Respuesta :

gmany

[tex] \dfrac{x+y}{3}=5\ \ \ \ |\cdot3\\\\x+y=15\ \ \ |-x\qquad\ or\qquad x+y=15\ \ \ \ |-y\\\\y=15-x\qquad\ \quad or\qquad x=15-y\\\\\left\{\begin{array}{ccc}x\in\mathbb{R}\\y=15-x\end{array}\right\ or\ \left\{\begin{array}{ccc}y\in\mathbb{R}\\x=15-y\end{array}\right [/tex]

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