Math Problem Solver

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

\[x = -2, y = 5\]\[x = 5, y = -2\]

Solve \(x + y = 3\) for \(x\). \[x = 3 - y\]Substitute \(3 - y\) for \(x\) in \(x^{2} + y^{2} = 29\) and simplify. $$\begin{aligned}x^{2} + y^{2} &= 29 \\ \left(3 - y\right)^{2} + y^{2} &= 29 \\ y^{2} - 3 y &= 10 \\y^{2} - 3 y - 10 &= 0 \\ \left(y - 5\right) \left(y + 2\right) &= 0 \\ y = -2&, y = 5\end{aligned}$$Substitute \(-2\) into \(x + y = 3\) to solve for \(x\). \[\begin{aligned}x - 2 &= 3\\x &= 5\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 5,\,y = -2\end{aligned}$$Substitute \(5\) into \(x + y = 3\) to solve for \(x\). \[\begin{aligned}x + 5 &= 3\\x &= -2\end{aligned}\]This yields the following solution. $$\begin{aligned}x = -2,\,y = 5\end{aligned}$$