Respuesta :

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[tex]\displaystyle\\ \left \{ {{1x-2y=-60~~~\Big| \cdot3} \atop {5x+6y=20~~~~~~~~~~}} \right\\\\\\ \left \{ {{3x-6y=-180} \atop {5x+6y=20~~~~}} \right\\ \text{- - - - - - - - - - - - - -}\\\\ 8x ~~~~~~/~~~=-160\\\\ 8x=-160\\\\ x= \frac{-160}{8} =\boxed{-20} \\\\ 5x+6y=20\\\\ 6y = 20 - 5x\\\\ y = \frac{20 - 5x}{6} = \frac{20 - 5\cdot (-20)}{6}=\frac{20+ 100}{6}=\frac{120}{6}=\boxed{20}[/tex]