Integrated Algebra
Notes: Solving systems by elimination
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Changing one equation
Another algebraic method of solving a system is by elimination.
Remember that equations need to be in one variable in order to be solved.
Linear equations are often in two variables, x and y. In order to solve for one,
we can eliminate the other.
To use elimination, we set up one variable to have opposite coefficients.
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identify the variable we want to eliminate by looking at the coefficients
identify the term we need to eliminate
multiply one or both of the equations by the necessary factor
combine the equations through addition
To eliminate a variable, we need to decide which one would be the
easiest to eliminate by looking at the coefficient of each. Looking at the
equations:
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is either pair opposite coefficients already?
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Is either pair the same coefficient?
Change one equation
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Is one a multiple of the other?
Change one equation
EX: Solve the system by elimination:
No work needed
x + 3y = 7
x - 4y = 14
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The x coefficients are already the same number, 1. I need to change
one equation to have an x coefficient of -1, then they will be opposites.
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I am going to leave the first equation alone, and multiply the second
equation by a –1 to get my –1x.
x + 3y = 7
x - 4y = 14
x + 3y = 7
-1 ( x - 4y = 14)
Now, substitute -1 for y and solve for x.
x + 3y = 7
x + 3(-1) = 7
x–3=7
x = 10
+
x + 3y = 7
-x + 4y = -14
7y = -7
y = -1
solution: (10, -1)
Integrated Algebra
EX: Solve by elimination
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x + 2y = 7
3x - 2y = -3
What variable should we eliminate? (Look at coefficients) ____
Since the coefficients of the y are already opposites, I can get rid of them
as they are. I do not need to change either equation.
+
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x + 2y = 7
3x - 2y = -3
4x
= 4
x=1
Substitute the x value in to find y.
EX: Solve by elimination.
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1 + 2y = 7
2y = 6
y=3
Solution: (1,3)
2x + 3y = 7
4x - 2y = -6
What variable should we eliminate? (Look at coefficients)
Since the x’s have coefficients where one is a multiple of the other, I am
going to eliminate the x’s.
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What coefficients do we need? ____ and _____
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What equation do we change, first or second? _________
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How do we change it? Multiply by _________
2x + 3y = 7
4x - 2y = -6
-2(2x + 3y = 7)
4x – 2y = -6
Substitute 5 for y and find x: 4x – 2(5) = 6
4x – 10 = 6
4x = 16
x=4
-4x – 6y = -14
+ 4x + 2y = - 6
-4y = -20
y = 5
Solution: (4,5)
Integrated Algebra
TRY: Solve by elimination
2x - 2y = 2
-3x + y = - 9
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What variable should we eliminate? (Look at coefficients) ____
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What coefficients do we need? ____ and _____
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What equation do we change, first or second? _________

How do we change it? Multiply by _________
2x - 2y = 2
-3x + y = -9
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Substitute the x value in to find y.
TRY: Solve by elimination
4x - 7y = -15
4x + 3y = 15
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What variable should we eliminate? (Look at coefficients) ____

What coefficients do we need? ____ and _____

What equation do we change, first or second? _________

How do we change it? Multiply by _________
4x - 7y = -15
4x + 3y = 15

Substitute the x value in to find y.
Integrated Algebra
HOMEWORK: Solving systems by elimination
Changing one equation
Solve each system by substitution.
1.
x+y=3
x+y=3
2.
3x + y = 13
2x – y = 2
3.
2x – 3y = 5
3x + 6y = 4
4.
x + 7y = 3
2x + y = 4
5.
3x – 2y = 10
2x – 2y = 5
6.
2x + 5y = 13
4x – 3y = -13
7.
-5x + 8y = 21
10x + 3y = 15
8.
11x – 3y = 10
-2x + 3y = 8
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