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 $x = 6$,

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

{$(1, 3), (6, 13)$}

## 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 $x = -4$,

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

{$(-3,9), (-4, 12)$}

## 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 linear 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 $(-2, -6)$

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 $x = -2$,

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

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