Math Problem Solver

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

\[x = -5, y = 8\]\[x = 2, y = 1\]

Solve \(x + y = 3\) for \(x\). \[x = 3 - y\]Substitute \(3 - y\) for \(x\) in \(x^{2} - 3 y = 1\) and simplify. $$\begin{aligned}x^{2} - 3 y &= 1 \\ \left(3 - y\right)^{2} - 3 y &= 1 \\ y^{2} - 9 y &= -8 \\y^{2} - 9 y + 8 &= 0 \\ \left(y - 8\right) \left(y - 1\right) &= 0 \\ y = 1&, y = 8\end{aligned}$$Substitute \(1\) into \(x + y = 3\) to solve for \(x\). \[\begin{aligned}x + 1 &= 3\\x &= 2\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 2,\,y = 1\end{aligned}$$Substitute \(8\) into \(x + y = 3\) to solve for \(x\). \[\begin{aligned}x + 8 &= 3\\x &= -5\end{aligned}\]This yields the following solution. $$\begin{aligned}x = -5,\,y = 8\end{aligned}$$