Respuesta :
[tex] \bf \begin{cases}
-7x-2y=13\\[-0.5em]
\hrulefill\\
x-2y=11\\
\boxed{x}=11+2y
\end{cases}~\hspace{5em}\stackrel{\textit{substitution \boxed{x} on the second equation}}{-7\left( \boxed{11+2y} \right)-2y=13}
\\\\\\
-77-14y-2y=13\implies -77-16y=13\implies -90=16y
\\\\\\
\cfrac{-90}{16}=y\implies \blacktriangleright -\cfrac{45}{8}=y \blacktriangleleft [/tex]
[tex] \bf ~\dotfill\\\\
\stackrel{\textit{substituting that \underline{y} in the second equation}}{x-2\left( \boxed{-\frac{45}{8}} \right)=11}\implies x+\cfrac{45}{4}=11
\\\\\\
\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{4}}{4x+45=44}\implies 4x=-1\implies \blacktriangleright x=-\cfrac{1}{4} \blacktriangleleft \\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
~\hspace{15em}\left(-\frac{1}{4}~,~-\frac{45}{8} \right) [/tex]
[tex] \bf \begin{cases}
y=7x-10\\
y=-3
\end{cases}~\hspace{1em}\stackrel{\textit{since we know y = -3, we plugged that in the first equation}}{-3=7x-10\implies 7=7x\implies \cfrac{7}{7}=x\implies \blacktriangleright 1=x \blacktriangleleft}\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
~\hspace{15em}(1,-3) [/tex]
