Math Problem Solver

The system of equations has \(4\) solutions.

\[x = -4, y = -3\]\[x = -4, y = 3\]\[x = 4, y = -3\]\[x = 4, y = 3\]

Solve \(x^{2} + y^{2} = 25\) for \(x\). \[x = - \sqrt{25 - y^{2}}, x = \sqrt{25 - y^{2}}\]Substitute \(- \sqrt{25 - y^{2}}\) for \(x\) in \(x^{2} - y^{2} = 7\) and simplify. $$\begin{aligned}x^{2} - y^{2} &amp= 7 \\ \left(- \sqrt{25 - y^{2}}\right)^{2} - y^{2} &= 7 \\ y^{2} &= 9 \\ \sqrt{y^{2}} &= \sqrt{9} \\ y = -3&, y = 3\end{aligned}$$Substitute \(-3\) into \(x^{2} + y^{2} = 25\) to solve for \(x\). $$\begin{aligned}x^{2} + \left(-3\right)^{2} &= 25 \\ x^{2} + 9 &= 25 \\x^{2} - 16 &= 0 \\ \left(x - 4\right) \left(x + 4\right) &= 0 \\ x = -4&, x = 4\end{aligned}$$This yields the following solutions. $$\begin{aligned}x = -4,\,y = -3\\ x = 4,\,y = -3\end{aligned}$$Substitute \(3\) into \(x^{2} + y^{2} = 25\) to solve for \(x\). $$\begin{aligned}x^{2} + 3^{2} &= 25 \\ x^{2} + 9 &= 25 \\x^{2} - 16 &= 0 \\ \left(x - 4\right) \left(x + 4\right) &= 0 \\ x = -4&, x = 4\end{aligned}$$This yields the following solutions. $$\begin{aligned}x = -4,\,y = 3\\ x = 4,\,y = 3\end{aligned}$$