2 = −r1 + r2     R3 = −r1 + r3 

1 2 45  1 −1 2.5 ( R2 = −r2 )  2 0 17  1 0 3 42.5  R1 = − r2 + r1  = 0 1 −1 2.5      R3 = −2r2 + r3  0 0 2 12  1 0 3 42.5 R3 = 12 r3 = 0 1 −1 2.5    6  0 0 1 The last row represents the equation z = 6 . Substituting this result into y − z = 2.5 (from the

825 Copyright © 2016 Pearson Education, Inc.

(

)

Chapter 12: Systems of Equations and Inequalities y − z = 2.5 y − 6 = 2.5 y = 8.5 Substituting z = 6 into x + 3 z = 42.5 (from the first row) gives x + 3z = 42.5

second row) gives

32

−3 6

2

3

−8 9

−3 3

1 3

0 2

Therefore, 2 x − 3x − 8 x − 3 = 0

( x − 3) ( 2 x 2 + 3x + 1) = 0 ( x − 3)( 2 x + 1)( x + 1) = 0

x + 3 ( 6 ) = 42.5 x = 24.5 Thus, flare jeans cost $24.50, camisoles cost $8.50, and t-shirts cost $6.00.

1 or x = −1 2 1   The solution set is −1, − ,3 . 2   x = 3 or x = −

4. 3x = 9 x+1

( )

3x = 32

Chapter 12 Cumulative Review

3x = 32 x + 2 x = 2x + 2

1. 2 x 2 − x = 0 x ( 2 x − 1) = 0

x = −2 The solution set is {−2} .

x = 0 or 2 x − 1 = 0 2x = 1 1 2  1 The solution set is 0,  .  2

5. log 3 ( x − 1) + log 3 ( 2 x + 1) = 2

x=

2.

(

x +1

log 3 ( ( x − 1)( 2 x + 1) ) = 2

( x − 1)( 2 x + 1) = 32 2 x2 − x − 1 = 9

3x + 1 = 4 3x + 1

)

2

=4

2 x 2 − x − 10 = 0

2

( 2 x − 5)( x + 2 ) = 0

3 x + 1 = 16 3x = 15 x=5 Check:

5 or x = −2 2 Since x = −2 makes the original logarithms 5 undefined, the solution set is   . 2 x=

3⋅5 +1 = 4 15 + 1 = 4

6.

16 = 4 4=4 The solution set is {5} . 3

2

3x = e

( )

ln 3x = ln e x ln 3 = 1

3. 2 x − 3x − 8 x − 3 = 0 The graph of Y1 = 2 x3 − 3x 2 − 8 x − 3 appears to have an x-intercept at x = 3 .

1 ≈ 0.910 ln 3  1  ≈ 0.910  . The solution set is   ln 3  x=

Using synthetic division:

826 Copyright © 2016 Pearson Education, Inc.

Chapter 12 Cumulative Review

7. g ( x) =

2 x3 x4 + 1

g (− x) =

10.

2(−x)

(−x)

4

3

=

−2 x3 = −g ( x) x4 + 1

+1 Thus, g is an odd function and its graph is symmetric with respect to the origin.

8.

x 2 + y 2 − 2 x + 4 y − 11 = 0

x 2 − 2 x + y 2 + 4 y = 11 ( x 2 − 2 x + 1) + ( y 2 + 4 y + 4) = 11 + 1 + 4 ( x − 1) 2 + ( y + 2) 2 = 16 Center: (1,–2); Radius: 4

f ( x) =

5 x+2

5 x+2 5 x= Inverse y+2 x( y + 2) = 5 xy + 2 x = 5 xy = 5 − 2 x 5 − 2x 5 = −2 y= x x 5 −1 Thus, f ( x) = − 2 x y=

Domain of f = {x | x ≠ −2} Range of f = { y | y ≠ 0} Domain of f −1 = {x | x ≠ 0} Range of f −1 = { y | y ≠ −2} . 11. a.

9.

f ( x ) = 3x − 2 + 1 x

Using the graph of y = 3 , shift the graph horizontally 2 units to the right, then shift the graph vertically upward 1 unit.

y = 3x + 6 The graph is a line. x-intercept: 0 = 3x + 6 3x = −6 x = −2

Domain: (−∞, ∞) Range: (1, ∞) Horizontal Asymptote: y = 1

827 Copyright © 2016 Pearson Education, Inc.

y-intercept: y = 3( 0) + 6 =6

Chapter 12: Systems of Equations and Inequalities

b.

c.

d.

x2 + y 2 = 4 The graph is a circle with center (0, 0) and radius 2.

f. y = e x

g.

y = ln x

h.

2 x2 + 5 y 2 = 1 The graph is an ellipse. x2 y2 + =1

y = x3

y=

1 x

1 2

2

 x   y   +  2   5  2   5

e.

y= x

828 Copyright © 2016 Pearson Education, Inc.

1 5

2

  =1  

Chapter 12 Projects

i.

−6

3 2

9

b.

x2 − 3 y 2 = 1 The graph is a hyperbola x2 y 2 − =1 1 1

6 −3

2

 x  y  =1   −  1   3     3 

9

−6

6 −3

f has a local maximum of 7 at x = −1 and a local minimum of 3 at x = 1 . c. j.

f is increasing on the intervals (−∞, −1) and (1, ∞) .

x2 − 2 x − 4 y + 1 = 0 x2 − 2 x + 1 = 4 y 4 y = ( x − 1) 2 1 y = ( x − 1) 2 4

Chapter 12 Projects Project I – Internet Based Project 1. 80% = 0.80 40% = 0.40 20% = 0.20

18% = 0.18 50% = 0.50 60% = 0.60

2% = 0.02 10% = 0.10 20% = 0.20

0.80 0.18 0.02  2. 0.40 0.50 0.10  0.20 0.60 0.20 

12.

3. 0.80 + 0.18 + 0.02 = 1.00 0.40 + 0.50 + 0.10 = 1.00 0.20 + 0.60 + 0.20 = 1.00 The sum of each row is 1 (or 100%). These represent the three possibilities of educational achievement for a parent of a child, unless someone does not attend school at all. Since these are rounded percents, chances are the other possibilities are negligible.

f ( x) = x3 − 3x + 5

a.

Let Y1 = x3 − 3x + 5 . 9

0.8 0.18 0.02 −6

4. P 2 = 0.4 0.2

6 −3

The zero of f is approximately −2.28 .

0.5

0.1

0.6

0.2

2

0.716 0.246 0.038 = 0.54 0.382 0.078 0.44 0.456 0.104 Grandchild of a college graduate is a college

829 Copyright © 2016 Pearson Education, Inc.

Chapter 12: Systems of Equations and Inequalities

graduate: entry (1, 1): 0.716. The probability is 71.6% 5. Grandchild of a high school graduate finishes college: entry (2,1): 0.54. The probability is 54%. 6. grandchildren → k = 2. v (2) = v (0) P 2 0.716 0.246 0.038 = [0.317 0.565 0.118] 0.54 0.382 0.078 0.44 0.456 0.104 = [0.583992 0.34762 0.068388]

College: ≈ 58.4% High School: ≈ 34.8% Elementary: ≈ 6.8% 7. The matrix totally stops changing at 0.64885496 0.29770992 0.05343511 30 P ≈ 0.64885496 0.29770992 0.05343511 0.64885496 0.29770992 0.05343511

v = uG

0 0  0  0 0  0 0  0 v= 1 1  1 1  1 1  1  1

0 0 0 0 0 0 0 0 1 1 1 0  0 1 0 1 0 1  0 1 1 0 1 1 1 0 0 0 1 1  1 0 1 1 0 1 1 1 0 1 1 0  1 1 1 0 0 0  0 0 0 1 1 1 0 0 1 0 0 1  0 1 0 0 1 0 0 1 1 1 0 0  1 0 0 1 0 0 1 0 1 0 1 0  1 1 0 0 0 1  1 1 1 1 1 1

c. Answers will vary, but if we choose the 6th row and the 10th row: 0101101 1001001 1102102 → 1100100 (13th row)

Project II a. 2 × 2 × 2 × 2 = 16 codewords. b. v = uG u will be the matrix representing all of the 4-digit information bit sequences. 0 0 0 0  0 0 0 1    0 0 1 0    0 0 1 1  0 1 0 0    0 1 0 1  0 1 1 0    0 1 1 1  u=  1 0 0 0  1 0 0 1    1 0 1 0  1 0 1 1    1 1 0 0  1 1 0 1    1 1 1 0    1 1 1 1  (Remember, this is mod two. That means that you only write down the remainder when dividing by 2. )

d. v = uG VH = uGH 0 0 0 0 GH =  0 0  0 0

0 0  0  0

1 0  1  e. rH = [0 1 0 1 0 0 0] 1 1  0 0  = [1 0 1]

error code: 0010 000 r : 0101 000 0111 000 This is in the codeword list.

830 Copyright © 2016 Pearson Education, Inc.

1 1 1 1  0 1  1 0 0 0  1 0 0 1 

Chapter 12 Projects Project III 1 1 1 1 1 1 1  a. AT =   3 4 5 6 7 8 9 

b. B = ( AT A) −1 AT Y  −2.357  B=   2.0357 

c. y = 2.0357 x − 2.357 d. y = 2.0357 x − 2.357

Project IV

Answers will vary.

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