[tex]\bf \begin{cases}
-2x-7y=22\\
-7x-5y=-1
\end{cases}\qquad \textit{let's solve the \underline{second one} for "x"}
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-7x-5y=-1\implies -7x=5y-1\implies x=\cfrac{5y-1}{-7}
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\boxed{x=\cfrac{1-5y}{7}}\impliedby \textit{now, let's substitute it in the \underline{first one}}
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[tex]\bf -2\left( \frac{1-5y}{7} \right)-7y=22\implies \cfrac{10y-2}{7}-7y=22
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\textit{now, let's multiply both sides by 7, to toss the denominator}
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7\left( \cfrac{10y-2}{7}-7y \right)=7(22)\implies 10y-2-49y=154
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-39y=156\implies y=\cfrac{154}{-39}\implies \boxed{y=-4}
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\textit{now, let's use that in the \underline{first one}}
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-2x-7(-4)=22\implies -2x+28=22\implies 6=2x\implies \cfrac{6}{2}=x
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\boxed{3=x}[/tex]
