Math Problem Solver

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

\[x = 34 - 2 \sqrt{285}, y = 4 - \frac{\sqrt{285}}{3}\]\[x = 2 \sqrt{285} + 34, y = 4 + \frac{\sqrt{285}}{3}\]

Solve \(x - 6 y = 10\) for \(x\). \[x = 6 y + 10\]Substitute \(6 y + 10\) for \(x\) in \(3 y^{2} = 4 x + 7\) and simplify. $$\begin{aligned}3 y^{2} &amp= 4 x + 7 \\ 3 y^{2} &= 4 \left(6 y + 10\right) + 7 \\ y^{2} - 8 y &= \frac{47}{3} \\y^{2} - 8 y - \frac{47}{3} &= 0 \\ \frac{3 y^{2} - 24 y - 47}{3} &= 0 \\ y = 4 - \frac{\sqrt{285}}{3}&, y = 4 + \frac{\sqrt{285}}{3}\end{aligned}$$Substitute \(4 - \frac{\sqrt{285}}{3}\) into \(x - 6 y = 10\) to solve for \(x\). \[\begin{aligned}x - 24 + 2 \cdot \sqrt{285} &= 10\\x - \left(24 - 2 \sqrt{285}\right) &= 10\\x &= 34 - 2 \sqrt{285}\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 34 - 2 \sqrt{285},\,y = 4 - \frac{\sqrt{285}}{3}\end{aligned}$$Substitute \(4 + \frac{\sqrt{285}}{3}\) into \(x - 6 y = 10\) to solve for \(x\). \[\begin{aligned}x - 2 \cdot \sqrt{285} - 24 &= 10\\x - \left(24 + 2 \sqrt{285}\right) &= 10\\x &= 2 \sqrt{285} + 34\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 2 \sqrt{285} + 34,\,y = 4 + \frac{\sqrt{285}}{3}\end{aligned}$$