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Find the coordinates of the points of intersection of the graphs without building them: 5x–4y=16 and x–2y=6

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\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(1)/(3) \blacktriangleleft \\\\[-0.35em] ~\dotfill


\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(1)/(3) \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \left( 1(1)/(3) ~,~-2(1)/(3)\right)~\hfill

User Darko Djuric
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