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
Step-by-step explanation:
