The following steps will be useful to solve system of linear equations using method of substitution.
Step 1 :
In the given two equations, solve one of the equations either for x or y.
Step 2 :
Substitute the result of step 1 into other equation and solve for the second variable.
Step 3 :
Using the result of step 2 and step 1, solve for the first variable.
Solve the following systems of equations by substitution.
Example 1 :
x - 5y + 17 = 0
2x + y + 1 = 0
Solution :
x - 5y + 17 = 0 -----(1)
2x + y + 1 = 0 -----(2)
Step 1 :
Solve (1) for x.
x - 5y + 17 = 0
Subtract 17 from each side.
x - 5y = -17
Add 5y to each side.
x = 5y - 17 -----(3)
Step 2 :
Substitute (5y - 17) for x in (2).
(2)-----> 2(5y - 17) + y + 1 = 0
10y - 34 + y + 1 = 0
11y - 33 = 0
Add 33 to each side.
11y = 33
Divide each side by 11.
y = 3
Step 3 :
Substitute 3 for y in (3).
(3)-----> x = 5(3) - 17
x = 15 - 17
x = -2
Therefore, the solution is
(x, y) = (-2, 3)
Example 2 :
5x + 3y - 8 = 0
2x - 3y + 1 = 0
Solution :
5x + 3y - 8 = 0 -----(1)
2x - 3y + 1 = 0 -----(2)
Step 1 :
Solve (1) for 3y.
5x + 3y - 8 = 0
Add 8 to each side.
5x + 3y = 8
Subtract 5x from each side.
3y = 8 - 5x -----(3)
Step 2 :
Substitute (8 - 5x) for 3y in (2).
(2)-----> 2x - (8 - 5x) + 1 = 0
2x - 8 + 5x + 1 = 0
7x - 7 = 0
Add 7 to each side.
7x = 7
Divide each side by 7.
x = 1
Step 3 :
Substitute 1 for x in (3).
(3)-----> 3y = 8 - 5(1)
3y = 8 - 5
3y = 3
Divide each side by 3.
y = 1
Therefore, the solution is
(x, y) = (1, 1)
Example 3 :
4x - 7y = 0
8x - y - 26 = 0
Solution :
4x - 7y = 0 -----(1)
8x - y - 26 = 0 -----(2)
Step 1 :
Solve (1) for 4x.
4x - 7y = 0
Add 7y to each side.
4x = 7y -----(3)
Step 2 :
Substitute 7y for 4x in (2).
(2)-----> 8x - y - 26 = 0
2(4x) - y - 26 = 0
2(7y) - y - 26 = 0
Simplify.
14y - y - 26 = 0
13y - 26 = 0
Add 26 to each side.
13y = 26
Divide each side by 13.
y = 2
Step 3 :
Substitute 2 for y in (3).
(3)-----> 4x = 7(2)
4x = 14
Divide each side by 4.
x = 3.5
Therefore, the solution is
(x, y) = (3.5, 2)
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