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Using Multiplication for Elimination in Systems of Linear Equations

Earlier, we saw how to solve systems of linear equations with graphing (looking at the points where the two functions intersect) and substitution (plugging one equation into the other). This lesson will focus on elimination as a method of solving these systems of equations.

The goal of elimination is exactly what it sounds like. we want to manipulate a system of equations so that one of the variables is eliminated. This is easiest when both equations have identical terms or terms that are the same except for opposite signs. In these cases, we can use the addition/subtraction elimination method.

But what are you supposed to do when none of the terms look similar? Not to worry — you can handle it. The approach is to make terms look similar by using multiplication in one or both equations. We're essentially going to manipulate the equations so that we can eliminate terms easily.

Let's see how that might look in action. Consider the following system of linear equations: $$\begin{align} 2x + y &= 10\\ 5x - 4y &= -1 \end{align}$$

Notice how both equations are in standard form. Standard form makes it easier to notice how we can solve the system.

Good job! In the equation $2x+y=10,$ the $x$ coefficient is $2$ and the $y$ coefficient is $1.$ In the equation $5x-4y=-1,$ the $x$ coefficient is $5$ and the $y$ coefficient is $-4.$

We want to make our job as easy as possible when we use the multiplication method.

Here's what we mean by that:

1) We could multiply terms in our equations by a constant to eliminate $x.$ That means we need the same coefficient in front of the $x$ in each equation.

If we want the same coefficient for $x$ in each equation, we could multiply terms in the first equation by $5$ and the second equation by $2$. This would give us $10x$ in each equation. The downside is that this requires two steps. (Two steps means more work and more opportunities for calculation errors.)

2) The other option is to multiply terms in our equations to eliminate $y$. That means we need the same coefficient in front of $y$ in each equation.

We could multiply terms in the first equation by $4$ to get the same $y$ coefficient in both equations . This would give us $4y$ in the first equation and a $-4y$ in the second equation. Let's do the multiplication now. Remember to do the same thing to both sides of the equation!

$$\begin{align} 2x + y &= 10\\ 4(2x+y)&=4(10)\\ 8x+4y&=40 \end{align}$$

Remember, whatever you do on one side of the equation, you must do on the other.

Now, we have a new equation: $8x+4y=40.$ We can put it in our linear system in place of $2x + y = 10$, then solve just like we do with normal addition elimination!

$$\begin{align} 8x+\cancel{4y}&=40\\ 5x \cancel{- 4y} &= -1\\ \hline 13x&=39 \end{align}$$

Remember to align the equations like you'd align decimals when you're adding and subtracting!

$$\begin{align} 13x &= 39\\\ x &= 3 \end{align}$$

Awesome! One variable down, one to go. Now all we need to do is plug $x=3$ into one of the two original equations and solve for $y$.

$$\begin{align} 2x + y &= 10 \\\ 2(3) + y &= 10 \\\ 6 + y &= 10 \\\ y &= 10 - 6 \\\ y &= 4 \end{align}$$

It is a VERY good idea to check your work by plugging your values for $x$ and $y$ into both of the original equations.

Multiply for Elimination

Some systems can't be solved with only adding and subtracting for elimination. These systems require our old friend, multiplication. You can multiply terms in one or both equations by a constant to set up your elimination. Then you can add or subtract terms in one equation from the other and eliminate a variable.

Example of Multiplying for Elimination

Say we have the following system of equations: $$\begin{align} -7x + y &= 2\\ 3x - 3y &= 3 \end{align}$$

The first thing we check for is the possibility of adding/subtracting terms to eliminate one of the variables. Unfortunately, we cannot do that since none of the $x$ or $y$ terms look similar, so we'll look at multiplication!

The next thing we're going to do is determine which variable we want to eliminate. Typically, you want to pick one that can easily be removed by using multiplication on only one equation. For our example, we can see that multiplying $-7x + y = 2$ by $3$ can help us out!

$$\begin{align} \color{#8925AE}{3}(-7x + y) &= \color{#8925AE}{3}(2)\\ 3x - 3y &= 3 \end{align}$$

So our system of equations now looks like:

$$\begin{align} -21x + 3y &= 6\\ 3x - 3y &= 3 \end{align}$$

Great! Now we can add terms in the second equation to the first to eliminate the $y$ variable!

$$\begin{align} -21x + 3y &= 6\\ 3x - 3y &= 3\\ \hline -18x +0y&= 9 \\ -18x&= 9 \\ x&=\boxed{-\frac{1}{2}} \end{align}$$

Great! One of our mysteries is solved. Now, we can substitute $x = -\frac{1}{2}$ into one of our original equations. Let's plug it into $3x - 3y = 3.$

$$\begin{align} 3x - 3y = 3\\ 3\Bigg(\frac{-1}{2}\Bigg) -3y&=3\\ \Bigg(\frac{-3}{2}\Bigg) -3y&=3\\ -3y&=\frac{9}{2}\\ y&=\boxed{-\frac{3}{2}} \end{align}$$

And there's our answer! $x = -\frac{1}{2}, y = -\frac{3}{2}.$

It's a great idea to double check your answer by plugging in your solved coordinates back into the original equation.

Here's a graphic that goes over this problem!

(Video) Solving a System of Equations Using the Elimination Method

by mathman1024

This video by mathman1024 looks at elimination with multiplication.

Summary

He looks at the following system: $$\begin{align} 3x+4y&=14\\ -x+3y&=17 \end{align}$$

The normal addition/subtraction elimination method wouldn't cancel any variables. Luckily, we have the multiplication property. This allows us to multiply both sides of an equation by the same constant while still having an equivalent expression. If you multiply by $3$ we can then cancel out the $x$ term. Here's how it works.

First, we can multiply all terms in the second equation by $3.$ $$\begin{align} {\color{#8925AE}{3}}(-x+3y)&={\color{#8925AE}{3}}(17)\\ -3x+9y&=51 \end{align}$$

Now, $-3x+9y=51$ is our new equation in the system.

$$\begin{align} \cancel{3x}+4y&=14\\ +\ \ \ \cancel{-3x}+9y&=51\\ \hline 0x+13y&=65\\ 13y&=65\\ y&=\boxed{5} \end{align}$$

So now that we have $y=5,$ we can plug it back in to either of the original equations and find $x$.

$$\begin{align} 3x+4y&=14\\ 3x+4(5)&=14\\ 3x+20&=14\\ 3x&=-6\\ x&=\boxed{-2} \end{align}$$

The solution for this system of equations is $(-2,5)$.

(Video) Solving a System of Equations Using the Elimination Method

by mathman1024

Some systems require you to multiply terms in both equations. Watch the video to learn about this method, then solve the example problem below.

Solution

We are trying to solve the linear system: $$\begin{align} 3x-2y&=2\\ 4x+3y&=31 \end{align}$$

First, notice that we can't use the addition/subtraction elimination method right away. Next notice that you can't just multiply terms in one equation by a constant to make any progress. You're going to have to multiply terms in both equations by a constant. Let's say we want to eliminate the $x$ variable first.

The $x$ coefficient in the first equation is $3,$ so we must multiply all terms in the second equation by $3.$

$$\begin{align} 4x+3y&=31\\ {\color{#8925AE}{3}}(4x+3y)&={\color{#8925AE}{3}}(31)\\ 12x+9y&=93 \end{align}$$

The $x$ coefficient in the second equation is $4,$ so we must multiply all terms in the first equation by $4.$

$$\begin{align} 3x-2y&=2\\ {\color{#8925AE}{4}}(3x-2y)&={\color{#8925AE}{4}}(2)\\ 12x-8y&=8 \end{align}$$

Now, our new linear system is: $$\begin{align} 12x-8y&=8\\ 12x+9y&=93 \end{align}$$

Now, we have two $12x$ terms, so we can use subtraction to eliminate the variable!

$$\begin{align} \cancel{12x}-8y&=8\\ \cancel{12x}+9y&=93\\ \hline 0x-17y&=-85\\ -17y&=-85\\ y&=\boxed{5} \end{align}$$

Now, we can plug $y=5$ back into either of the original equations to solve for $x.$

$$\begin{align} 3x-2y&=2\\ 3x-2(5)&=2\\ 3x-10&=2\\ 3x&=12\\ x&=\boxed{4} \end{align}$$

So our solution is $(4, 5),$ or $x=4, y=5.$