1. Solving Linear Systems by Substitution
7.2
Solving Linear Systems by Substitution
Objective 1 – Use substitution to solve a linear system.
Steps to solve by substitution
Isolate a variable in one of the equations.
2. Substitute the expression into the other equation.
3. Solve for one of the variables.
4. Use the variable in step 3 to solve the other variable

2. Y = 2x
7x + y = 36
Example
Since y = 2x, substitute the y in the second equation with 2x
7x + 2x = 36
9x = 36
x = 4
Since x= 4, Use it to solve for y
Y = 2(4)
Y = 8
Solution is (4,8)

3. Example
y = 3 – 3x
7x + 2y = 1
Since y = 3 – 3x, substitute y with 3 – 3x into the 2nd equation
7x + 2(3 – 3x) = 1
7x + 6 – 6x = 1
x + 6 = 1
x = – 5
y = 3 – 3(-5)
Since x = – 5, use it to solve for y
y = 3 – -15
y = 18
Solution (-5, 18)

4. Solve the linear system
-x + y = 1
2x + y = -2
EXAMPLE
Solve the linear system
Solve for y in Equation 1
y = x + 1
Substitute x + 1 for y in equation 2 and solve for x
2x + y = -2
2x + x + 1 = -2
3x + 1 = -2
3x = -3
x = -1
Substitute -1 for x to find y
y = x + 1
Solution is ( -1, 0)
y =
y = 0

5. Solve the linear system
2x + 2y = 3
x – 4y = -1
EXAMPLE
Solve the linear system
Solve for x in Equation 2
x = 4y - 1
Substitute 4y - 1 for x in equation 1 and solve for y
2x + 2y = 3
2(4y – 1) + 2y = 3
8y – 2 + 2y = 3
10y = 5
Substitute ½ for y to find x
Solution is (1, ½)
x = 2 - 1
x = 1