# Solving a System of Equations Using Elimination

Learn how to solve a system of equations using elimination.

To solve a system of equations using elimination, you start by adding them together to form one equation. This is done by combining like terms.

However, you have to set the equations so that a variable cancels out when you add the 2 equations together.

This can be done by multiplying each equation by a common factor so that a variable in both equations can be canceled out.

You should only be left with one variable and real numbers.

After that, you simplify to find what that variable is equal to.

You then plug your answer into one of the original equations to find the other variable, and to find the coordinate.

To check your answer, plug both x and y-values into an original equation to see if it holds true.

Example:

$4x + 2y = 8$
$\underline{-2x - 2y = 6}$
$2x = 14$
$x = 7$
$\downarrow$
$4(7) + 2y = 8$
$28 + 2y = 8$
$20 = -2y$
$y = -10$

So the coordinates for the point of intersection would be (7,-10).

Check:

$-2x - 2y = 6$
$-2(7) - 2(10) = 6$
$-14 + 20 = 6$
$6 = 6$

## Video-Lesson Transcript

In this lesson, we’ll review elimination which is a method used to solve systems of equation.

Here, we have a system of equation

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

We’ll do elimination by adding the equations together in such a way that one of the variables cancels out.

Let’s add these two and the answer is

$2x = 14$

Here we have an equation with only one variable.

We can easily solve for $x$.

But let’s take a look back at how it worked.

First, equations should be written in the same format.

$x$-term plus $y$-term equals a number

Make sure that they are lined up correctly when we add them.

Another important aspect is that when we add them, one of the terms has to cancel out.

Now, let’s continue solving for $x$

$2x = 14$
$\dfrac{2x}{2} = \dfrac{14}{2}$
$x = 7$

Now, let’s use the value of $x$ to solve for $y$.

Let’s take the first equation and substitute $x = 7$

$4x +2y = 8$
$4(7) +2y = 8$
$28 +2y = 8$

Subtract $28$ on both sides

$28 - 28 + 2y = 8 - 28$
$2y = -20$
$\dfrac{2y}{2} = \dfrac{-20}{2}$
$y = -10$

Our solution is $(7, -10)$

$x = 7$
$y = -10$

Let’s check our answer.

Since we used the first equation to solve for the value of $y$, now let’s use the second equation for checking.

$-2x - 2y = 6$
$-2(7) - 2(-10) = 6$
$-14 + 20 = 6$
$6 = 6$

That checks out.

In this system of equation, the equations are not written in the same format.

$3x + 4y = 5$
$4x = 4y + 9$

So, we will rewrite the bottom equation to follow the format of the top equation.

$4x = 4y + 9$

Let’s bring the $y$-term to the left by subtracting on both sides.

$4x - 4y = 4y - 4y + 9$
$4x - 4y = 9$

Now our two equations are

$3x + 4y = 5$
$4x - 4y = 9$

At this point, we can them together using elimination. The answer is

$7x = 14$

Then let’s solve for $x$

$\dfrac{7x}{7} = \dfrac{14}{7}$
$x = 2$

Now, let’s use this to solve for $y$.

Let’s substitute $x = 2$ on the first equation

$3x + 4y = 5$
$3(2) + 4y = 5$
$6 + 4y = 5$
$6 - 6 + 4y = 5 - 6$
$4y = -1$
$\dfrac{4y}{4} = \dfrac{-1}{4}$
$y = - \dfrac{1}{4}$

So our solution is $(2, - \dfrac{1}{4})$

It’s okay to get a fraction as one of the variables.

When two equations are not written in the same format, just manipulate the function or equation by bringing over terms in the order you want them to be.

And then we can use elimination.

Let’s look at this system of equation and try to solve using elimination.

$3x - 4y = 2$
$2x + y = 5$

It’s in the correct format and lined up correctly.

But none of the variables can be canceled out if we add them up.

Now, we have to change one of the equations so we can use elimination.

To change it, we just have to multiply the equation by a constant.

We can multiply the bottom equation by $4$ so our $y$-term will be the same.

So the process, if a system of equation has no variable that can be canceled out, is to pick an equation that you can multiply a constant number. Make sure that the answer will result in an equation that is canceled out by the other equation.

So we’ll have

$4 (2x + y = 5)$
$8x + 4y = 20$

Now, we’ll continue solving. Let’s add our two equations

$3x - 4y = 2$
$8x + 4y = 20$

$11x = 22$

Then solve for $x$

$\dfrac{11x}{11} = \dfrac{22}{11}$
$x = 2$

Now we can use this value to solve for $y$.

Let’s plug it in the top equation.

$3x - 4y = 2$
$3(2) - 4y = 2$
$6 - 4y = 2$

Subtract $6$ on both sides

$6 - 6 - 4y = 2 - 6$
$- 4y = -4$

Then divide both sides by $-1$

$\dfrac{-4y}{-1} = \dfrac{-4}{-1}$
$y = 1$

So our solution is $(2, 1)$

To recap, we have to multiply one equation by a number so one of the variable will cancel out.

Let’s look at a second way of doing this.

Let’s have the same example

$3x - 4y = 2$
$2x + y = 5$

On the first method, we cancelled $y$-term.

But what if you want to get rid of the $x$?

Here, it’s not easy to multiply so the $x$ will cancel out.

We have to multiply the two equations by two different numbers so we’ll get the same $x$-terms.

We have to think what is the common multiple of $3$ and $2$.

Well it’s $6$.

We’ll multiply the top equation by $2$ and multiply the bottom equation by $3$ so we’ll have $6x$ on both equations.

But that’s not it!

We will be adding the equations so if we add both positive numbers, we’ll have positive too.

So, let’s make one of the multipliers be negative.

Just make sure to make one of them positive and the other negative so we can cancel out.

$2 (3x - 4y = 2)$
$-3 (2x + y = 5)$

Now, ready to solve

$6x - 8y = 4$
$-6x - 3y = -15$

Let’s add them and the answer is

$-11y = -11$
$\dfrac{-11y}{-11} = \dfrac{-11}{-11}$
$y = 1$

Our answer is the same as our answer using the first method.

Let’s use $y = 1$ to solve for $x$ using the top equation.

$3x - 4y = 2$
$3x - 4(1) = 2$
$3x - 4 = 2$
$3x - 4 + 4 = 2 + 4$
$3x = 6$
$\dfrac{3x}{3} = \dfrac{6}{3}$
$x = 2$

Again our solution is $(2, 1)$

Sometimes it possible to multiply one equation by a number to get one of the variables to cancel out.

Or you can also multiply both equations by two different numbers such that one variable will cancel out.

Just make sure that you’re doing it on the variable you want to cancel out.

In this case, we want to cancel out the $x$-terms, the coefficients were $2$ and $3$ to get a common multiple of $6$.

Make sure that one is positive and the other one is negative so they cancel out when we add them.