Math Problem Solver

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

\[x = -1, y = -3\]\[x = 3, y = 5\]

Solve \(y = 2 x - 1\) for \(x\). \[x = \frac{y}{2} + \frac{1}{2}\]Substitute \(\frac{y}{2} + \frac{1}{2}\) for \(x\) in \(y^{2} = 4 x + 13\) and simplify. $$\begin{aligned}y^{2} &= 4 x + 13 \\ y^{2} &= 4 \left(\frac{y}{2} + \frac{1}{2}\right) + 13 \\ y^{2} - 2 y &= 15 \\y^{2} - 2 y - 15 &= 0 \\ \left(y - 5\right) \left(y + 3\right) &= 0 \\ y = -3&, y = 5\end{aligned}$$Substitute \(-3\) into \(y = 2 x - 1\) to solve for \(x\). \[\begin{aligned}-3 &= 2 x - 1\\- 2 x &= 2\\x &= -1\end{aligned}\]This yields the following solution. $$\begin{aligned}x = -1,\,y = -3\end{aligned}$$Substitute \(5\) into \(y = 2 x - 1\) to solve for \(x\). \[\begin{aligned}5 &= 2 x - 1\\- 2 x &= -6\\x &= 3\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 3,\,y = 5\end{aligned}$$