Substitution Method of Systems of Linear Equations

1. SOLVING
SYSTEMS OF
LINEAR
EQUATIONS BY
SUBSTITUTION

2. Solving systems of linear equation by substitution
Solving a system of linear equations by substitution is another
way of finding the solution set of the system wherein one of
the equations is transformed into the form y = ax + c.

3. EXAMPLE 1:
Solve the system:
2x + 3y = 12 (1st equation)
x + y = 5 (2nd equation)
Step 1:
Pick one of the equation that you will change in the form
y = ax+b.
Change 2nd equation in the form y = ax+b
x + y = 5
y = -x + 5

4. EXAMPLE 1:
Solve the system:
2x + 3y = 12 (1st equation)
x + y = 5 (2nd equation)
Step 2:
Substitute the value of the 2nd equation (y) to the 1st equation.
y=-x+5
2x+3(-x+5) =12
2x -3x +15 = 12
-x = 12-15
x=3

5. EXAMPLE 1:
Solve the system:
2x + 3y = 12 (1st equation)
x + y = 5 (2nd equation)
Step 3:
Substitute the value of the x to the 1st equation.
2x + 3y = 12
2(3) + 3y = 12
6 + 3y = 12
3y = 12 - 6
3y = 6
y = 2

6. EXAMPLE 1:
Solve the system:
2x + 3y = 12 (1st equation)
x + y = 5 (2nd equation)
Step 4:
The solution set is (3, 2).
Check if you got the correct solution set by substituting it to
the equations.
2(3) + 3(2) = 12
6+6 = 12
12 = 12
x + y = 5
3 + 2 = 5
5 = 5

8. EXAMPLE 2:
Solve the system:
x + 3y = 7 (1st equation)
4x - 2y = 0 (2nd equation)
Step 1:
Pick one of the equation that you will change in the form
y = ax+b.
Change 2nd equation in the form y = ax+b
4x - 2y = 0
-2y = -4x
y = 2x

9. EXAMPLE 2:
Solve the system:
x + 3y = 7 (1st equation)
4x - 2y = 0 (2nd equation)
Step 2:
Substitute the value of the 2nd equation (y) to the 1st equation.
y=2x
x + 3y = 7
x + 3(2x) = 7
x + 6x = 7
7x = 7
x = 1

10. EXAMPLE 2:
Solve the system:
x + 3y = 7 (1st equation)
4x - 2y = 0 (2nd equation)
Step 3:
Substitute the value of the x to the 1st equation.
x + 3y = 7
1 + 3y = 7
3y = 7-1
3y = 6
y = 2

11. EXAMPLE 2:
Solve the system:
x + 3y = 7 (1st equation)
4x - 2y = 0 (2nd equation)
Step 4:
The solution set is (1, 2).
Check if you got the correct solution set by substituting it to
the equations.
x+3y=7
1+3(2)=7
1+6=7
7=7
4x-2y=0
4(1)-2(2)=0
4-4=0
0=0

13. EXAMPLE 3:
Solve the system:
y = 4x (1st equation)
3x + y = -21 (2nd equation)
Step 1:
Since the 1st equation is already in the form y = ax+b, no need
to choose which equation should be changed.
y=4x

14. EXAMPLE 3:
Solve the system:
y = 4x (1st equation)
3x + y = -21 (2nd equation)
Step 2:
Substitute the value of the 1st equation (y) to the 2nd equation.
y=4x
3x+y=-21
3x+4x=-21
7x=-21
x=-3

15. EXAMPLE 3:
Solve the system:
y = 4x (1st equation)
3x + y = -21 (2nd equation)
Step 3:
Substitute the value of the x to the 1st equation.
y=4x
y=4(-3)
y=-12

16. EXAMPLE 3:
Solve the system:
y = 4x (1st equation)
3x + y = -21 (2nd equation)
Step 4:
The solution set is (-3, -12).
Check if you got the correct solution set by substituting it to
the equations.
y=4x
-12=4(-3)
-12=-12
3x+y=-21
3(-3) + (-12) = -21
-9 - 12 = -21
-21=-21

18. EXAMPLE 4:
Solve the system:
x + y = 5(1st equation)
y = x + 3(2nd equation)
Step 1:
Since the 1st equation is already in the form y = ax+b, no need
to choose which equation should be changed.
y = x + 3

19. EXAMPLE 4:
Solve the system:
x + y = 5(1st equation)
y = x + 3 (2nd equation)
Step 2:
Substitute the value of the 2nd equation (y) to the 1st equation.
x + y = 5
x + x+3 = 5
2x=5-3
2x = 2
x = 1

20. EXAMPLE 4:
Solve the system:
x + y = 5(1st equation)
y = x + 3 (2nd equation)
Step 3:
Substitute the value of the x to the 1st equation.
x + y = 5
1 + y = 5
y = 5 - 1
y = 4

21. EXAMPLE 4:
Solve the system:
x + y = 5(1st equation)
y = x + 3 (2nd equation)
Step 4:
The solution set is (1, 4).
Check if you got the correct solution set by substituting it to
the equations.
x + y = 5
1 + 4 = 5
5 = 5
y = x + 3
4 = 1 + 3
4 = 4