Math Problem Solver

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

\[x = -3, y = 2\]\[x = -2, y = 6\]

Solve \(y = 4 x + 14\) for \(x\). \[x = \frac{y}{4} - \frac{7}{2}\]Substitute \(\frac{y}{4} - \frac{7}{2}\) for \(x\) in \(y = x^{2} + 9 x + 20\) and simplify. $$\begin{aligned}y &amp= x^{2} + 9 x + 20 \\ y &= \left(\frac{y}{4} - \frac{7}{2}\right)^{2} + 9 \left(\frac{y}{4} - \frac{7}{2}\right) + 20 \\ y^{2} - 8 y &= -12 \\y^{2} - 8 y + 12 &= 0 \\ \left(y - 6\right) \left(y - 2\right) &= 0 \\ y = 2&, y = 6\end{aligned}$$Substitute \(2\) into \(y = 4 x + 14\) to solve for \(x\). \[\begin{aligned}2 &= 4 x + 14\\- 4 x &= 12\\x &= -3\end{aligned}\]This yields the following solution. $$\begin{aligned}x = -3,\,y = 2\end{aligned}$$Substitute \(6\) into \(y = 4 x + 14\) to solve for \(x\). \[\begin{aligned}6 &= 4 x + 14\\- 4 x &= 8\\x &= -2\end{aligned}\]This yields the following solution. $$\begin{aligned}x = -2,\,y = 6\end{aligned}$$