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Helppppppp me plsss fast due at 7

Helppppppp me plsss fast due at 7-example-1
User Morandg
by
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1 Answer

7 votes

Answer: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

Explanation:

User Yzorg
by
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