Math Problem Solver

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

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

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