Math Problem Solver

The system of equations has \(2\) solutions.

\[x = 2, y = 10\]\[x = 15, y = -3\]

Solve \(x + y = 12\) for \(x\). \[x = 12 - y\]Substitute \(12 - y\) for \(x\) in \(\left(x + 3\right) \left(y - 2\right) = 2 x y\) and simplify. $$\begin{aligned}\left(x + 3\right) \left(y - 2\right) &= 2 x y \\ \left(\left(12 - y\right) + 3\right) \left(y - 2\right) &= 2 \left(12 - y\right) y \\ y^{2} - 7 y &= 30 \\y^{2} - 7 y - 30 &= 0 \\ \left(y - 10\right) \left(y + 3\right) &= 0 \\ y = -3&, y = 10\end{aligned}$$Substitute \(-3\) into \(x + y = 12\) to solve for \(x\). \[\begin{aligned}x - 3 &= 12\\x &= 15\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 15,\,y = -3\end{aligned}$$Substitute \(10\) into \(x + y = 12\) to solve for \(x\). \[\begin{aligned}x + 10 &= 12\\x &= 2\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 2,\,y = 10\end{aligned}$$