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# Systems of equations with elimination

Walk through examples of solving systems of equations with elimination.
In this article, we're going to be solving systems of linear equations using a strategy called elimination. First, we need to understand that it's okay to add equations together.
Key idea: Anytime we have two true equations, we can add or subtract them to create another true equation.
For example, here are two very basic true equations:
$2=2$
$5=5$
We can add these equations together to create another true equation:
Or we can subtract these equations to create another true equation:
Here's another example with more complicated equations:
Great, now that we see that it's okay to add or subtract equations, we can move onto solving a system of equations using elimination.

## Solving a system of equations using elimination

We'll solve this system of equations as an example:
The hard thing about solving is that there are two variables $x$ and $y$. If only we could get rid of one of the variables...
Here's an idea! Let's add the two equations together to cancel out the $y$ variable:
Brilliant! Now we have an equation with just the $x$ variable that we know how to solve:
Baller! Let's use the first equation to find $y$ when $x$ equals $5$:
$\begin{array}{rlr}x+3y& =8& \text{Equation 1}\\ \\ 5+3y& =8& \text{Substitute 5 for x}\\ \\ 3y& =3& \text{Subtract 5 from each side}\\ \\ y& =1& \text{Divide each side by 3}\end{array}$
Sweet! So the solution to the system of equations is $\left(5,1\right)$.
Use elimination to solve the following system of equations.
$4y-2x=4$
$5y+2x=23$
$x=$
$y=$

## Multiplying one of the equations by a constant, then using elimination

That last example worked out great because the $y$ variable was eliminated when we added the equations. Sometimes it isn't quite that easy.
Take this system of equation as an example:
If we add these equations, neither the $x$ or $y$ variable will be eliminated, so that won't work. Here are the steps for problems like this:
Step 1: Multiply one of the equations by a constant so that when we add it to the other equation, one of the variables is eliminated.
$\begin{array}{rlr}-2\left(3x-4y\right)& =-2\left(1\right)& \text{Multiply the second equation by}-2\\ \\ -6x+8y& =-2& \text{Simplify to get a new equation}\end{array}$
Step 2: Add the new equation to the equation we didn't use in step 1 in order to eliminate one of the variables.
Step 3: Solve for $y$.
$\begin{array}{rl}13y& =26\\ \\ y& =2& \text{Divide each side by 13}\end{array}$
Step 4: Substitute $y=2$ into one of the original equations, and solve for $x$.
$\begin{array}{rlr}3x-4y& =1& \text{Equation 2}\\ \\ 3x-4\left(2\right)& =1& \text{Substitute 2 for y}\\ \\ 3x-8& =1\\ \\ 3x& =9& \text{Add 8 to each side}\\ \\ x& =3& \text{Divide each side by 3}\end{array}$
So our solution is $\left(3,2\right)$.
Use elimination to solve the following system of equations.
$8x+14y=12$
$-6x-7y=-16$
$x=$
$y=$

## Multiplying both of the equations by a constant, then using elimination

Sometimes we'll need to multiply both equations by a constant when using elimination.
For example, consider this system of equations:
Here are the steps to solve a system of equations like this one:
Step 1: Multiply each equation by a constant so we can eliminate one variable.
$5x+3y=14⇒\text{Multiply by 2}⇒10x+6y=28$
$3x+2y=8⇒\text{Multiply by 3}⇒9x+6y=24$
Step 2: Combine the new equations to eliminate one variable.
$\begin{array}{rl}10x+6y& =28\\ 9x+6y& =24\\ \\ x+0& =4& \text{Subtract the equations}\end{array}$
Step 3: Substitute $x=4$ into one of the original equations, and solve for $y$.
$\begin{array}{rlr}3x+2y& =8& \text{Equation 2}\\ \\ 3\left(4\right)+2y& =8& \text{Substitute 4 for x}\\ \\ 12+2y& =8\\ \\ 2y& =-4& \text{Subtract 12 from each side}\\ \\ y& =-2& \text{Divide each side by 2}\end{array}$
So our solution is $\left(4,-2\right)$.
Use elimination to solve the following system of equations.
$5x+4y=-14$
$3x+6y=6$
$x=$
$y=$

## Let's practice!

1) Use elimination to solve the following system of equations.
$3y+x=7$
$2y-x=-2$
$x=$
$y=$

2) Use elimination to solve the following system of equations.
$\phantom{\rule{2em}{0ex}}-7y-4x=1$
$\phantom{\rule{2em}{0ex}}7y-2x=53$
$x=$
$y=$

3) Use elimination to solve the following system of equations.
$-9y+4x-20=0$
$-7y+16x-80=0$
$x=$
$y=$

4) Use elimination to solve the following system of equations.
$3x-11y=-1$
$2x-5y=-3$
$x=$
$y=$

## Challenge problem

A school is selling tickets to a play. On the first day of ticket sales, the school sold $6$ adult tickets and $10$ student tickets for a total of $\mathrm{}140$. On the second day of ticket sales, the school sold $7$ adult tickets and $3$ student tickets for a total of $\mathrm{}94$.
Write and solve a system of equations to find the cost of an adult ticket and the cost of a student ticket.
The cost of an adult ticket is $\mathrm{}$
.
The cost of a student ticket is $\mathrm{}$
.