Math Problem Solver

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

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

Solve \(\left(x + 1\right)^{2} + \left(y + 1\right)^{2} = 5\) for \(x\). \[x = - \sqrt{- y^{2} - 2 y + 4} - 1, x = \sqrt{- y^{2} - 2 y + 4} - 1\]Substitute \(- \sqrt{- y^{2} - 2 y + 4} - 1\) for \(x\) in \(y^{2} + \left(x + 2\right)^{2} = 5\) and simplify. $$\begin{aligned}y^{2} + \left(x + 2\right)^{2} &= 5 \\ y^{2} + \left(\left(- \sqrt{- y^{2} - 2 y + 4} - 1\right) + 2\right)^{2} &= 5 \\ y^{2} + \left(\sqrt{- y^{2} - 2 y + 4} - 1\right)^{2} &= 5 \end{aligned}$$Substitute \(-2\) into \(\left(x + 1\right)^{2} + \left(y + 1\right)^{2} = 5\) to solve for \(x\). $$\begin{aligned}\left(x + 1\right)^{2} + \left(-2 + 1\right)^{2} &= 5 \\ \left(x + 1\right)^{2} + 1 &= 5 \\\left(x + 1\right)^{2} - 4 &= 0 \\ \left(x - 1\right) \left(x + 3\right) &= 0 \\ x = -3&, x = 1\end{aligned}$$This yields the following solution. $$\begin{aligned}x = -3,\,y = -2\end{aligned}$$Substitute \(- \sqrt{- y^{2} - 2 y + 4} - 1\) for \(x\) in \(y^{2} + \left(x + 2\right)^{2} = 5\) and simplify. $$\begin{aligned}y^{2} + \left(x + 2\right)^{2} &= 5 \\ y^{2} + \left(\left(- \sqrt{- y^{2} - 2 y + 4} - 1\right) + 2\right)^{2} &= 5 \\ y^{2} + \left(\sqrt{- y^{2} - 2 y + 4} - 1\right)^{2} &= 5 \end{aligned}$$Substitute \(\sqrt{- y^{2} - 2 y + 4} - 1\) for \(x\) in \(y^{2} + \left(x + 2\right)^{2} = 5\) and simplify. $$\begin{aligned}y^{2} + \left(x + 2\right)^{2} &= 5 \\ y^{2} + \left(\left(\sqrt{- y^{2} - 2 y + 4} - 1\right) + 2\right)^{2} &= 5 \\ y^{2} + \left(\sqrt{- y^{2} - 2 y + 4} + 1\right)^{2} &= 5 \end{aligned}$$Substitute \(1\) into \(\left(x + 1\right)^{2} + \left(y + 1\right)^{2} = 5\) to solve for \(x\). $$\begin{aligned}\left(x + 1\right)^{2} + \left(1 + 1\right)^{2} &= 5 \\ \left(x + 1\right)^{2} + 4 &= 5 \\\left(x + 1\right)^{2} - 1 &= 0 \\ x \left(x + 2\right) &= 0 \\ x = -2&, x = 0\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 0,\,y = 1\end{aligned}$$