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Answer:
We're asked to solve this system of equations:
\begin{aligned} 2y+7x &= -5\\\\ 5y-7x &= 12 \end{aligned}
2y+7x
5y−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−7x
7y+0
=−5
=12
=7
Solving for yyy, we get:
\begin{aligned} 7y+0 &=7\\\\ 7y &=7\\\\ y &=\goldD{1} \end{aligned}
7y+0
7y
y
=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+7x
2⋅1+7x
2+7x
7x
x
=−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−7x
5⋅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+16x−80
=0
=0
We can multiply the first equation by -4−4minus, 4 to get an equivalent equation that has a \purpleD{-16x}−16xstart color #7854ab, minus, 16, x, end color #7854ab term. Our new (but equivalent!) system of equations looks like this:
\begin{aligned} 36y\purpleD{-16x}+80&=0\\\\ -7y+16x-80&=0 \end{aligned}
36y−16x+80
−7y+16x−80
=0
=0
Adding the equations to eliminate the xxx terms, we get:
\begin{aligned} 36y-\redD{16x} +80&=0 \\ {+}~-7y+\redD{16x}-80&=0\\ \hline\\ 29y+0 -0&=0 \end{aligned}
36y−16x+80
+ −7y+16x−80
29y+0−0
=0
=0
=0
Solving for yyy, we get:
\begin{aligned} 29y+0 -0&=0 \\\\ 29y&=0 \\\\ y&=\goldD 0 \end{aligned}
29y+0−0
29y
y
=0
=0
=0
Plugging this value back into our first equation, we solve for the other variable:
\begin{aligned} 36y-16x+80&=0\\\\ 36\cdot 0-16x+80&=0\\\\ -16x+80&=0\\\\ -16x&=-80\\\\ x&=\blueD{5} \end{aligned}
36y−16x+80
36⋅0−16x+80
−16x+80
−16x
x
=0
=0
=0
=−80
=5
The solution to the system is x=\blueD{5}x=5x, equals, start color #11accd, 5, end color #11accd, y=\goldD{0}y=0y, equals, start color #e07d10, 0, end color #e07d10.
Want to see another example of solving a complicated problem with the elimination method? Check out this video(Opens in a new window).
Step-by-step explanation:
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