[tex]\bf \begin{cases}y-2x=3\\3x-2y=5\end{cases}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\\stackrel{\textit{we solve the first equation for \underline{y}}}{\boxed{y}=3+2x}~\hspace{3em}\stackrel{\textit{we \underline{substitute} the found \boxed{y} in the second equation}}{3x-2\left( \boxed{3+2x} \right)=5}[/tex]
[tex]\bf 3x-6-4x=5\implies -x=11\implies x=\cfrac{11}{-1}\implies \blacktriangleright x=-11 \blacktriangleleft\\\\\\\stackrel{\textit{and \underline{substituting} the found \underline{x} in the 1st equation}}{y-2(-11)=3\implies y+22=3}\implies \blacktriangleright y=-19 \blacktriangleleft \\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill (-11,-19)~\hfill[/tex]
