Math Problem Solver

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

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

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