Question

What is the solution of the system?

Use the elimination method.

2x+y=20
6x−5y=12

Enter your answer in the boxes.


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Answer 1

We have
`\begin{bmatrix}2x+y=20\\ 6x-5y=12\end{bmatrix}`


`\mathrm{Multiply\:}2x+y=20\mathrm{\:by\:}3:\ 6x+3y=60`


`\begin{bmatrix}6x+3y=60\\ 6x-5y=12\end{bmatrix}`

6x - 5y = 12
-
6x + 3y = 60
/
-8y = -48


`-8y=-48 \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-8 \ \textgreater \ (-8y)/(-8)=(-48)/(-8) \ \textgreater \ y=6`


`\mathrm{For\:}6x+3y=60\mathrm{\:plug\:in\:}\ \:y=6`


`6x+3\cdot \:6=60 \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:3\cdot \:6=18 \ \textgreater \ 6x+18=60`


`\mathrm{Subtract\:}18\mathrm{\:from\:both\:sides} \ \textgreater \ 6x+18-18=60-18 \ \textgreater \ 6x=42`


`\mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \ (6x)/(6)=(42)/(6) \ \textgreater \ x = 7`

Therefore....

`The\:solutions\:to\:the\:system\:of\:equationts\:are \ \textgreater \ y=6,\:x=7`