Math Problem Solver

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

\[x = -2, y = -3\]\[x = 0, y = 1\]

Solve \(x^{2} + 6 x + y^{2} = 1\) for \(x\). \[x = - \sqrt{10 - y^{2}} - 3, x = \sqrt{10 - y^{2}} - 3\]Substitute \(- \sqrt{10 - y^{2}} - 3\) for \(x\) in \(x^{2} + 2 x + y^{2} + 2 y = 3\) and simplify. $$\begin{aligned}x^{2} + 2 x + y^{2} + 2 y &= 3 \\ \left(- \sqrt{10 - y^{2}} - 3\right)^{2} + 2 \left(- \sqrt{10 - y^{2}} - 3\right) + y^{2} + 2 y &= 3 \\ 2 y + 4 \sqrt{10 - y^{2}} + 13 &= 3 \end{aligned}$$Substitute \(- \sqrt{10 - y^{2}} - 3\) for \(x\) in \(x^{2} + 2 x + y^{2} + 2 y = 3\) and simplify. $$\begin{aligned}x^{2} + 2 x + y^{2} + 2 y &= 3 \\ \left(- \sqrt{10 - y^{2}} - 3\right)^{2} + 2 \left(- \sqrt{10 - y^{2}} - 3\right) + y^{2} + 2 y &= 3 \\ 2 y + 4 \sqrt{10 - y^{2}} + 13 &= 3 \end{aligned}$$Substitute \(\sqrt{10 - y^{2}} - 3\) for \(x\) in \(x^{2} + 2 x + y^{2} + 2 y = 3\) and simplify. $$\begin{aligned}x^{2} + 2 x + y^{2} + 2 y &= 3 \\ \left(\sqrt{10 - y^{2}} - 3\right)^{2} + 2 \left(\sqrt{10 - y^{2}} - 3\right) + y^{2} + 2 y &= 3 \\ 2 y - 4 \sqrt{10 - y^{2}} + 13 &= 3 \end{aligned}$$Substitute \(-3\) into \(x^{2} + 6 x + y^{2} = 1\) to solve for \(x\). $$\begin{aligned}x^{2} + 6 x + \left(-3\right)^{2} &= 1 \\ x^{2} + 6 x + 9 &= 1 \\x^{2} + 6 x + 8 &= 0 \\ \left(x + 2\right) \left(x + 4\right) &= 0 \\ x = -4&, x = -2\end{aligned}$$This yields the following solution. $$\begin{aligned}x = -2,\,y = -3\end{aligned}$$Substitute \(1\) into \(x^{2} + 6 x + y^{2} = 1\) to solve for \(x\). $$\begin{aligned}x^{2} + 6 x + 1^{2} &= 1 \\ x^{2} + 6 x + 1 &= 1 \\x^{2} + 6 x &= 0 \\ x \left(x + 6\right) &= 0 \\ x = -6&, x = 0\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 0,\,y = 1\end{aligned}$$