An example of solving a system of linear equations using substitution.

1. Solve the following system of equations by substitution.
5𝑥 − 𝑦 = 5
4𝑥 − 𝑦 = 3

2. Solve the following system of equations by substitution.
5𝑥 − 𝑦 = 5
4𝑥 − 𝑦 = 3
First, solve one of the equations for
one variable.
It is a good idea to pick a variable
that has a coefficient of 1 or -1 to
avoid fractions.
Let’s solve the first equation for y.

3. Solve the following system of equations by substitution.
5𝑥 − 𝑦 = 5
4𝑥 − 𝑦 = 3
Subtract 5x from both sides
5𝑥 − 𝑦 = 5
−5𝑥 −5𝑥
−𝑦 = 5 − 5𝑥 Multiply both sides by -1
−1 × × −1
𝑦 = 5𝑥 − 5

4. Solve the following system of equations by substitution.
5𝑥 − 𝑦 = 5
4𝑥 − 𝑦 = 3
5𝑥 − 𝑦 = 5
−5𝑥 −5𝑥
−𝑦 = 5 − 5𝑥−1 × × −1
𝑦 = 5𝑥 − 5
Now that we have solve Eq 1 for y,
we substitute into the second
equation.
4𝑥 − 𝑦 = 3
4𝑥 − 5𝑥 − 5 = 3

5. Solve the following system of equations by substitution.
5𝑥 − 𝑦 = 5
4𝑥 − 𝑦 = 3
5𝑥 − 𝑦 = 5
−5𝑥 −5𝑥
−𝑦 = 5 − 5𝑥−1 × × −1
𝑦 = 5𝑥 − 5
Now, we have an equation in one
variable so we can solve for x.
4𝑥 − 𝑦 = 3
4𝑥 − 5𝑥 − 5 = 3
4𝑥 − 5𝑥 + 5 = 3
−𝑥 + 5 = 3
−𝑥 = −2
𝑥 = 2

6. Solve the following system of equations by substitution.
5𝑥 − 𝑦 = 5
4𝑥 − 𝑦 = 3
5𝑥 − 𝑦 = 5
−5𝑥 −5𝑥
−𝑦 = 5 − 5𝑥−1 × × −1
𝑦 = 5𝑥 − 5
Now that we know x = 2, we can
substitute to find y.
4𝑥 − 𝑦 = 3
4𝑥 − 5𝑥 − 5 = 3
4𝑥 − 5𝑥 + 5 = 3
−𝑥 + 5 = 3
−𝑥 = −2
𝑥 = 2
𝑦 = 5 2 − 5
𝑦 = 10 − 5 = 5

7. Solve the following system of equations by substitution.
5𝑥 − 𝑦 = 5
4𝑥 − 𝑦 = 3
5𝑥 − 𝑦 = 5
−5𝑥 −5𝑥
−𝑦 = 5 − 5𝑥−1 × × −1
𝑦 = 5𝑥 − 5
The solution is the ordered pair
2, 5
4𝑥 − 𝑦 = 3
4𝑥 − 5𝑥 − 5 = 3
4𝑥 − 5𝑥 + 5 = 3
−𝑥 + 5 = 3
−𝑥 = −2
𝑥 = 2
𝑦 = 5 2 − 5
𝑦 = 10 − 5 = 5