In this video, we are going to look at how to solve a system of equations using substitution – linear and quadratic. These are solved similarly to how we would normally solve any other system of equations. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.

For example:
y=x^2+4x-2
y=2x-2
To solve, we are going to take 2x-2, which equals y and substitute it into the first equation where we see y. Now we have

2x-2=x^2+4x-2

Since it is a quadratic, we want to get one side equal to 0 so we can factor. First, add 2 to both sides to get
2x=x^2+4x
Then subtract 2x from both sides to get
0=x^2+2x
When we factor, we get
0=x(x+2)
Set each factor equal to 0
x=0 and x+2=0
We find that
x=0,-2
Now to get the y values, we substitute these x values into an equation. If we substitute them into y=2x-2 then we have
y=2(0)-2 which simplifies to y=-2
y=2(-2)-2 which simplifies to y=-6
Therefore, (0,-2) and (-2,-6) are the solutions to this system of equations.

Example of Solving A System Of Equations Using Substitution (Linear & Quadratic)

Example 1

y = 2x + 1
y = x^2 - 5x +7

Let’s substitute the first equation as the value of y in the second equation.
2x+1 = x^2 -5x +7

Get one side to be zero.

2x+1 -1= x^2 -5x +7 -1
2x = x^2 - 5x +6
2x-2x = x^2 - 5x +6-2x
0 = x^2 - 7x +6

Now, let’s factor this.

0 = (x-1)(x-6)

Then

x - 1 = 0
x = 1

and

x - 6 = 0
x = 6

Now, let’s substitute the values of x to get the value of y.
Let’s use the first equation.
In x = 1,

y = 2x + 1
y = 2(1) + 1
y = 2 + 1
y = 3

In x = 6,

y = x^2 - 5x +7
y = (6)^2 - 5(6) +7
y = 36-30+7
y = 13

So our answers are:
{(1, 3), (6, 13)}

Example 2

y=-3x
y=x^2+4x+12

Let’s substitute the first equation as the value of y in the second equation.
-3x = x^2+4x+12

Get one side to be zero.

-3x = x^2+4x+12
-3x+3x = x^2+4x+12+3x
0 = x^2+7x+12

Now, let’s factor this.

0 = (x+3)(x+4)

Then

x +3 = 0
x = -3

and

x +4 = 0
x = -4

Now, let’s substitute the values of x to get the value of y.
Let’s use the first equation.
In x = -3,

y = -3x
y = -3(-3)
y = 9

In x = -4,

y=x^2+4x+12
y=(-4)^2+4(-4)+12
y = 16-16+12
y = 12

So our answers are:
{(-3,9), (-4, 12)}

Video-Lesson Transcript

Let’s go over solving a system of equations algebraically for linear and quadratic functions.

This is very similar to how you solve a system of equations except we have a variable raised to the second power.

For example:

y = x^2 + 4x - 2
y = 2x - 2

Now, let’s substitute one expression to the value of the other equation.

In this case, let’s get the second equation and substitute it in the first equation.

Now, we have

2x - 2 = x^2 + 4x - 2

Like normal, we’ll have one side of it to be equal zero.

2x - 2 + 2 = x^2 + 4x - 2 + 2
2x = x^2 + 4x
2x - 2x = x^2 + 4x - 2x
0 = x^2 + 2x

There’s no c-term so we’re going to factor this.

Identify the greatest common factor

0 = x (x + 2)

Now, we have two factors which is each equal to zero.

x = 0

and

x + 2 = 0
x + 2 - 2 = 0 - 2
x = -2

Now, let’s solve for the value of y by substituting the value of x in one of the equations.

In x = 0,

y = 2x - 2
y = 2(0) - 2
y = 0 - 2
y = - 2

Here, our answer is (0, -2).

In x = -2,

y = 2x - 2
y = 2(-2) - 2
y = -4 - 2
y = -6

Here, our answer is (-2, -6)

Solving A System Of Equations Using Substitution (Linear & Quadratic)

Let’s try another system of equations.

y = 4x + 2
y = x^2 + x - 8

Same thing, let’s substitute the first equation as the value of y in the second equation.

4x + 2 = x^2 + x - 8

Same strategy. Get one side to be zero.

4x - 4x + 2 = x^2 + x - 4x - 8
2 = x^2 - 3x - 8
2 - 2 = x^2 - 3x - 8 - 2
0 = x^2 - 3x - 10

Now, let’s factor this.

0 = (x - 5) (x + 2)

Then

x - 5 = 0
x - 5 + 5 = 0 + 5
x = 5

and

x + 2 = 0
x + 2 - 2 = 0 - 2
x = -2

Now, let’s substitute the values of x to get the value of y.

Let’s use the first equation.

In x = 5,

y = 4x + 2
y = 4(5) + 2
y = 20 + 2
y = 22

In x = -2,

y = 4x + 2
y = 4(-2) + 2
y = -8 + 2
y = -6

So our answers are:

{(5, 22), (-2, -6)}