Math Problem Solver

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

\[x = 1, y = -2\]\[x = 4, y = 4\]

Solve \(y^{2} = 4 x\) for \(x\). \[x = \frac{y^{2}}{4}\]Substitute \(\frac{y^{2}}{4}\) for \(x\) in \(y = 2 x - 4\) and simplify. $$\begin{aligned}y &amp= 2 x - 4 \\ y &= 2 \left(\frac{y^{2}}{4}\right) - 4 \\ y^{2} - 2 y &= 8 \\y^{2} - 2 y - 8 &= 0 \\ \left(y - 4\right) \left(y + 2\right) &= 0 \\ y = -2&, y = 4\end{aligned}$$Substitute \(-2\) into \(y^{2} = 4 x\) to solve for \(x\). \[\begin{aligned}4 &= 4 x\\- 4 x &= -4\\x &= 1\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 1,\,y = -2\end{aligned}$$Substitute \(4\) into \(y^{2} = 4 x\) to solve for \(x\). \[\begin{aligned}16 &= 4 x\\- 4 x &= -16\\x &= 4\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 4,\,y = 4\end{aligned}$$