Solving A System of Linear Equations by Elimination
Solving a system of linear equations by adding and subtracting (elimination).
To solve a system of linear equations by elimination requires one of the variables, in both
equations, to have opposite numerical coefficients. This can be achieved by multiplying one or
both of the equations by an appropriate value. When this is done, the equations are then
combined. This will eliminate the variable with the opposite coefficients and will leave one
variable in the answer equation. This equation must be solved. The solution will then be
substituted into one of the given equations of the system to determine the value of the variable
that was eliminated. To facilitate this procedure, the system should be aligned as
a1 x  b1 y  c1 


a2 x  b2 y  c2 
Example1: Solve the following system of linear equations by using the method of
elimination.
3x  5 y  12 


2 x  10 y  4  The system is properly aligned.
Choose one variable and perform the necessary multiplication to produce opposite coefficients
for that variable in both equations.
23x  5 y  12  6x – 10y = 24 This equation was multiplied by 2 in
2 x  10 y  4
2x + 10y = 4
order to change the coefficient of y
to a negative 10. This makes the
8x
= 28 coefficients of y opposites. When the
8
8
equations were combined (added),
the y’s were eliminated.
x = 3.5
Substitute this value into one of the given equations of the system to determine the value of y.
2(3.5) + 10y = 4
7 + 10y = 4
7 – 7 + 10y = 4 – 7
10y = -3
10
10
y = – 0.3
The common point or the solution is (3.5, – 0.3)
Solution:
Example 2:
Solve the following system of linear equations by using the method of
elimination.
2 x  9 y  1


4 x  y  15 

 22 x  9 y  1

4 x  y  15
 4 x  18 y  2
4 x  y  15
 17 y  17
 17
17
y
 17
 17
y  1
Substitute this value into one of the given equations of the system to determine the value of x.
4 x  y  15
4 x   1  15
4 x  1  15
4 x  1  1  15  1
The common point or the solution is (4,-1)
4 x  16
4
16
x
4
4
x4
Example 3: Solve the following system of linear equations by using the method of
elimination.
3x  2 y  9  0
4x  3y  5
3x  2 y  9  0

  3x  2 y  9  9  0  9 4 x  3 y  3 y  3 y  5
4 x  3 y  5 
3 x  2 y  9
4x  3y  5
3 x  2 y  9
The system is properly aligned.
4x  3y  5
9 x  6 y  27
33 x  2 y  9 

8 x  6 y  10
24 x  3 y  5
17 x  17
17 x  17

17
17
x  1
Substitute this value into one of the given equations of the system to determine the value of y.
3x  2 y  9  0
3 1  2 y  9  0
 3  2y  9  0
2y  6  0
2y  6  6  0  6
2 y  6
The common point or the solution is (-1,-3)
2
6
y
2
2
y  3
Example 4: Solve the following system of linear equations by using the method of
elimination.
2
3

20 3 x  20 2 y  202
 4 x  5 y  2 
4
5


 
1
3
113
1
3
113
 x y 

14 x  14 y  14
 7
7
2
7
2
7 
 215 x  8 y  40

152 x  21y  226
15 x  8 y  40
2 x  21y  226
 30 x  16 y  80
30 x  315 y  3390
331 y  3310
331
3310
y
331
331
y  10
Substitute this value into one of the given equations of the system to determine the value of x.
3
2
x  10   2
4
5
3
x4 2
4
3
x44  24
4
3
x6
4
4 3 x  46
4
3 x  24
The common point or the solution is (8,10)
3
24
x
3
3
x8
Exercises: Solve the following system of linear equations by using the method of
elimination.
16 x  y  181  0
1. 

19 x  y  214 
3
26 

x  y  
4. 
5
5
4 y  61  7 x 
 x  69  6 y 
2. 

3x  4 y  45
 x  7 y  38 
3. 

14 y   x  46
Solutions:
 x  3 y  15  0
1. 

7 y  3x  21

Comparison Method
 x  3 y  15  0
 x  x  3 y  15  0  x
3 y  15  15  x  15
7 y  3x  21
7 y  3x  3x  3 x  21
7 y  3x  21
7
3
21
y  x
7
7
7
3
y  x3
7
3 y  x  15
3
1
15
y  x
3
3
3
1
y  x5
3
1
3
x5 x3
3
7
1 
3 
21 x   215  21 x   213
3 
7 
7 x  105  9x  63
7 x  105  105  9x  63  105
7 x  9x  42
7 x  9x  9x  9x  42
 2x  42
2
 42
x
2
2
x  21
1
x5
3
1
y  21  5
3
y  75
y  12
y
The solution is (21, 12)
 x  3 y  11
2. 

3x  2 y  30
x  3 y  11
x  3 y  3 y  3 y  11
x  3 y  11
Substitution Method
x  3 y  11
x  3 9  11
x  27  11
x  16
3 x  2 y  30
3 3 y  11  2 y  30
 9 y  33  2 y  30
 7 y  33  30
 7 y  33  33  30  33
 7 y  63
7
63
y
7
7
y  9
The solution is (16,-9).
2 x  y  7  0
3. 

x  2 y  1  0 
2x  y  7  0
2x  2x  y  7  0  2x
y  7  2 x
y  7  7  2 x  7
y  2 x  7
x  2y 1  0
x  x  2y 1  0  x
 2 y 1  x
 2 y 1  1  x  1
 2 y  x  1
2
1
1
y
x
2
2
2
1
1
y  x
2
2
Graphing Method
Following is the graph of y  2 x  7 and y 
1
1
x .
2
2
The solution is the point of intersection (3, 1).
Solutions: Elimination Method
16 x  y  181  0
1. 

19 x  y  214 
16 x  y  181  0
16 x  y  181  181  0  181
16 x  y  181
16 x  y  181

19 x  y  214
 16 x  y  181
19 x  y  214
3x  33
3
33
x
3
3
x  11
The solution is (11, -5).
 1(16 x  y  181)
19 x  y  214
16 x  y  181
16(11)  y  181
176  y  181
176  176  y  181  176
1
5
y
1
1
y  5
 x  69  6 y 
2. 

3x  4 y  45
x  69  6 y
x  6 y  69  6 y  6 y
x  6 y  69
x  6 y  69

3 x  4 y  45
3 x  4 y  45
3 x  4 y  45  4 y
3 x  4 y  45
 3( x  6 y  69)

3 x  4 y  45
 3x  18 y  207
3x  4 y  45
14 y  252
14
 252
y
14
14
y  18

x  69  6 y
x  69  6(18)
x  69  108
x  39
The solution is (-39, -18)
 x  7 y  38 
3. 

14 y   x  46
x  7 y  38
x  7 y  7 y  7 y  38
x  7 y  38
14 y   x  46
14 y  x   x  x  46
x  14 y  46
x  7 y  38

x  14 y  46
2 x  14 y  76

x  14 y  46
2( x  7 y  38)

x  14 y  46
3x  30
3
30
x
3
3
x  10
x  7 y  38
10  7 y  38
10  38  7 y  38  38
 28  7 y
 28 7
 y
7
7
4 y
The solution is (10, -4).
3
26 

x  y  
4. 
5
5
4 y  61  7 x 
3
26
y
5
5
3 
 26 
5 x   5 y   5 
5 
 5 
5 x  3 y  26
x
5 x  3 y  26

7 x  4 y  61
4 y  61  7 x
4 y  7 x  61  7 x  7 x
7 x  4 y  61
4(5 x  3 y  26)

3(7 x  4 y  61)
20 x  12 y  104
21x  12 y  183
41x  287
41
287
x
41
41
x7
The solution is (7, 3)
4 y  61  7 x
4 y  61  7(7)
4 y  61  49
4 y  12
4
12
y
4
4
y3