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Solve both by substitution method (algebra 2) and show work please

-2x-7y=22
-7x-5y=-1

User Venessa
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\bf \begin{cases} -2x-7y=22\\ -7x-5y=-1 \end{cases}\qquad \textit{let's solve the \underline{second one} for


\bf -2\left( (1-5y)/(7) \right)-7y=22\implies \cfrac{10y-2}{7}-7y=22 \\\\\\ \textit{now, let's multiply both sides by 7, to toss the denominator} \\\\\\ 7\left( \cfrac{10y-2}{7}-7y \right)=7(22)\implies 10y-2-49y=154 \\\\\\ -39y=156\implies y=\cfrac{154}{-39}\implies \boxed{y=-4} \\\\\\ \textit{now, let's use that in the \underline{first one}} \\\\\\ -2x-7(-4)=22\implies -2x+28=22\implies 6=2x\implies \cfrac{6}{2}=x \\\\\\ \boxed{3=x}
User Cornwell
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