RAHUL Algebra 1_Grade 912 Equations and Inequalities
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Equations Equation s and Inequalities
Equations and Inequalities
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Copyright © 2009 3P Learning. All rights reserved. First edition printed 2009 in Australia. A catalogue record for this book is available from 3P Learning Ltd.
ISBN
978-1-921861-82-6
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Equations & Inequalities EQUATIONS & INEQUALITIES Somemes just wring variables or ‘pronumerals’ in mathemacal expressions expressions is not enough. Somemes the value of the variable is needed. An equaon is used to nd the value of a variable.
Answer these quesons, before working through the chapter.
I used to think: What does it mean to say that an equaon is ‘linear’?
What does it mean to make a variable the subject of an equaon?
What are the signs of inequality?
Answer these quesons, afer working working through the chapter.
But now I think: What does it mean to say that an equaon is ‘linear’?
What does it mean to make a variable the subject of an equaon?
How are equaons solved?
What do I know now that I didn’t know before?
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What is a Linear Linear Equation? Equation? An equaon is a mathemacal expression that has two sides separated by an equals sign ( =) and at least one variable (or ‘pronumeral’). If the highest power of the variable is 1 then the equaon is called a ‘linear equaon’. equaon’. For example, these are all equaons because ‘ =’ appears in each of them. a
x +
4 = 12
b
3 x = 15
2 + 3 x = 8
c
d
x
3
=5
e
6 x = 12 7
These are also all linear because the power of x of x (the (the variable) is 1 (there is no x no x2 or or x x3 etc).
How are Equations Solved? To solve ANY equaon the goal is always to get the variable by itself. But whatever is done to nd the variable by itself,, must also be done to the other side. Each linear equaon has only one soluon! itself Finding x by by itself
If you look at the linear equaon x equaon x + 4 = 12, it it’s ’s probably easy to see x see x = 8. Your brain has actually simplied the equaon to have x have x by by itself without you even realising it. This is how: Subtracng 4 from the le side of the equaon leaves x by itself. But the same must be done on the right side.
x +
4 = 12
4
4 = 12
x +
x =
4
8
For the equaon in b , to nd x by by itself, both sides must be divided by 3
3 x = 15 3 x = 15 3 3
We must do the same to both sides
x = 5
For the equaon in c , it takes two steps to get x by by itself
2 + 3 x = 8 Subtract 2 from both sides to leave the term with x by itself
2 + 3 x
2= 8 2 3 x = 6
Divide both sides by 3 to leave the x by itself
3 x = 6 3 3 x = 2
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Equations & Inequalities
Basics
If the variable is in the numerator of a fracon, like in
d
, then both sides must be mulplied by the denominator.
For the equaon in d , we need to mulply the le hand side by 3 to nd x by itself
x
3
#3
= 5 #3
x = 15 Somemes the variable will be in a f racon. Mulply and divide both sides to get the variable by itself. In the equaon in e
is the numerator of a fracon
x
Mulply both sides by the denominator (always remove the denominator rst!)
6 x 7
#7
= 12 # 7
6 x = 84 6 x = 84 6 6 x = 14
Divide both sides by 6 to leave the x by itself
Somemes the le side will have to be simplied by collecng like terms: Solve for x in the following linear equaon
2 x + 5 + 3 x = 30 5 x + 5 = 30
Simplify by collecng like terms
5 x + 5
5 = 30 5 x = 5 x = 5 x =
Divide both sides by 5 to leave x by itself
5
25 25 5 5
What happens if the Power of the Variable is 2? If the highest power of the variable is 2 (ie. x2 is in the equaon), then the equaon is not linear, but is called a Quadrac equaon. To get the variable by itself on one side, the square root is used to change x2 to x. However, the square root of both sides must be found. Each quadrac equaon has two soluons! Solve for x in the following quadrac equaon
x2 + x2 + Find the square root of both sides
5
5 = 21 5 = 21
x2
= 16
x2
= !4
5
Subtract 5 from both sides
^ h
2
x = ! 4
2
Because 4 = 16 and
^ h 4
2
= 16
So x = -4 or x = 4
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1. Choose whether the following equaons are linear or quadrac by circling the appropriate word (don’t solve): a
x + 3 = 6 is a
linear
/quadrac equaon
b
3 x2 = 12 is a linear / quadrac equaon
c
3 x = 12 is a linear / quadrac equaon
d
5 x = 3 is a linear / quadrac equaon 2
2. Solve these linear equaons: a
x + 4 = 7
b
x + 3 = 10
c
x - 2 = 13
d
x - 6 = 10
4
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Equations & Inequalities
Questions
Basics
3. Find the value of the variable in each of these linear equaons: a
2 x = 8
b
5 x = 35
c
x = 5
d
x = 8
e
3 x = 9 2
f
5 x = 5 3
4
3
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4. Solve for the variable in each linear equaon: a
4 x + 3 = 23
b
2a - 5 = 9
c
5m + 6 = 31
d
-4 + 7n = 24
e
-2k = 10
f
-m + 4 = 6
6
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(Hint: -m = -1m)
Equations & Inequalities
Questions
Basics
5. Solve for the variable in each linear equaon: a
y
4
=4
b
5 x = 15 4
c
2 p =6 9
d
4d = 8 11
e
7 x = 14 3
f
10m = 25 4
b
4 x2 = 100
6. Solve these quadrac equaons: a
y2 - 9 = 0
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What happens if the Variable is on BOTH sides of the Equation? If the variables appear on both sides of the equaon, one side needs to be changed to have no variables in it. REMEMBER! Whatever is done to one side must be done to the other side too. Solve this linear equaon:
8 x 3 x can be subtracted from both sides. Now the right hand side will have no variables.
8 x
4 = 3x + 6
4 3x = 3x + 6 3x 5 x
5 x
4= 6
4+ 4 = 6+ 4
Add 4 to both sides so that 5 x is by itself
5 x = 10 5 x = 10 5 5
Divide both sides by 5 so that x is by itself
x = 2
Solve this linear equaon:
3 p 12 = 16 4p 4 p can be added to both sides. Now the right hand side will have no variables.
3 p 12 + 4p = 16
4p + 4p
7 p 12 = 16 7 p 12 + 12 = 16 + 12
Add 12 to both sides so that 7 p is by itself
7 p = 28 7 p = 28 7 7
Divide both sides by 7 so that p is by itself
p = 4
If the equaon has brackets, they must be expanded rst. Then just solve the equaon the same way as before. Solve the following linear equaon: Expand both brackets
^
h ^
3 x 3 = 2 x 3 x 9 = 2x 3 x
9 2x x
Add 9 to both sides so that x is by itself
x
2
h
4
4 2x
9=
9+ 9 =
4 4+ 9
x = 5
8
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2 x can be subtracted from both sides so that the right side will have no variables
Equations & Inequalities
Knowing More
Equations with Fractions Some shortcuts can be used when the variable is part of a fracon in the equaon. Type 1: One fracon in the equaon – Mulply both sides of the equaon by the denominator. Solve the linear equaon: (mulply both sides by the denominator)
x
3
+4= 6
The denominator of the fraction is 3
3 # ` x + 4 j = 6 # 3 3 x + 12 = 18
Mulply both sides by the denominator
x = 18 12 x = 6 Here is another example with the variable in a fracon: Solve the linear equaon: (mulply both sides by the denominator)
^5 x 3h = 2 x 2
The denominator of the fraction is 2
2#
Mulply both sides by the denominator
c 5 x2 3 m = 2 5 x
5 x 3
# 2 x
3 = 4x 4x = 4x 4 x
x
3= 0 x = 3
Type 2: More than one fracon in the equaon – Mulply both sides of the equaon by the LCD of the fracons. Solve the linear equaon: (Find the LCD)
x
3
+5= x+4 4
The LCD of the two fractions is 12
Mulply both sides by the LCD
12 # ` x + 5j = 12 # ` x + 4 j 3 4 4 x + 60 = 3x + 48 4 x + 60 3x 4 x
60 = 3x + 48 3x
60
3x = 48 60 x = 12
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Word Problems Word problems can be transformed into equaons which can be solved. Transforming the words into an equaon is the most dicult part. The equaons can be solved using one of the methods learned in this chapter. When a number is halved the answer is 5, what is this number?
Let the number be x x
According to the word problem
2
=5
2 # x = 2 # 5 2
Mulply both sides by the denominator
x = 10 In the above example x = 5 is the equaon represented by the word problem.
2
To nd an equaon from a word problem, let a variable equal the missing value and use the informaon in the word problem to create an equaon. Daniel is 20 cm taller than Philip and the sum of their height is 320 cm. How tall is Philip?
•
Let Philip’s height be x cm
•
So Daniel’s height must be ( x + 20) cm PhilipísHeight + DanielísHeight = 320 cm
^
h
x + x + 20 = 320
2 x + 20 = 320 2 x = 320
20
2 x = 300 2 x = 300 2 2 x = 150 •
So Phillip is 150 cm tall (and Daniel is 170 cm tall)
Lauren is 5 years older than her sister Maria. If the sum of their ages is 25, how old is Lauren?
•
Let Lauren’s age be x
•
So Maria’s age is x - 5
(Maria) (Lauren)
^ x h
5 +
x
= 25
2 x
5 = 25 2 x = 25 + 5 2 x = 30
x =
10
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Equations & Inequalities
Questions
Knowing More
1. In these linear equaons the variable appears on boths sides. Solve for the missing value: a
2u - 10 = 3u
b
7 x - 18 = 3x + 10
c
3 x + 2 = -3
d
4 y + 18 = 12 - 2y
e
10 n - 6 = 2 10 + n
h
f
6m - 4 = - 5 m + 3
g
8 k - 4 - 5k + 3 = 4
h
5 2 y - 1 - 6 y - 2 + 3 = 6
i
8t - 2t - 18 = -12
j
2 a + 3 - 3 a + 4 = -10
^
h
^
h ^
^
h
^
h
^
h
^
h ^
h
^
h ^
h
2. Find Ivan's mistake when he tried to solve this equaon?
^
h ^
h
3 h+ 2 = 2 h+ 1 + 5 3h + 2 = 2h + 2 + 5 3h + 2 = 2h + 7 3h + 2 2 h 2 = 2h + 7 2h
2
h= 5
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Questions
3. Solve these linear equaons which contain fracons:
+ 4 = 5n + 5 2 12
a
x
c
3b + 4 = 4 5
d
c+ c
16r + 2 = 10 5
f
m
e
12
8
1= 4
b
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n
2
2
+ m =5 3
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= 12
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Knowing More
Equations & Inequalities
Questions
Knowing More
4. Solve these linear equaons which contain fracons: a
4q 30 7q + 5 = 3 6
b
5u + 5 + 5u + 10 + 2 = 7 6 9
c
2g 3g g + = 1 + 3 10 2 4
d
2= 6
(Hint: Find LCD of ALL fractions)
e
6 8d 2
x
(Hint: Multiply both sides by the denominator)
=1
f
3k
45
k
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= 2
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5. Three mes a number is 45. What is the number?
6. Claire, Leanne and Lindsay are sisters. Claire is two years older than Leanne and Leanne is 4 years older than Lindsay. The sum of all their ages is 54. How old is each sister?
7. Charlie has been collecng stamps which he keeps in two separate books. The second book has 7 more than triple the stamps of the rst book. If he has 35 stamps in total (from both books) then: a
How many stamps are in the rst book?
b
How many stamps are in the second book?
8. Victor has a bag lled with 2c and 5c coins. He has 2 more coins worth 2c than the coins worth 5c. How many 2c coins does Victor have if all his coins sum to 88c?
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Equations & Inequalities
Using Our Knowledge
What are Linear Inequalities? An inequality is a mathemacal expression with two sides separated by one of these inequality signs: • • • •
2 greater
than $ greater than or equal to
1 less than # less
than or equal to
For example 3 1 4 and 8 2 2. If there is a variable in the inequality and its highest power is 1, then the inequality is a linear inequality (unless the variable is in a denominator). These expressions are all linear inequalies because an inequality appears in each of them and the highest power of the variable is 1. a
x +
b
215
3 x $ 18
c
3 x $ 18
d
x
4
1
7
^
2 x
e
4
h
#
16
How are Inequalities Solved? Just like equaons, the aim is to simplify the inequality to get the variable by itself on one side. Whatever is done to one side must be done to both sides. Solve these inequalies a
y + 3 y + 3
$
10
3m
b
3m
3 $ 10 3
4 1 14
4 + 4 1 14 + 4 3m 1 18 3m 1 18 3 3 m16
y $ 7
Multiplying or Dividing by a Negative Number Everyone knows that 2 1 5 is true. If both sides are mulplied by -1 then -2 1 -5. This is NOT true. If both sides of an inequality are mulplied by a negave number then the inequality sign must be reversed. So 2 2 5. Solve these inequalies a
x
2 2 # x 2
2
5
2
2#5
b
5 3 x # 17 5 5 3 x
5 # 17 5 3 x # 12
x 2 10
x $ 4
x 1 10 Inequality sign is reversed aer mulplying both sides by -1
Inequality sign is reversed aer mulplying both sides by -3
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Using Our Knowledge
Other than reversing the inequality sign when mulplying or dividing by a negave number, inequalies are solved in the same way as equaons. Variable is on both sides
5 x + 25
a
$
2x + 4
5 x + 25 25 $ 2x + 4 25 5 x
2x $ 2x 3 x 3
$
3m + 3 1 5m 5
b
3m + 3 3 1 5m
21 2x
3m
5m 1 5m
21 3
5 3 8 5m
2m 1 8
x $ 7
m24 Inequality sign is reversed aer dividing both sides by -2
Inequalies with brackets a
3
^
q+
3
h
#
^
2 y + 5
b
27
9 # 27 3q 3
x #
2 y + 10 10 2 3y + 15 10
9
18 3
#
2
2 y + 10 2 3y + 15
3q + 9 # 27 3q + 9
h 3^ y + 5h
2 y
3y 2 3y + 5 3y y
6
25
y 1 5 Inequality sign is reversed aer dividing both sides by -1
Inequalies with fracons (mulply by the LCD of ALL the fracons in the inequality)
a
6k 8 5 6k 8 5# 5 6k
$
8
$
8 #5
8+ 8
$
40 + 8
6k 6
$
48 6
b
2t # 10 4 3 12 # ` t 2t j # 10 # 12 4 3 t
3t
LCD of the fracons
8t # 120 5t # 120 t $ 24
k$8
Inequality sign is reversed aer dividing both sides by -4
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Equations & Inequalities
Using Our Knowledge
Graphing Inequalities Soluons to inequalies can be represented on a number line. For example, look at the inequality x > 2. This means x can be any number greater than, but not equal to 2. On a number line x > 2 looks like this:
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
2
3
4
5
6
On a number line x ≥ 2 looks like this:
-6
-5
-4
-3
-2
-1
0
1
Can you spot the dierence in the graphs above? In the rst graph 2 is not included in the inequality (>), so the circle on the number line is hollow. In the second graph the inequality includes the number 2 (≥) so the circle is solid. Here are some more examples:
a
x < - 1
-5 -4 -3 -2 -1
c
2
3
4
d
0
1
2
3
4
5
1
2
3
4
5
0
1
2
3
4
5
x < -3 or x ≥ 0
-5 -4 -3 -2 -1
0
1
2
3
4
5
3
4
5
0
1
2
3
4
5
0
1
2
3
4
5
0
1
2
3
4
5
0
1
2
3
4
5
x ≤ 0 or x ≥ 4
-5 -4 -3 -2 -1
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2
0 ≤ x < 5
-5 -4 -3 -2 -1
j
1
-5 ≤ x ≤ 1
-5 -4 -3 -2 -1
h
0
x < 5
-5 -4 -3 -2 -1
f
0
x ≤ -1
-5 -4 -3 -2 -1
5
-3 - 4
-5 -4 -3 -2 -1
e
b
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Questions
Using Our Knowledge
1. Idenfy if the following are true or false: a
623
b
c
328
d
21 5
f
81 4
e
424
518
2. Solve these inequalies: a
314
x +
b
x
4$5
d
p
10 #
8
c
m+
e
5q # 35
f
4h + 3 2 51
g
5 x 2 24 + x
h
3 x 1
i
4h 8 2
j
3 x 4
18
7$
1
4
2
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^
2x 5
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h 3^1 xh #
$
14
Equations & Inequalities
Questions
Using Our Knowledge
3. Graph these inequalies: a
x 2 3
b
-5 -4 -3 -2 -1
c
2
3
4
5
0
1
2
3
4
5
5 # x 1 5
x 1 0 and x
-5 -4 -3 -2 -1
d
0
1
2
3
4
5
$3
-5 -4 -3 -2 -1
1
2
3
4
1
2
3
4
5
0
1
2
3
4
5
0
1
2
3
4
5
0
1
2
3
4
5
-5 -4 -3 -2 -1
0
1
2
3
4
5
-5 -4 -3 -2 -1
0
1
2
3
4
5
-5 -4 -3 -2 -1
0
1
2
3
4
5
1 1 x 1 3
2 1 x # 4
-5 -4 -3 -2 -1
h
0
0
-5 -4 -3 -2 -1
f
-5 -4 -3 -2 -1
g
1
x # 0
-5 -4 -3 -2 -1
e
0
x $ 2
5
x 2 4 and x
#
1
-5 -4 -3 -2 -1
4. Write down the inequality represented by each of the following graphs:
a
b
-5 -4 -3 -2 -1
0
1
2
3
4
5
c
d
-5 -4 -3 -2 -1
0
1
2
3
4
5
f
e
-5 -4 -3 -2 -1
0
1
2
3
4
5
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Questions
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5. Solve these more complicated linear inequalies, then graph their soluon:
^
a
5 x + 6 2 9 x + 2
c
4 3d #2 8
e
20
a
2
$
h
4a + 3 10
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^
b
7 y
4 y+ 4
d
b
b
8
3
f
6c 2 48
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#
5
1
1 2
h
2
5
c
12
Equations & Inequalities
Thinking More
Simultaneous Equations What happens when there are two variables in equaons?
The equaon x + y = 7 has possible soluons:
= 1 and
y
=6
= 2 and
y
=9
x x
x x
= 3 and
= 10 and
y= y
The equaon 2 x + 3 y = 18 has possible soluons: x = -
3
and
y=
8
x =
3
and
y=
4
x =
6
and
y=
2
10
and
y =-
Common Soluon
4
= 3
x =
4
and many other possibilities
and many other possibilities
The equaons have the common soluon x = 3 and y = 4. This soluon solves both equaons simultaneously. Since two equaons are solved at the same me they are called ‘simultaneous equaons.’ There are three methods to solve simultaneous equaons. All of them work with any queson. It’s up to you to choose the method you think is easiest.
Method 1: Substitution In this method, one of the variables is made the subject of the formula and then substuted into the other equaon. It has three easy steps. Solve the simultaneous equaons using substuon
x + y = 7
1
2 x + 3 y = 18
2
Step 1: Make one of the variables the subject in the equaon, for example, using 1
y = 7 - x Step 2: Substute this expression for y into the 2 to make a new equaon, and solve this new equaon:
2 x + 3(7 - x) = 18 2 x + 21 - 3 x = 18 - x = -3 x = 3 Step 3: Substute this x-value into either of the equaons to nd the y-value:
Substute into
1
3 + y = 7 y = 7
Second equaon
3
2
^h
2 3 + 3 y = 18 3 y = 12
y = 4
y = 4
So the soluons are x = 3 and y = 4. As you can see in Step 3, it doesn’t maer which original equaon you substute the value of the rst variable into. The same answer is found for both. So choose the equaon you think would be easier to use.
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Method 2: Elimination If the coecients of one of the pronumerals are the same in both equaons, then the equaons can be subtracted from one another to eliminate one of the variables. If the coecients of both variables are dierent in each equaon we can make them the same by mulplying one of the equaons by an appropriate number. Solve the simultaneous equaons using eliminaon
x + y = 7
1
2 x + 3 y = 18
2
Step 1: Make sure one of the variables has the same coecient in both equaons
At this stage both x and y have dierent coecients in the two equaons. If
1
is mulplied by 2 then the coecient of x will be the same as in
2#
2 x + 2y = 14
1
same coefficient for x as
2
3
2
Step 2: Subtract equaons with the same coecients
Subtract
3
from
2
2 x + 3y = 18
^2
x +
2y = 14 y
2
h
3
=4
The x has been eliminated and we only need to solve for y
Step 3: Substute the value of the solved variable into any equaon to nd the value of the variable which is
sll unknown
Substute y = 4 into
1
x + 4 = 7 x = 3 So the soluons are x = 3 and y = 4.
In Step 3, y = 4 could have also been substuted into choose the equaon you think would be easier to use.
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and the same value for x would have been found. So
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Method 3: Graphical Method If the straight line graphs are drawn for each linear equaon, then the point where the lines intersect ( x, y) will be the soluon to x and y respecvely. Solve the simultaneous equaons using the graphical method
x + y = 7
1
2 x + 3 y = 18
2
Step 1: Make y the subject of the equaon for 1 and 2
y = - x + 7
1
y = - 2 x + 6
2
3
Step 2: Draw the graphs of these two equaons on the same axes y
8 7
2
6
Lines intersect
5 4 3 2 1
1
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
x
-1 -2
Step 3: Read the point where the lines intersect
The lines intersect at the point ( x, y) = (3, 4) So x = 3 and y = 4 Noce that all three methods have the soluon x = 3 and y = 4. All three methods work all the me so just use the one you feel is the easiest. The next page demonstrates all 3 methods with another example.
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Find the soluon to these simultaneous equaons:
2 y + x = 3 Using a
a
and 3 y + 4 x = 2
1
substuon,
b
2
eliminaon and
c
graphical method.
Substuon
Step 1: Make one of the variables the subject of
1
x = 3 - 2 y
^
h
3 y + 4 3 2y = 2
Step 2: Substute this expression into 2 and solve:
3 y + 12
8y = 2 5 y = 10 y = 2
^h
Step 3: Substute this value into 1 or 2 to solve for the remaining variable: 2 2 + x = 3
x = 3
4= 1
So x = -1 and y = 2
b
Eliminaon
Step 1: Make sure one of the variables has the same coecient in both equaons
4# 3
1
= 8 y + 4x = 12
3
has the same coecient for x as
2
Step 2: Subtract equaons with the same coecients to eliminate a variable
8 y + 4x = 12
^3
y +
4x = 2
h
3 2
5 y = 10 So 5 y ` y
= 10 =2
Step 3: Substute the value of the solved variable into any equaon to nd the value of the variable which is sll unknown
Substute y = 2 into
1
to obtain:
^h
2 2 + x = 3 ` x
=3 4
x = 1 So x = -1 and y = 2
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Graphical
Step 1: Make y the subject of both equaons
y =
1 2
x+
3 2
1
y =
4 3
x+
2 3
2
y
2 1 1
Step 2: Draw the graphs of these two equaons on the same axes Step 3: Read the point where the lines intersect
-1
0
1
2
-1
3
x
2
The lines intersect at (-1, 2) so x = -1 and y = 2
As you can see, all three methods produce the same soluon.
Simultaneous Equations Word Problems As with single linear equaons, word problems can be translated into simultaneous equaons. Determine which TWO missing values are required and choose variables to represent these. Write two equaons using these variables and then use Substuon, Eliminaon or the Graphical Method to solve the equaons. The sum of two numbers is 12 and their dierence is 6. Find the two numbers:
Let x and y represent the numbers. So
x + y = 12
1
and x - y = 6
2
These simultaneous equaons can be solved using subsituon, eliminaon or the graphical method.
Juan is twelve years older than his sister Jamila. In two years Jamila will be half Juan’s age. Find Juan and Jamila’s age:
Let x = Juan’s age Let y = Jamila’s age x - y = 12
1
x + 2 =
y
2
2
These are simultaneous equaons which can be solved using subsituon, eliminaon or the graphical method.
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1. Write down 2 possible soluons for the variables in these equaons: a
x + y = 4
b
2a + b = 6
c
3 x - 4 y = 10
2. Solve for the variables in these simultaneous equaons using the substuon method: a
2 x + y = -1
b
x - 2 y = -4
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2 p + 3q = 10 2q - 4 p = 44
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3. Use the graphical method to solve for these equaons: a
3 x + 2 y = 2 2 x - y = 6
b
3 y - 4 x = 24 2 y + 2 x = 2
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4. Solve for the variables in these simultaneous equaons using the eliminaon method: a
3 x - y = -15 y + 2 x = 0
b
b - 4a = -12
3a - 2b = -1
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5. Solve these simultaneous equaons using any method:
8c - 8d = 2 9c - 8d = 5
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6. Solve these simultaneous equaons using any method:
7m = 16 - 6 n 2n = 3m + 16
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Equations & Inequalities
Questions
Thinking More
7. Solve these simultaneous equaons using any method:
6a - 2b - 10 = 0 2a + 3b - 29 = 0
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8. Find equaons and solve them for these word problems (using any method): a
The sum of two numbers is 12. The sum of the rst number, and double the second number is 16. What are the numbers?
b
Ari is three years older than Eric. In three years from now, Ari will be twice as old as Eric will be. How old are they now?
c
A resturaunt sells two kinds of meals: pizza and pasta. A pizza costs $14 and a pasta costs $10. In a single day the resturaunt sold 79 meals. If they earned $994 on this day, how many of each meal was sold?
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Equations & Inequalities
Answers
Using Our Knowledge: 3.
Thinking More: 1.
e
a
Anything where x and y add up to 4 Eg x = 3, y = 1 or x = 2, y = 2 or x = 5, y = -1 x = 4, y = 0 x = 3.5, y = 0.5 etc
or or
f b
Anything where twice a, plus b, makes 6. Eg a = 1, b = 4, or a = 2, b = 2 or a = 3, b = 0 or a = 4, b = -2 or a = 0.5, b = 5 etc
c
Anything where 3 mes the rst number, minus three mes the second number, makes 10. Eg x = 6, y = 4, or x = -2, y = -4 or x = 30, y = 20 or x = 2, y = -1 or x = 1, y = -1.75 etc
a
x = -1.2 and y = 1.4
g
h
4.
5.
a
x 1 4
b
x # 4
c
x $ -1
d
-2 # x 1 4
e
-4 1 x # 0
f
x # -3
a
x 1 3
b
y 2 7
c
d $ 4
2.
b
3.
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b$
a
b
x = 2 and y = -2 4.
d
p = -7 and q = 8
a
x = -3 and y = 6
b
a = 5 and b = 8
24
e
a#
1 3
5.
a
c=
f
c 1 2.6
6.
a
n
7.
a
a=
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3 and
d =
= 5
4,
b=
7
11 4
x = -3 and y = 4
Equations & Inequalities
Answers
Thinking More: 8.
a
a = 8 and b = 4
b
Eric is 0; Ari is 3.
c
51 pizzas and 28 pasta meals were sold
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