Mathematics Algebra of Linear Equations...
TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION IN TWO VARIABLES) 1. A collection of 105 coins consists of 1.00 peso and 5.00 pesos. If the total value is 205.00 pesos, find the number of coins of each denomination in the collection. 2. A merchant paid P48,000 for some dresses and shoes. He paid P400.00 for each dress and P1,000.00 for each pair of shoes. He sold the dresses at a profit of 20% and the shoes at a profit of 50%. If his total profit was P18,000. How many dresses and shoes did he buy? 3. The sum of two numbers is 115. The difference is -21. Find the numbers. 4. Two angles are complementary, one angle is 42 degrees more than half the other. Find the angles. 5. A two digit number is 6 times the sum of its digits. The tens digit is 1 more than the units digit. Find the original number. 6. The sum of two numbers is 29. One number is 7 less than twice the other number. Find the two numbers. 7. Eugene and Erick went to Jollibee to buy french-fries and hamburgers. Eugene bought 6 hamburgers and 4 french-fries French fries for P330.00. Erick bought 4 hamburgers and 9 french fries for P505.00. What is the cost of each french-fries and each hamburgers? 8. Jeny has P1,490.00 in P20 bills and P50 bills. If she has 40 pieces in all, how many of each denomination does she have? 9. A vending machine takes only nickels and dimes. There are 5 times as many dimes as nickels in the machine. The face value of the coins is $4.40. How many of each coin are in the machine. 10. A chemistry experiment calls for a 30% sulfuric acid solution. If the lab supply room has only 50% and 20% sulfuric acid solution on hand, how much of each should be mixed to obtain twelve liters of a 30% solution.
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TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION IN TWO VARIABLES)
SOLUTION 1. A collection of 105 coins consists of 1.00 peso and 5.00 pesos. If the total value is 205.00 pesos, find the number of coins of each denomination in the collection.
Solution by elimination: Let x = number of 1 peso coins Y= number of 5 peso coins
x + y = 105 ( equation 1) x + 5 y = 205 ( equation 2 ) Subtract the first equation from the second equation( equation 2 minus equation 1) , the result is 4 y =100
⇒ y = 25 , substitute in the first equation
X + 25 = 105
⇒ x = 80 Answer # 1: There are 80 →1 peso coins and 25 →5 peso coins. 2. A merchant paid P48,000 for some dresses and shoes. He paid P400.00 for each dress and P1,000.00 for each pair of shoes. He sold the dresses at a profit of 20% and the shoes at a profit of 50%. If his total profit was P18,000. How many dresses and shoes did he buy?
Solution by elimination method: Let x = number of dresses Y = number of shoes equation 1 : 400 x + 1000 y = 48000 Equation: equation 2 : ( 0.2 )( 400 x ) + ( 0.5)(1000 y ) = 18000 Divide equation 1 by 100 the result is 4 x +10 y = 480 , call this equation 3
Simplify equation 2, the result is 80 x + 500 y =18000 , divide this by 20, the result is 4 x + 25 y = 900 , call this equation 4
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TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION IN TWO VARIABLES)
Subtract equation 3 from equation 4
4 x + 25 y = 900 − ( 4 x +10 y = 480 ) 15 y = 420 ⇒ y = 28 , substitute in equation 3 and
solve for x. 4 x +10 y = 480 becomes
4 x +10( 28) = 480
⇒ 4 x = 200 ⇒ x = 50 Answer #2: He buy 50 dresses and 28 shoes 3. The sum of two numbers is 115. The difference is -21. Find the numbers.
Solution by elimination method: Let x and y be the numbers equation1 : x + y = 115 equation 2 : x − y = −21 Subtract equation 2 from equation 1, the result is x + y =115
−( x − y = −21) 2 y =136
⇒ y = 68 , substitute this in equation 1 and solve for x. x + y = 115 becomes
x + 68 = 115 ⇒ x = 47 Answer #3: The numbers are 68 and 47 4. Two angles are complementary, one angle is 42 degrees more than half the other. Find the angles.
Solution by substitution method: Let x and y be the measure of the angles
equation1 : x + y = 90 equation2 : x = 42 + 1 y 2 Substitute equation 2 in equation 1, the result is 1 42 + y + y = 90 , multiply both sides by 2 it becomes 2 84 + y + 2 y = 180
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TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION IN TWO VARIABLES) ⇒ 84 + 3 y = 180 ⇒ 3 y = 96
⇒ y = 32 substitute this in equation 2
1 ( 32) = 58 2 Answer # 4: The measure of the angles are 32 and 58 degrees. x = 42 +
5. A two digit number is 6 times the sum of its digits. The tens digit is 1 more than the units digit. Find the original number.
Solution by elimination method: Let x = units digit Y = ten’s digit The value of the number is x+10y . equation 1 : x + 10 y = 6 ( x + y ) equation 2 : y = 1 + x Simplify equation 1 it becomes, x +10 y = 6 x + 6 y ⇒ 5x − 4 y = 0
⇒ 5x = 4 y
⇒ 5 x = 4(1 + x ) we substitute equation 2.
⇒ 5x = 4 + 4x ⇒ x = 4 , therefore y=5. Answer: The original number is 54.
6. The sum of two numbers is 29. One number is 7 less than twice the other number. Find the two numbers.
Solution by substitution method: Let x and y be the numbers equation 1 : x + y = 29 equation 2 : x = 2 y − 7 Substitute equation 2 in equation 1, the result is
( 2 y − 7 ) + y = 29
⇒ 3 y = 36
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TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION IN TWO VARIABLES) ⇒ y = 12 , substitute this in equation 2 to solve for x.
x = 2(12 ) − 7
⇒ x = 17 Answer # 6 : The numbers are 17 and 12. 7. Eugene and Erick went to Jollibee to buy french-fries and hamburgers. Eugene bought 6 hamburgers and 4 french-fries French fries for P330.00. Erick bought 4 hamburgers and 9 frenchfriyes for P505.00. What is the cost of each french-fries and each hamburgers?
Solution by elimination method: Let x = number of hamburgers Y = number of French fries equation 1 : 6 x + 4 y = 330 × 2 ⇒12 x + 8 y = 660, call this equation 3 equation 2 : 4 x + 9 y = 505 × 3 ⇒12 x + 27 y = 1515, call this equation 4
12 x + 27 y = 1515 − Subtract equation 3 from equation 4, the result is (12 x + 8 y = 660 ) 19 y = 855 ⇒ y = 45 , substitute this in equation 1 and solve for x,
6 x + 4( 45) = 330
⇒ 6 x = 150 ⇒ x = 25 Answer #7: The cost of each hamburger is P25.00 and the cost of each French fries is P45.00
8. Jeny has P1,490.00 in P20 bills and P50 bills. If she has 40 pieces in all, how many of each denomination does she have?
Solution by elimination method: Let x= number of P20 bills Y=number of P50 bills equation 1 : x + y = 40 equation 2 : 20 x + 50 y = 1490
Multiply equation 1 by 20, the result is 20 x + 20 y = 800 , subtract this from equation 2, the result is
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TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION IN TWO VARIABLES) 20 x + 50 y = 1490 − ( 20 x + 20 y = 800 ) 30 y = 690 ⇒ y = 23 , substitute this in equation 1, and solve for x.
x + 23 = 40 ⇒ x = 17 Answer #8: There are 17 P20 bills and 23 P50 bills 9. A vending machine takes only nickels and dimes. There are 5 times as many dimes as nickels in the machine. The face value of the coins is $4.40. How many of each coin are in the machine.
Solution by elimination method: Let x = number of nickels Y= number of dimes y = 5x equation1 : equation2 : .05 x + .10 y = 4.40
Multiply equation 2 by 100, the result is, 5 x +10 y = 440 , then use equation 1 the result is
5 x +10( 5 x ) = 440
⇒ 55 x = 440 ⇒ x = 8 , substitute in equation 1 and solve for y y = 5( 8)
⇒ y = 40
Answer #9: There are 8 nickels and 40 dimes. 10. A chemistry experiment calls for a 30% sulfuric acid solution. If the lab supply room has only 50% and 20% sulfuric acid solution on hand, how much of each should be mixed to obtain twelve liters of a 30% solution. Solution by elimination method: Let x = amount of 50% sulfuric acid solution in liters Y = amount of 20% sulfuric acid solution in liters
x + y = 12 equation1 : equation2 : 0.5( x ) + 0.2( y ) = 0.3(12 ) Multiply equation 2 by 10, the result is, 5 x + 2 y = 36 , call this equation 4 6 To suggest a topic e-mail
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TOPIC: WORD PROBLEMS (SYSTEM OF LINEAR EQUATION IN TWO VARIABLES)
Multiply equation 1 by 2, the result is 2 x + 2 y = 24 , call this equation 5 Subtract equation 5 from equation 4, the result is 5 x + 2 y = 36 − ( 2 x + 2 y = 24 ) 3 x = 12
⇒ x = 4 , substitute in equation 1 and solve for y, the result is 4 + y = 12
⇒ y =8
Answer #10: They need 4 liters of 50% sulfuric acid and 8 liters of 20% sulfuric acid.
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