Math Problem Solver

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

\[x = -6, y = -2\]\[x = 2, y = 6\]

Solve \(x^{2} + y^{2} = 40\) for \(x\). \[x = - \sqrt{40 - y^{2}}, x = \sqrt{40 - y^{2}}\]Substitute \(- \sqrt{40 - y^{2}}\) for \(x\) in \(y = x + 4\) and simplify. $$\begin{aligned}y &= x + 4 \\ y &= \left(- \sqrt{40 - y^{2}}\right) + 4 \\ y &= 4 - \sqrt{40 - y^{2}} \end{aligned}$$Substitute \(-2\) into \(x^{2} + y^{2} = 40\) to solve for \(x\). $$\begin{aligned}x^{2} + \left(-2\right)^{2} &= 40 \\ x^{2} + 4 &= 40 \\x^{2} - 36 &= 0 \\ \left(x - 6\right) \left(x + 6\right) &= 0 \\ x = -6&, x = 6\end{aligned}$$This yields the following solution. $$\begin{aligned}x = -6,\,y = -2\end{aligned}$$Substitute \(- \sqrt{40 - y^{2}}\) for \(x\) in \(y = x + 4\) and simplify. $$\begin{aligned}y &= x + 4 \\ y &= \left(- \sqrt{40 - y^{2}}\right) + 4 \\ y &= 4 - \sqrt{40 - y^{2}} \end{aligned}$$Substitute \(\sqrt{40 - y^{2}}\) for \(x\) in \(y = x + 4\) and simplify. $$\begin{aligned}y &= x + 4 \\ y &= \left(\sqrt{40 - y^{2}}\right) + 4 \\ y &= \sqrt{40 - y^{2}} + 4 \end{aligned}$$Substitute \(6\) into \(x^{2} + y^{2} = 40\) to solve for \(x\). $$\begin{aligned}x^{2} + 6^{2} &= 40 \\ x^{2} + 36 &= 40 \\x^{2} - 4 &= 0 \\ \left(x - 2\right) \left(x + 2\right) &= 0 \\ x = -2&, x = 2\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 2,\,y = 6\end{aligned}$$