![\bf \begin{cases} -2x+y=0\\ \boxed{y} = 2x\\[-0.5em] \hrulefill\\ 5x+3y=-11 \end{cases}\qquad \qquad \implies \stackrel{\textit{substituting on the 2nd equation}}{5x+3\left( \boxed{2x} \right)=-11} \\\\\\ 5x+6x=-11\implies 11x=-11\implies x = \cfrac{-11}{11}\implies \blacktriangleright x = -1 \blacktriangleleft \\\\\\ \stackrel{\textit{we know that}}{y = 2x}\implies \blacktriangleright y = -2 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (-1~~,~~-2)~\hfill](/tpl/images/0471/9099/c9c2c.png)
{-2x+y=0
5x+3y=-11
solve the substitution...
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