Elimination by Addition and Subtraction - Examples & Practice

(Video) The Elimination Method: Adding

by ThinkwellVids

(Video) The Elimination Method: Subtracting

by ThinkwellVids

These videos by ThinkwellVids cover elimination through addition and subtraction.

Summary

Here are the basic steps you can take:

Step One: Decide whether to add or subtract.

What do the terms with variables look like in the equations? If there are two that are identical, you should subtract. If there are two that have the same absolute value but different signs, you should add.

Step Two: Add or subtract terms in one equation from terms in the other.

This way, they will cancel out! That's called eliminating the variable. Remember, only add or subtract like terms!

Step Three: Solve the equation for the remaining variable.

Since you've eliminated one variable, there should only be one more left. Solve for it!

Step Four: Plug the new number into one of the starting equations to solve for the other variable.

Step Five: Check your numbers with both equations to make sure they work.

And you're done!

Solution

We want to use elimination to find the solution to the following system of linear equations:

$$\begin{align} 5x-6y&=30\\ 7x+6y&=42 \end{align}$$

Step One: Decide whether to add or subtract.

The first equation has a $-6y$ and the second equation has a $6y,$ so we want to add in order to cancel out the variable. In the next step, we'll see how this works

Step Two: Add terms in one equation from like terms in the other.

$$\begin{align} 5x-\cancel{6y}&=30\\ 7x+\cancel{6y}&=42\\ \hline 12x + 0 &=72 \end{align}$$

Notice how when we add $-6y$ and $6y$ we get $0.$ Now we have a new equation, $12x + 0 =72,$ with only the $x$ variable!

Step Three: Solve the equation for the remaining variable.

$$\begin{align} 12x + 0 &=72\\ 12x&=72\\ x&=\boxed{6} \end{align}$$

Step Four: Plug the new number into one of the starting equations to solve for the other variable.

Let's plug $x=6$ back into the first original equation (we could also have picked the second original equation.

$$\begin{align} 5x-6y&=30\\ 5(6)-6y&=30\\ 30-6y&=30\\ -6y&=0\\ y&=\boxed{0} \end{align}$$

Step Five: Check your numbers with both equations to make sure they work.

$$\begin{align} 5x-6y&=30\\ 5(6)-6(0)&=30\\ 30-0&=30\\ 30&=30 \end{align}$$ $$\begin{align} 7x+6y&=42\\ 7(6)+6(0)&=42\\ 42+0&=42\\ 42&=42 \end{align}$$

Both equations are valid, so $x=6, y=0$ is the solution to our system of linear equations!

Add or Subtract for Elimination in Systems of Linear Equations

Earlier we saw how to solve a system of linear equations via grouping and via substitution.

The main idea of the elimination method is to add or subtract like terms from the equations in order to eliminate one of the variables.

Let's look at an example. $$\begin{align} 2x+5y&=11\\ 3x-5y&=4 \end{align}$$

We can see that both of these equations are in standard form.

Before you start, make sure the equations are lined up. Notice how in our example the $x$ variables are aligned and so are the $y$ variables, the equal signs, and the constants. This is similar to setting up addition and subtraction of decimals!

Now look more closely at the terms in the equations. Notice that in the first equation there is a $5y$ and in the second equation there is a $-5y.$ These terms are identical, just with opposite signs! If we add them together, they will cancel out and become zero.

When you want to use the elimination method to solve a system of equations, check to see if any of the terms are either identical or just have switched signs. If the terms are identical, you want to use subtraction to eliminate the variable, and if the terms just have opposite signs, you want to use addition to eliminate the variable.

Since $5y$ and $-5y$ have opposite signs, we want to add in order to eliminate the variable. But what are we adding?

We want to add like terms in both equations. That means we want to add $2x+3x, 5y+(-5y),$ and $11+4.$ Here's how it works: $$\begin{align} 2x+\cancel{5y}&=11\\ +\quad3x-\cancel{5y}&=4\\ \hline 5x+0y&=15\\ 5x&=15 \end{align}$$

Notice, we eliminated the $y$ variable and we now have an equation containing only the $x$ variable: $5x=15.$ Now we can solve for $x!$ $$\begin{align} 5x&=15\\ x&=\boxed{3} \end{align}$$

The last step is to plug $x=3$ back into either of the original equations equation and solve for $y.$ Let's pick the first equation. $$\begin{align} 2x+5y&=11\\ 2(3)+5y&=11\\ 6+5y&=11\\ 5y&=5\\ y&=\boxed{1} \end{align}$$

So the answer to the system is $(3,1),$ or $x=3, y=1.$ If we plug these values back into each of the original equations, they will be valid!

Try another example!

Solution

We want to find the solution to the linear system

$$\begin{align} 3x+4y&=37\\ 3x-2y&=13 \end{align}$$

First notice that both equations have an identical $3x.$ This means we want to subtract terms in the second equation from terms in the first equation. That way, the $3x's$ will cancel out.

$$\begin{align} \cancel{3x}+4y&=37\\ -\quad\cancel{3x}-2y&=13\\ \hline 0x+6y&=24\\ 6y&=24 \end{align}$$

$($Notice that $4y-(-2y)=4y+2y=6y.)$

Now, we need to solve for $y$ in the equation $6y=24.$

$$\begin{align} 6y&=24\\ y&=\boxed{4} \end{align}$$

The last step is to plug $y=4$ back into either of the original equations equation and solve for $x.$ Let's pick the second equation. $$\begin{align} 3x-2y&=13\\ 3x-2(4)&=13\\ 3x-8&=13\\ 3x&=21\\ x&=\boxed{7} \end{align}$$

So the answer to the system is $(7,4),$ or $x=7, y=4.$ Let's plug these values back into each of the original equations to check that these solutions are valid!

$$\begin{align} 3x+4y&=37\\ 3(7)+4(4)&=37\\ 21+16&=37\\ 37&=37 \end{align}$$ $$\begin{align} 3x-2y&=13\\ 3(7)-2(4)&=13\\ 21-8&=13\\ 13&=13 \end{align}$$

So $x=7, y=4$ is the correct solution!

But what if none of the variables will cancel out?

Add or Subtract for Elimination in Systems of Linear Equations

Here is a graphic with an example of how to ADD terms in a system of equations in order to eliminate the terms with the $y$ variable. After the elimination, solve for the $x$ variable, then plug $x$ into one of the equations to solve for $y.$

Using subtraction to eliminate terms in a system works in a similar way, except you use the subtraction method when you have two identical terms with the same sign! For example, if the bottom equation in our example had a $3y$ instead of a $-3y,$ you could use subtraction instead of addition!