Math Problem Solver

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

\[x = -5, y = 3\]\[x = -3, y = 5\]

Solve \(x^{2} + y^{2} = 34\) for \(x\). \[x = - \sqrt{34 - y^{2}}, x = \sqrt{34 - y^{2}}\]Substitute \(- \sqrt{34 - y^{2}}\) for \(x\) in \(y = x + 8\) and simplify. $$\begin{aligned}y &amp= x + 8 \\ y &= \left(- \sqrt{34 - y^{2}}\right) + 8 \\ y &= 8 - \sqrt{34 - y^{2}} \end{aligned}$$Substitute \(3\) into \(x^{2} + y^{2} = 34\) to solve for \(x\). $$\begin{aligned}x^{2} + 3^{2} &= 34 \\ x^{2} + 9 &= 34 \\x^{2} - 25 &= 0 \\ \left(x - 5\right) \left(x + 5\right) &= 0 \\ x = -5&, x = 5\end{aligned}$$This yields the following solution. $$\begin{aligned}x = -5,\,y = 3\end{aligned}$$Substitute \(5\) into \(x^{2} + y^{2} = 34\) to solve for \(x\). $$\begin{aligned}x^{2} + 5^{2} &= 34 \\ x^{2} + 25 &= 34 \\x^{2} - 9 &= 0 \\ \left(x - 3\right) \left(x + 3\right) &= 0 \\ x = -3&, x = 3\end{aligned}$$This yields the following solution. $$\begin{aligned}x = -3,\,y = 5\end{aligned}$$