Solve a System of Equations Using Substitution

One way to solve a system of equations algebraically is to use the substitution method. The substitution method consists of substituting one equation into another and solving for one variable. Then use this known variable to solve for the other unknown.

Let’s go through an example of how the substitution method works.

Suppose you want to solve:

x + 3y = 6
2x – y = 5

You can use the substitution method to find the values of x and y.

First, rewrite equation x + 3y = 6 as x = -3y + 6.

Next, substitute x = - 3y + 6 into equation 2x – y = 5.

2(-3y + 6) - y = 5

Solve for y.

-6y + 12 – y = 5
-7y + 12 = 5
-7y = -7
y = 1

The last step is to use the value of y = 1 to solve for x. Use either equation to find x.

x + 3y = 6
x + 3(1) = 6
x + 3 = 6
x = 3

The final solution is (3, 1).

Solving a System of Equations: With one linear function and one quadratic function.

Find the solution to the following system: (Substitution works best!)

Given:  (1) y = x² + 2x + 1

 (2) x + y = 1

(3)    x = 1 – y                           →  (Rewrite equation (2) by subtracting y from both sides.)

(4)    y = (1 – y)² + 2(1 – y) + 1   →    (Substitute (1 – y) into equation (1))

(5)    y = 1 – 2y + y² + 2 – 2y + 1    →  (Simplify equation (4))

(6)    y = y² – 4y + 4 = 0             →    (Simplify equation (5))

(7)    0 = y² – 5y + 4                   →    (Subtract y from both sides of equation (6).)

(8)    0 = (y – 1)(y – 4)                 →    (Factor equation (7).)

(9)    y₁ = 1    y₂ = 4                   →    (Solve for each y.)

(10)  y₁ = 1: x₁ = 1 – (1) = 0;

        y₂ = 4: x₂ = 1 – (4) = -3      → (Substitute y-values from step (9) into equation (3) and solve.)

(11) Solutions are (0, 1) and (-3, 4).

Think & Click Systems with Linear and Quadratic Equations

Now, you try. Complete the Think & Click activity by looking at each problem below, thinking about it, and then clicking on the question to reveal the solution.