Math Problem Solver

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

\[x = - \frac{1}{2}, y = \frac{3}{2}\]\[x = 4, y = 33\]

Solve \(y = 2 x^{2} + 1\) for \(x\). \[x = - \frac{\sqrt{2 y - 2}}{2}, x = \frac{\sqrt{2 y - 2}}{2}\]Substitute \(- \frac{\sqrt{2 y - 2}}{2}\) for \(x\) in \(y = 7 x + 5\) and simplify. $$\begin{aligned}y &amp= 7 x + 5 \\ y &= 7 \left(- \frac{\sqrt{2 y - 2}}{2}\right) + 5 \\ y &= 5 - \frac{7 \sqrt{2 y - 2}}{2} \end{aligned}$$Substitute \(\frac{3}{2}\) into \(y = 2 x^{2} + 1\) to solve for \(x\). $$\begin{aligned}\frac{3}{2} &= 2 x^{2} + 1 \\\frac{1}{2} - 2 x^{2} &= 0 \\ - \frac{\left(2 x - 1\right) \left(2 x + 1\right)}{2} &= 0 \\ x = - \frac{1}{2}&, x = \frac{1}{2}\end{aligned}$$This yields the following solution. $$\begin{aligned}x = - \frac{1}{2},\,y = \frac{3}{2}\end{aligned}$$Substitute \(- \frac{\sqrt{2 y - 2}}{2}\) for \(x\) in \(y = 7 x + 5\) and simplify. $$\begin{aligned}y &amp= 7 x + 5 \\ y &= 7 \left(- \frac{\sqrt{2 y - 2}}{2}\right) + 5 \\ y &= 5 - \frac{7 \sqrt{2 y - 2}}{2} \end{aligned}$$Substitute \(\frac{\sqrt{2 y - 2}}{2}\) for \(x\) in \(y = 7 x + 5\) and simplify. $$\begin{aligned}y &amp= 7 x + 5 \\ y &= 7 \left(\frac{\sqrt{2 y - 2}}{2}\right) + 5 \\ y &= \frac{7 \sqrt{2 y - 2}}{2} + 5 \end{aligned}$$Substitute \(33\) into \(y = 2 x^{2} + 1\) to solve for \(x\). $$\begin{aligned}33 &= 2 x^{2} + 1 \\32 - 2 x^{2} &= 0 \\ - 2 \left(x - 4\right) \left(x + 4\right) &= 0 \\ x = -4&, x = 4\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 4,\,y = 33\end{aligned}$$