Math Problem Solver

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

\[x = \frac{13}{4}, y = - \frac{7}{2}\]\[x = 6, y = 2\]

Solve \(3 x + y^{2} = 22\) for \(x\). \[x = \frac{22}{3} - \frac{y^{2}}{3}\]Substitute \(\frac{22}{3} - \frac{y^{2}}{3}\) for \(x\) in \(2 x - y = 10\) and simplify. $$\begin{aligned}2 x - y &amp= 10 \\ 2 \left(\frac{22}{3} - \frac{y^{2}}{3}\right) - y &= 10 \\ 2 y^{2} + 3 y &= 14 \\2 y^{2} + 3 y - 14 &= 0 \\ \left(y - 2\right) \left(2 y + 7\right) &= 0 \\ y = - \frac{7}{2}&, y = 2\end{aligned}$$Substitute \(- \frac{7}{2}\) into \(3 x + y^{2} = 22\) to solve for \(x\). \[\begin{aligned}3 x + \frac{49}{4} &= 22\\3 x &= \frac{39}{4}\\x &= \frac{13}{4}\end{aligned}\]This yields the following solution. $$\begin{aligned}x = \frac{13}{4},\,y = - \frac{7}{2}\end{aligned}$$Substitute \(2\) into \(3 x + y^{2} = 22\) to solve for \(x\). \[\begin{aligned}3 x + 4 &= 22\\3 x &= 18\\x &= 6\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 6,\,y = 2\end{aligned}$$