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When a variable is eliminated from the equation during the equation solving process, what are the two possible solutions and what do they mean? Give an example of your own and explain what each answer would look like.

User Steven Vachon
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Answer:

What is the elimination method?

The elimination method is a technique for solving systems of linear equations. Let's walk through a couple of examples.

Example 1

We're asked to solve this system of equations:

\begin{aligned} 2y+7x &= -5\\\\ 5y-7x &= 12 \end{aligned}2y+7x5y−7x=−5=12

We notice that the first equation has a 7x7x7, x term and the second equation has a -7x−7xminus, 7, x term. These terms will cancel if we add the equations together—that is, we'll eliminate the xxx terms:

\begin{aligned} 2y+\redD{7x} &= -5 \\ +~5y\redD{-7x}&=12\\ \hline\\ 7y+0 &=7 \end{aligned}2y+7x+ 5y−7x7y+0=−5=12=7

Solving for yyy, we get:

\begin{aligned} 7y+0 &=7\\\\ 7y &=7\\\\ y &=\goldD{1} \end{aligned}7y+07yy=7=7=1

Plugging this value back into our first equation, we solve for the other variable:

\begin{aligned} 2y+7x &= -5\\\\ 2\cdot \goldD{1}+7x &= -5\\\\ 2+7x&=-5\\\\ 7x&=-7\\\\ x&=\blueD{-1} \end{aligned}2y+7x2⋅1+7x2+7x7xx=−5=−5=−5=−7=−1

The solution to the system is x=\blueD{-1}x=−1x, equals, start color #11accd, minus, 1, end color #11accd, y=\goldD{1}y=1y, equals, start color #e07d10, 1, end color #e07d10.

We can check our solution by plugging these values back into the original equations. Let's try the second equation:

\begin{aligned} 5y-7x &= 12\\\\ 5\cdot\goldD{1}-7(\blueD{-1}) &\stackrel ?= 12\\\\ 5+7 &= 12 \end{aligned}5y−7x5⋅1−7(−1)5+7=12=?12=12

Yes, the solution checks out.

If you feel uncertain why this process works, check out this intro video for an in-depth walkthrough.

Example 2

We're asked to solve this system of equations:

\begin{aligned} -9y+4x - 20&=0\\\\ -7y+16x-80&=0 \end{aligned}−9y+4x−20−7y+

User Etienne Arthur
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