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Math 11 Principles
Lesson 4 - 5.4: Solving Systems by Substitution
First, let’s review the skill of “isolating a variable”
Isolate/ Solve for “x”.
a) x  6 y  2
b) 3x  2 y  8
Isolate/ Solve for “y”.
a)  4 x  y  11
b) 3x  2 y  8
In Section 5.2, a linear system was solved by adding or subtracting equations to eliminate
one variable. In this section, variables will eliminated by isolating/solving one equation for
ONE variable and then “SUBSTITUTING the result” into the other equation.
Example 1: Solve the following system by substitution.
4x  y  8
x  2 y  7
1 - Decide on a variable to ISOLATE in one of the equations.
2 - Then “substitute the result” into the other equation
3 - Then solve the one variable equation you just created.
Example 2: Solve by substitution.
5x  3 y  5
2 x y  8
Example 3: Solve by substitution.
Example 4: Solve by substitution.
2x – 4y = 10
y = 2x + 2
4x – 3y = -10
x
y

1
4
3
Example 5: Solve by substitution.
y  9  x2
y  x3
Example 7: Solve by substitution.
y x
y  x3
Assignment – page 325 #1a, 4, 5
Example 6: Solve by substitution.
y x
y  x4