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

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.

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.

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 $(-2, -6)$ 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:

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