• To eliminate means to get rid of or remove.
• You solve equations by eliminating one of the variables (x or y) using addition or subtraction.

Solve the following system of linear equations by elimination.
Add equation (1) to equation (2)

2x – 3y = 15
5x + 3y = 27
7x + 0y = 42
7x = 42 x = 6
(1)
(2)
 By eliminating y, we can now solve for x

Substitute x= 6 into equation (1) to solve for y
Check your solution x = 6 and y = -1 in equation (2)
2x – 3y = 15
2(6) – 3y = 15
12 – 3y = 15
– 3y = 15 – 12
– 3y = 3 y = -1
5x + 3y = 27
5(6) + 3(-1) = 27
30 – 3 = 27
27 = 27
LS = RS
Therefore, the solution set = (6,-1)

5x – 6y = -32
3x + 6y = 48

7x + 2y = −19
−x + 2y = 21

If you have noticed in the last few examples that to eliminate a variable, it’s coefficients must have a sum or difference of zero.
Sometimes you may need to multiply one or both of the equations by a nonzero number first so that you can then add or subtract the equations to eliminate one of the variables.
We can add these two equations together to eliminate the y variable.
We can add these two equations together to eliminate the x variable.
2x + 5y = 17
6x – 5y = -19
7x + 2y = 10
-7x + y = -16
2x + 5y = -22
10x + 3y = 22
What are we going to do with these equations, can’t eliminate a variable the way they are written?

Multiplying One Equation
Solve by Elimination
2x + 5y = -22
10x + 3y = 22
2x + 5y = -22
10x + 3y = 22
5 (2x + 5y = -22)
10x + 3y = 22
10x + 25y = -110
(10x + 3y = 22)
0 + 22y = -132 y = -6

Step 2 y = -6
Solve for the eliminated variable using either of the original equations.
2x + 5y = -22
2x + 5( -6 ) = -22
2x – 30 = -22
2x = 8 x = 4
The solution is (4, -6).
Choose the first equation.
Substitute -6 for y.
Solve for x.

Solve by elimination.
-2x + 5y = -32
7x – 5y = 17
2x – 3y = 61
2x + y = -7
3x – 10y = -25
4x + 40y = 20

5x + 4y = -28
3x + 10y = -13

Multiplying Both
Equations
To eliminate a variable, you may need to multiply both equations in a system by a nonzero number. Multiply each equation by values such that when you write equivalent equations, you can then add or subtract to eliminate a variable.
4x + 2y = 14
7x + 3y = -8 In these two equations you cannot use graphing or substitution very easily. However ever if we multiply the first equation by 3 and the second by
2, we can eliminate the y variable .
Find the least common multiple LCM of the coefficients of one variable, since working with smaller numbers tends to reduce the likelihood of errors .
4 x 7 = 28
2 x 3 = 6

4x +
7x
2y
– 3y
= 14
= - 8
3
2
(4x +
(7x
2y
– 3y
= 14)
= -8)
12x + 6y = 42
14x – 6y = -16
26x + 0 = 26
26x = 26 x = 1
Solve for the eliminated variable y using either of the original equations.
4x + 2y = 14
4(1) + 2y = 14
4 + 2y = 14
2y = 10 y = 5 The solution is ( 1, 5 ).

Solving Systems of Equations
The Best Time to Use Which Method
Graphing:
Used to estimate the solution, since graphing usually does not give an exact solution. Y = 2x - 3 y = x - 1
Substitution:
If one of the variables in either equation has a
coefficient of 1 or –1 3y + 2x = 4
-6x + y = -7
Elimination
Using
Addition:
If one of the variables has opposite coefficients in the
Elimination two equations 5x – 6y = -32
3x + 6y = 48
Using
Subtraction:
If one of the variables has the same coefficient in the two equations 2x + 3y = 11
2x + 9y = 1
Elimination
Using
Multiplication: If none of the coefficients are 1 or –1 and neither of the variables can be eliminated by simply adding and subtracting the equations. 5 (2x + 5y = -22)
10x + 3y = 22
3 (4x + 2y = 14)
2 (7x = 3y = -8)

Closure
Solve by your method of choice:
1) 2x + 5y = 17
6x + 5y = -9
2) 7x + 2y = 10
-7x + y = -16
3) 2x – 3y = 61
2x + y = -7
5) y = 2x y = x – 1
4) 24x + 2y = 52
6x – 3y = -36
6) 9x + 5y = 34
8x – 2y = -2

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