Math Problem Solver

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

\[x = -2, y = -8\]\[x = 4, y = 4\]

Solve \(x^{2} - y = 12\) for \(x\). \[x = - \sqrt{y + 12}, x = \sqrt{y + 12}\]Substitute \(- \sqrt{y + 12}\) for \(x\) in \(x y = 16\) and simplify. $$\begin{aligned}x y &amp= 16 \\ \left(- \sqrt{y + 12}\right) y &= 16 \\ y \sqrt{y + 12} &= -16 \end{aligned}$$Substitute \(-8\) into \(x^{2} - y = 12\) to solve for \(x\). $$\begin{aligned}x^{2} - -8 &= 12 \\ x^{2} + 8 &= 12 \\x^{2} - 4 &= 0 \\ \left(x - 2\right) \left(x + 2\right) &= 0 \\ x = -2&, x = 2\end{aligned}$$This yields the following solution. $$\begin{aligned}x = -2,\,y = -8\end{aligned}$$Substitute \(- \sqrt{y + 12}\) for \(x\) in \(x y = 16\) and simplify. $$\begin{aligned}x y &amp= 16 \\ \left(- \sqrt{y + 12}\right) y &= 16 \\ y \sqrt{y + 12} &= -16 \end{aligned}$$Substitute \(\sqrt{y + 12}\) for \(x\) in \(x y = 16\) and simplify. $$\begin{aligned}x y &amp= 16 \\ \left(\sqrt{y + 12}\right) y &= 16 \\ y \sqrt{y + 12} &= 16 \end{aligned}$$Substitute \(4\) into \(x^{2} - y = 12\) to solve for \(x\). $$\begin{aligned}x^{2} - 4 &= 12 \\ x^{2} - 4 &= 12 \\x^{2} - 16 &= 0 \\ \left(x - 4\right) \left(x + 4\right) &= 0 \\ x = -4&, x = 4\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 4,\,y = 4\end{aligned}$$