Respuesta :
let's do so using substitution
[tex]\bf \begin{cases} -3x-4y=-14\\ 9x+2y=22 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{solving for "y" in the 1st equation}~\hfill }{-3x-4y=-14\implies -3x+14-4y=0} \\\\\\ -3x+14=4y\implies \cfrac{14-3x}{4}=\boxed{y} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{substituting "y" in the 2nd equation}}{9x+2\left( \boxed{\cfrac{14-3x}{4}} \right)=22}\implies 9x+\cfrac{14-3x}{2}=22[/tex]
[tex]\bf \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2\left( 9x+\cfrac{14-3x}{2}\right)=2(22)}\implies 18x+(14-3x)=44\implies 15x+14=44 \\\\\\ 15x = 30\implies x = \cfrac{30}{15}\implies \blacktriangleright x = 2\blacktriangleleft \\\\\\ \stackrel{\textit{since we know that }}{\cfrac{14-3x}{4}=y}\implies \cfrac{14-3(2)}{4}=y\implies \cfrac{14-6}{4}=y\implies \blacktriangleright 2 = y \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (2,2)~\hfill[/tex]
