[tex]\bf \begin{cases} 5x-4y=16\\ \cline{1-1} x-2y=6\\ \boxed{x}=6+2y \end{cases}~\hspace{7em}\stackrel{\textit{substituting \underline{x} in the first equation}}{5\left( \boxed{6+2y} \right)-4y=16} \\\\\\ 30+10y-4y=16\implies 30+6y=16\implies 6y=-14 \\\\\\ y=-\cfrac{14}{6}\implies y=-\cfrac{7}{3}\implies \blacktriangleright y=-2\frac{1}{3} \blacktriangleleft \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \stackrel{\textit{since we know that}}{x=6+2y}\implies x=6+2\left( -\cfrac{7}{3} \right)\implies x=6+\left(-\cfrac{14}{3} \right) \\\\\\ x=6-\cfrac{14}{3}\implies x=\cfrac{18-14}{3}\implies x=\cfrac{4}{3}\implies \blacktriangleright x=1\frac{1}{3} \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \left( 1\frac{1}{3} ~,~-2\frac{1}{3}\right)~\hfill[/tex]
