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# The Elimination Method With Multiplication

### Learning Outcomes

• Use the elimination method with multiplication
• Express the solution of a dependent system of equations containing two variables

## Solve a system of equations when multiplication is necessary to eliminate a variable

Many times adding the equations or adding the opposite of one of the equations will not result in eliminating a variable. Look at the system below. [latex-display]\begin{array}{r}3x+4y=52\\5x+y=30\end{array}[/latex-display] If you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables. So let’s now use the multiplication property of equality first. You can multiply both sides of one of the equations by a number that will allow you to eliminate the same variable in the other equation. We do this with multiplication.  Notice that the first equation contains the term $4y$, and the second equation contains the term $y$. If you multiply the second equation by $−4$, when you add both equations the y variables will add up to $0$. The following example takes you through all the steps to find a solution to this system.

### Example

Solve for $x$ and $y$. Equation A: $3x+4y=52$ Equation B: $5x+y=30$

Answer: Look for terms that can be eliminated. The equations do not have any x or y terms with the same coefficients. [latex-display]\begin{array}{r}3x+4y=52\\5x+y=30\end{array}[/latex-display] Multiply the second equation by $−4$ so they do have the same coefficient. [latex-display]\begin{array}{l}\,\,\,\,\,\,\,\,\,3x+4y=52\\−4\left(5x+y\right)=−4\left(30\right)\end{array}[/latex-display] Rewrite the system and add the equations. [latex-display]\begin{array}{r}3x+4y=52\,\,\,\,\,\,\,\\−20x–4y=−120\end{array}[/latex-display] Solve for $x$. [latex-display]\begin{array}{l}−17x=-68\\\,\,\,\,\,\,\,\,\,\,x=4\end{array}[/latex-display] Substitute $x=4$ into one of the original equations to find y. [latex-display]\begin{array}{r}3x+4y=52\\3\left(4\right)+4y=52\\12+4y=52\\4y=40\\y=10\end{array}[/latex-display] Check your answer. [latex-display]\begin{array}{r}3x+4y=52\\3\left(4\right)+4\left(10\right)=52\\12+40=52\\52=52\\\text{TRUE}\\\\5x+y=30\\5\left(4\right)+10=30\\20+10=30\\30=30\\\text{TRUE}\end{array}[/latex-display] The answers check.

The solution is $(4, 10)$.

Caution!  When you use multiplication to eliminate a variable, you must multiply EACH term in the equation by the number you choose.  Forgetting to multiply every term is a common mistake.

### Example

Solve the given system of equations by the elimination method.

$\begin{array}{l}3x+5y=-11\hfill \\ x - 2y=11\hfill \end{array}$

Answer: Adding these equations as presented will not eliminate a variable. However, we see that the first equation has $3x$ in it and the second equation has $x$. So if we multiply the second equation by $-3,\text{}$ the x-terms will add to zero.

$\begin{array}{llll}\text{ }x - 2y=11\hfill & \hfill & \hfill & \hfill \\ -3\left(x - 2y\right)=-3\left(11\right)\hfill & \hfill & \hfill & \text{Multiply both sides by }-3.\hfill \\ -3x+6y=-33\hfill & \hfill & \hfill & \text{Use the distributive property}.\hfill \end{array}$

$\begin{array}\ \hfill 3x+5y=−11 \\ \hfill −3x+6y=−33 \\ \text{_____________} \\ \hfill 11y=−44 \\ \hfill y=−4 \end{array}$

For the last step, we substitute $y=-4$ into one of the original equations and solve for $x$.

$\begin{array}{c}3x+5y=-11\\ 3x+5\left(-4\right)=-11\\ 3x - 20=-11\\ 3x=9\\ x=3\end{array}$

Our solution is the ordered pair $\left(3,-4\right)$. Check the solution in the original second equation.
$\begin{array}{llll}\text{ }x - 2y=11\hfill & \hfill & \hfill & \hfill \\ \left(3\right)-2\left(-4\right)=11\hfill & \hfill & \hfill & \hfill \\ 11=11\hfill & \hfill & \hfill & \text{True}\hfill \end{array}$

Below is another video example of using the elimination method to solve a system of linear equations in which we multiply one of the equations be a constant. https://youtu.be/_liDhKops2w It is worth demonstrating that there is more than one way to solve a system.  Consider our first example. Instead of multiplying one equation in order to eliminate a variable when the equations were added, we could have multiplied both equations by different numbers. Let’s remove the variable $x$ this time. Multiply Equation A by $5$ and Equation B by $−3$.

### Example

Solve for $x$ and $y$. [latex-display]\begin{array}{r}3x+4y=52\\5x+y=30\end{array}[/latex-display]

Answer: Look for terms that can be eliminated. The equations do not have any x or y terms with the same coefficient. [latex-display]\begin{array}{r}3x+4y=52\\5x+y=30\end{array}[/latex-display] In order to use the elimination method, you have to create variables that have the same coefficient—then you can eliminate them. Multiply the top equation by $5$. [latex-display]\begin{array}{r}5\left(3x+4y\right)=5\left(52\right)\\5x+y =30\,\,\,\,\,\,\,\,\,\,\,\,\\15x+20y=260\,\,\,\,\,\,\\5x+y=30\,\,\,\,\,\,\,\,\,\end{array}[/latex-display] Now multiply the bottom equation by $−3$. [latex-display]\begin{array}{r}15x+20y=260\,\,\,\,\,\,\,\,\\-3(5x+y)=−3(30)\\15x+20y=260\,\,\,\,\,\,\,\,\\−15x–3y=−90\,\,\,\,\,\,\,\end{array}[/latex-display] Next add the equations, and solve for $y$. [latex-display]\begin{array}{r}15x+20y=260\\−15x–3y=\,–90\\17y=170\\y=\,\,\,10\end{array}[/latex-display] Substitute $y=10$ into one of the original equations to find $x$. [latex-display]\begin{array}{r}3x+4y=52\\3x+4\left(10\right)=52\\3x+40=52\\3x=12\\x=4\,\,\,\end{array}[/latex-display] You arrive at the same solution as before.

The solution is $(4, 10)$.

These equations were multiplied by $5$ and $−3$ respectively, because that gave you terms that would add up to $0$. Be sure to multiply all of the terms of the equation. In the next example, we will see that sometimes we have to multiply both numbers by different numbers in order for one variable to be eliminated.

### Example

Solve the given system of equations in two variables by elimination.
$\begin{array}{c}2x+3y=-16\\ 5x - 10y=30\end{array}$

Answer: One equation has $2x$ and the other has $5x$. The least common multiple is $10x$, so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate $x$ by multiplying the first equation by $-5$ and the second equation by $2$.

$\begin{array}{l} -5\left(2x+3y\right)=-5\left(-16\right)\hfill \\ \text{ }-10x - 15y=80\hfill \\ \text{ }2\left(5x - 10y\right)=2\left(30\right)\hfill \\ \text{ }10x - 20y=60\hfill \end{array}$
Then, we add the two equations together.

$\begin{array}\ −10x−15y=80 \\ \:\:10x−20y=60 \\ \text{______________} \\ \text{ }\:\:\:\:\:\:\:\:\:\:−35y=140 \\ \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:y=−4 \end{array}$

Substitute $y=-4$ into the original first equation.

$\begin{array}{c}2x+3\left(-4\right)=-16\\ 2x - 12=-16\\ 2x=-4\\ x=-2\end{array}$

The solution is $\left(-2,-4\right)$. Check it in the second original equation.
$\begin{array}{r}\hfill \text{ }5x - 10y=30\\ \hfill 5\left(-2\right)-10\left(-4\right)=30\\ \hfill \text{ }-10+40=30\\ \hfill \text{ }30=30\end{array}$

Below is a summary of the general steps for using the elimination method to solve a system of equations.

### How To: Given a system of equations, solve using the elimination method

1. Write both equations with and y-variables on the left side of the equal sign and constants on the right.
2. Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
3. Solve the resulting equation for the remaining variable.
4. Substitute that value into one of the original equations and solve for the second variable.
5. Check the solution by substituting the values into the other equation.

### Try It

[ohm_question]38343[/ohm_question]
In the next example, we will show how to solve a system with fractions. As with single linear equations, the easiest way to solve is to clear the fractions first with the least common denominator.

### Example

Solve the given system of equations in two variables by elimination.

$\begin{array}{l}\dfrac{x}{3}+\dfrac{y}{6}=3\hfill \\ \dfrac{x}{2}-\dfrac{y}{4}=\text{ }1\hfill \end{array}$

Answer: First clear each equation of fractions by multiplying both sides of the equation by the least common denominator

$\begin{array}{l}6\left(\dfrac{x}{3}+\dfrac{y}{6}\right)=6\left(3\right)\hfill \\ \text{ }2x+y=18\hfill \\ 4\left(\dfrac{x}{2}-\dfrac{y}{4}\right)=4\left(1\right)\hfill \\ \text{ }2x-y=4\hfill \end{array}$

Now multiply the second equation by $-1$ so that we can eliminate the x-variable.

$\begin{array}{l}-1\left(2x-y\right)=-1\left(4\right)\hfill \\ \text{ }-2x+y=-4\hfill \end{array}$

Add the two equations to eliminate the x-variable and solve the resulting equation.

$\begin{array}\ \hfill 2x+y=18 \\ \hfill−2x+y=−4 \\ \text{_____________} \\ \hfill 2y=14 \\ \hfill y=7 \end{array}$

Substitute $y=7$ into the first equation.

$\begin{array}{l}2x+\left(7\right)=18\hfill \\ \text{ }2x=11\hfill \\ \text{ }x=\dfrac{11}{2}\hfill \end{array}$

The solution is $\left(\dfrac{11}{2},7\right)$. Check it in the other equation.

$\begin{array}{c}2x-y=4\\ 2(\dfrac{11}{2})-7=4\\ 11-7=4 \\ 4=4\end{array}$

In the following video, you will find one more example of using the elimination method to solve a system; this one has coefficients that are fractions. https://youtu.be/s3S64b1DrtQ It is possible to use the elimination method with multiplication and get a result that indicates no solutions or infinitely many solutions, just as with the other methods we have learned for finding solutions to systems.  Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or elimination, the resulting equation will be an identity such as $0=0$. The last example includes two equations that represent the same line and are therefore dependent.

### Example

Find a solution to the system of equations using the elimination method.

$\begin{array}{c}x+3y=2\\ 3x+9y=6\end{array}$

Answer: With the elimination method, we want to eliminate one of the variables by adding the equations. In this case, focus on eliminating $x$. If we multiply both sides of the first equation by $-3$, then we will be able to eliminate the $x$ -variable.

$\begin{array}{l}\text{ }x+3y=2\hfill \\ \left(-3\right)\left(x+3y\right)=\left(-3\right)\left(2\right)\hfill \\ \text{ }-3x - 9y=-6\hfill \end{array}$

$\begin{array} \hfill−3x−9y=−6 \\ \hfill3x+9y=6 \\ \hfill \text{_____________} \\ \hfill 0=0 \end{array}$

We can see that there will be an infinite number of solutions that satisfy both equations. If we rewrote both equations in slope-intercept form, we might know what the solution would look like before adding. Look at what happens when we convert the system to slope-intercept form.

$\begin{array}{l}\text{ }x+3y=2\hfill \\ \text{ }3y=-x+2\hfill \\ \text{ }y=-\dfrac{1}{3}x+\dfrac{2}{3}\hfill \\ 3x+9y=6\hfill \\ \text{ }9y=-3x+6\hfill \\ \text{ }y=-\dfrac{3}{9}x+\dfrac{6}{9}\hfill \\ \text{ }y=-\dfrac{1}{3}x+\dfrac{2}{3}\hfill \end{array}$

See the graph below. Notice the results are the same. The general solution to the system is $\left(x, -\dfrac{1}{3}x+\dfrac{2}{3}\right)$.

In the following video, the elimination method is used to solve a system of equations. Notice that one of the equations needs to be multiplied by a negative one first.  Additionally, this system has an infinite number of solutions. https://youtu.be/NRxh9Q16Ulk In our last video example, we present a system that is inconsistent; it has no solutions which means the lines the equations represent are parallel to each other. https://youtu.be/z5_ACYtzW98

### Try It

[ohm_question]115192[/ohm_question]

## Summary

Multiplication can be used to set up matching terms in equations before they are combined to aid in finding a solution to a system. When using the multiplication method, it is important to multiply all the terms on both sides of the equation—not just the one term you are trying to eliminate.

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