## How to Use the Elimination Method

- Add, subtract, or multiply both sides of the equations until the variables are on the same side of the equal sign and the coefficients of one of the variables are equal and opposite.
- Add the equations together.
- Solve for the non-eliminated variable.
- Solve for the eliminated variable by plugging the result from Step 3 into either one of the original equations.

## Examples

Use the elimination method to solve this system of equations:

\[3x-5y=25\]

\[-3x+7y=-29\]

*Which variable is the easiest to eliminate?*

\[3x-5y=25\]

\[-3x+7y=-29\]

The \(x\) variables already have equal and opposite coefficients (\(3\) and \(-3\)), so it will be easiest to eliminate \(x\).

All the variables are on the left side of the equation so I don’t need to rearrange the equations at all.

*What is the result when you add the equations together?*

\[\begin{array}{cccc} 3x & -5y & = & 25 \\ -3x & +7y & = & -29 \\ \hline & \yellow 2y & \yellow = & \yellow -4 \end {array}\]

When I add the equations together, the \(x\)’s eliminate each other, leaving me with \(\yellow 2y=-4\).

*What does y equal?*

\[\yellow 2y=-4\]

To find the y-coordinate of the solution, I will divide both sides of the eliminated equation by \(\yellow 2\) to undo the multiplication.

\[\yellow y=-2\]

The y-coordinate of my solution is \(\yellow -2\).

*What does y equal?*

Now that I know that \(\yellow y=-2\), I can plug \(\yellow -2\) into either one of the original equations to find the value of \(x\).

\[3x-5y=25\]

\[-3x+7y=-29\]

I think it looks slightly easier to plug \(\yellow y=-2\) into the first equation, so I will do that.

\[3x-5({\yellow -2})=25\]

I will simplify the right side of the equation by multiplying \(-5\) and \(\yellow -2\).

\[3x+10=25\]

Next, I will subtract \(10\) from both sides of the equation to undo the addition.

\[3x=15\]

Then I will divide both sides of the equation by \(3\) to undo the multiplication.

\[x=5\]

The x-coordinate of my solution is \(\yellow 5\).

\(\yellow (5, -2)\) is the solution to this system of equations.

\[3x-5y=25\]

\[-3x+7y=-29\]

I can check my answer by plugging \(\yellow x=5\) and \(\yellow y=-2\) into both equations.

**First Equation**

\[3{\yellow (5)}-5{\yellow(-2)}=25\]

\(3\) times \(5\) is \(15\).

And \(5\) times \(-2\) is \(-10\).

\(15\) minus \(-10\) does equal \(25\).

So, \(\yellow (5, -2)\) is a solution to the first equation.

**Second Equation**

\[-3{\yellow (5)}+7{\yellow(-2)}=-29\]

\(-3\) times \(5\) is \(-15\).

And \(7\) times \(-2\) is \(-14\).

\(-15\) plus \(-14\) does equal \(-29\).

So, \(\yellow (5, -2)\) is a solution to the second equation.

Use the elimination method to solve this system of equations:

\[11x-7y=35\]

\[9x+4y=87\]

*Which variable is the easiest to eliminate?*

\[11x-7y=35\]

\[9x+4y=87\]

The coefficients for the \(y\) variables already have opposite signs, so I think it will be slightly easier to eliminate \(y\).

The least common multiple of \(7\) and \(4\) is \(28\), so I need to get one of the coefficients of \(y\) to equal \(28\) and one of them to equal \(-28\).

To do that, I am going to multiply the first equation by \(4\) so that the \(-7y\) becomes a \(-28y\).

\[{\blue 4}[11x-7y=35]\]

\[\blue 44x-28y=140\]

Then I will multiply the second equation by \(7\) so that the \(4y\) becomes \(28y\).

\[{\blue 7}[9x+4y=87]\]

\[\blue 63x+28y=609\]

All the variables are on the left side of the equations so I don’t need to rearrange the equations at all.

The new versions of the equations are:

\[\blue 44x-28y=140\]

\[\blue63x+28y=609\]

The coefficients of the \(y\) variables are equal and opposite, which is what I want so they eliminate each other in Step 2.

*What is the result when you add the equations together?*

\[\begin{array}{cccc} 44x & -28y & = 140\\ 63x & +28y & = & 609 \\ \hline \blue 107x & & \blue= & \blue 749 \end {array}\]

When I add the equations together, the \(y\)’s eliminate each other, leaving me with \(\blue 107x=749\).

*What does y equal?*

\[\blue 107x=749\]

To find the x-coordinate of the solution, I will divide both sides of the eliminated equation by \(\blue 107\) to undo the multiplication.

\[\blue x=7\]

The x-coordinate of my solution is \(\blue 7\).

*What does y equal?*

Now that I know that \(\blue x-=7\), I can plug \(\blue 7\) into either one of the original equations to find the value of \(y\).

\[11x-7y=35\]

\[9x+4y=87\]

I think it looks slightly easier to plug \(\blue x=7\) into the first equation, so I will do that.

\[9({\blue 7})+4y=87\]

I will simplify the right side of the equation by multiplying \(9\) and \(\blue 7\).

\[63+4y=87\]

Next, I will subtract \(63\) from both sides of the equation to undo the addition.

\[4y=24\]

Then I will divide both sides of the equation by \(4\) to undo the multiplication.

\[y=6\]

The y-coordinate of my solution is \(\blue 6\).

\(\blue (7, 6)\) is the solution to this system of equations.

\[11x-7y=35\]

\[9x+4y=87\]

I can check my answer by plugging \(\blue x=7\) and \(\blue y=6\) into both equations.

**First Equation**

\[11{\blue (7)}-7{\blue (6)}=35\]

\(11\) times \(7\) is \(77\).

And \(7\) times \(6\) is \(42\).

\(77\) minus \(42\) does equal \(35\).

So, \(\blue (7, 6)\) is a solution to the first equation.

**Second Equation**

\[9{\blue (7)}+4{\blue (6)}=87\]

\(9\) times \(7\) is \(63\).

And \(4\) times \(6\) is \(24\).

\(63\) plus \(24\) does equal \(87\).

So, \(\blue (7, 6)\) is a solution to the second equation.

Use the elimination method to solve this system of equations:

\[\frac{2}{3}x-\frac{5}{2}y=-\frac{41}{6}\]

\[y=4x-1\]

*Which variable is the easiest to eliminate?*

\[\frac{2}{3}x-\frac{5}{2}y=-\frac{41}{6}\]

\[y=4x-1\]

The \(y\) variable in the second equation has a coefficient of 1, so I think it will be easiest to eliminate \(y\).

First of all, I will multiply the first equation by \(6\) so I won’t have to deal with the fractions.

\[{\green 6}[\frac{2}{3}x-\frac{5}{2}y=-\frac{41}{6}]\]

\[\green 4x-15y=-41\]

The coefficient for \(y\) in the first equation is \(\green -15\) now, so I need to make the coefficient of \(y\) in the second equation \(\green 15\), so they will eliminate each other in Step 2.

To do that, I will multiply the second equation by \(15\).

\[{\green 15}[y=4x-1]\]

\[\green 15y=60x-15\]

I will also subtract \(\green 60x\) from both sides of the equation so that the variables are all on the same side.

\[\green -60x+15y=-15\]

The new versions of the equations are:

\[\green 4x-15y=-41\]

\[\green -60x+15y=-15\]

The coefficients of the \(y\) variables are equal and opposite, which is what I want so they eliminate each other in Step 2.

*What is the result when you add the equations together?*

\[\begin{array}{cccc} 4x & -15y & = -41\\ -60x & +15y & = & -15 \\ \hline \green -56x & & \green= & \green -56 \end {array}\]

When I add the equations together, the \(y\)’s eliminate each other, leaving me with \(\green -56x=-56\).

*What does y equal?*

\[\green -56x=-56\]

To find the x-coordinate of the solution, I will divide both sides of the eliminated equation by \(\green -56\) to undo the multiplication.

\[\green x=1\]

The x-coordinate of my solution is \(\green 1\).

*What does y equal?*

Now that I know that \(\green x=1\), I can plug \(\green 1\) into either one of the original equations to find the value of \(y\).

\[\frac{2}{3}x-\frac{5}{2}y=-\frac{41}{6}\]

\[y=4x-1\]

It looks much easier to plug \(\green x=1\) into the second equation, so I will do that.

\[y=4({\green 1})-1\]

I will simplify the right side of the equation by multiplying \(4\) and \(\green 1\).

\[y=4-1\]

Next, I will subtract \(1\) from \(4\).

\[y=3\]

The y-coordinate of my solution is \(\green 3\).

\(\green (1,3)\) is the solution to this system of equations.

\[\frac{2}{3}x-\frac{5}{2}y=-\frac{41}{6}\]

\[y=4x-1\]

I can check my answer by plugging \(\green x=1\) and \(\green y=3\) into both equations.

**First Equation**

\[\frac{2}{3}{\green (1)}-\frac{5}{2}{\green (3)}=-\frac{41}{6}\]

\(\frac{2}{3}\) times \(1\) is \(\frac{2}{3}\).

And \(\frac{5}{2}\) times \(3\) is \(\frac{15}{2}\).

\(\frac{2}{3}\) minus \(\frac{15}{2}\) does equal \(-\frac{41}{6}\).

So, \(\green (1, 3)\) is a solution to the first equation.

**Second Equation**

\[{\green (3)}=4{\green (1)}-1\]

\(4\) times \(1\) is \(4\).

And \(4\) minus \(1\) is \(3\).

So, \(\green (1, 3)\) is a solution to the second equation.