Respuesta :

[tex]\begin{cases} y = -\cfrac{1}{3}x+2\\[1em] y = x + 2 \end{cases}\qquad \implies \stackrel{y}{-\cfrac{1}{3}x+2}~~=~~\stackrel{y}{\cfrac{}{}x + 2} \\\\\\ -\cfrac{1}{3}x=x\implies -x=3x\implies 0=4x\implies \cfrac{0}{4}=x\implies \boxed{0=x} \\\\\\ \stackrel{\textit{substituting on the 2nd equation}}{y = x + 2\implies y = 0+2\implies }\boxed{y = 2} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (0~~,~~2)~\hfill[/tex]

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