Math Problem Solver

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

\[x = -2, y = -4\]\[x = 2, y = 4\]

Solve \(- x^{2} + x y = 4\) for \(x\). \[x = \frac{y}{2} - \frac{\sqrt{y^{2} - 16}}{2}, x = \frac{y}{2} + \frac{\sqrt{y^{2} - 16}}{2}\]Substitute \(\frac{y}{2} - \frac{\sqrt{y^{2} - 16}}{2}\) for \(x\) in \(4 x^{2} - 4 x y = 16 - 2 y^{2}\) and simplify. $$\begin{aligned}4 x^{2} - 4 x y &amp= 16 - 2 y^{2} \\ 4 \left(\frac{y}{2} - \frac{\sqrt{y^{2} - 16}}{2}\right)^{2} - 4 \left(\frac{y}{2} - \frac{\sqrt{y^{2} - 16}}{2}\right) y &= 16 - 2 y^{2} \\ y^{2} &= 16 \\ \sqrt{y^{2}} &= \sqrt{16} \\ y = -4&, y = 4\end{aligned}$$Substitute \(-4\) into \(- x^{2} + x y = 4\) to solve for \(x\). $$\begin{aligned}- x^{2} + x \left(-4\right) &= 4 \\ - x^{2} - 4 x &= 4 \\- x^{2} - 4 x - 4 &= 0 \\ - \left(x + 2\right)^{2} &= 0 \\ x = -2\end{aligned}$$This yields the following solution. $$\begin{aligned}x = -2,\,y = -4\end{aligned}$$Substitute \(4\) into \(- x^{2} + x y = 4\) to solve for \(x\). $$\begin{aligned}- x^{2} + x 4 &= 4 \\ - x^{2} + 4 x &= 4 \\- x^{2} + 4 x - 4 &= 0 \\ - \left(x - 2\right)^{2} &= 0 \\ x = 2\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 2,\,y = 4\end{aligned}$$