Math Problem Solver

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

\[x = 3, y = 7\]\[x = 10, y = -7\]

Solve \(2 x + y = 13\) for \(x\). \[x = \frac{13}{2} - \frac{y}{2}\]Substitute \(\frac{13}{2} - \frac{y}{2}\) for \(x\) in \(x^{2} + x y - 30 = 0\) and simplify. $$\begin{aligned}x^{2} + x y - 30 &= 0 \\ \left(\frac{13}{2} - \frac{y}{2}\right)^{2} + \left(\frac{13}{2} - \frac{y}{2}\right) y - 30 &= 0 \\ y^{2} &= 49 \\ \sqrt{y^{2}} &= \sqrt{49} \\ y = -7&, y = 7\end{aligned}$$Substitute \(-7\) into \(2 x + y = 13\) to solve for \(x\). \[\begin{aligned}2 x - 7 &= 13\\2 x &= 20\\x &= 10\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 10,\,y = -7\end{aligned}$$Substitute \(7\) into \(2 x + y = 13\) to solve for \(x\). \[\begin{aligned}2 x + 7 &= 13\\2 x &= 6\\x &= 3\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 3,\,y = 7\end{aligned}$$