Math Problem Solver

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

\[x = \sqrt{2}, y = - 3 \sqrt{2}\]\[x = - \sqrt{2}, y = 3 \sqrt{2}\]

Solve \(- 3 x^{2} + y^{2} = 12\) for \(x\). \[x = - \frac{\sqrt{3 y^{2} - 36}}{3}, x = \frac{\sqrt{3 y^{2} - 36}}{3}\]Substitute \(- \frac{\sqrt{3 y^{2} - 36}}{3}\) for \(x\) in \(y = - 3 x\) and simplify. $$\begin{aligned}y &= - 3 x \\ y &= - 3 \left(- \frac{\sqrt{3 y^{2} - 36}}{3}\right) \\ y &= \sqrt{3 y^{2} - 36} \end{aligned}$$Substitute \(3 \sqrt{2}\) into \(- 3 x^{2} + y^{2} = 12\) to solve for \(x\). $$\begin{aligned}- 3 x^{2} + \left(3 \sqrt{2}\right)^{2} &= 12 \\ 18 - 3 x^{2} &= 12 \\6 - 3 x^{2} &= 0 \\ - 3 \left(x^{2} - 2\right) &= 0 \\ x = - \sqrt{2}&, x = \sqrt{2}\end{aligned}$$This yields the following solution. $$\begin{aligned}x = - \sqrt{2},\,y = 3 \sqrt{2}\end{aligned}$$Substitute \(- \frac{\sqrt{3 y^{2} - 36}}{3}\) for \(x\) in \(y = - 3 x\) and simplify. $$\begin{aligned}y &= - 3 x \\ y &= - 3 \left(- \frac{\sqrt{3 y^{2} - 36}}{3}\right) \\ y &= \sqrt{3 y^{2} - 36} \end{aligned}$$Substitute \(\frac{\sqrt{3 y^{2} - 36}}{3}\) for \(x\) in \(y = - 3 x\) and simplify. $$\begin{aligned}y &= - 3 x \\ y &= - 3 \left(\frac{\sqrt{3 y^{2} - 36}}{3}\right) \\ y &= - \sqrt{3 y^{2} - 36} \end{aligned}$$Substitute \(- 3 \sqrt{2}\) into \(- 3 x^{2} + y^{2} = 12\) to solve for \(x\). $$\begin{aligned}- 3 x^{2} + \left(- 3 \sqrt{2}\right)^{2} &= 12 \\ 18 - 3 x^{2} &= 12 \\6 - 3 x^{2} &= 0 \\ - 3 \left(x^{2} - 2\right) &= 0 \\ x = - \sqrt{2}&, x = \sqrt{2}\end{aligned}$$This yields the following solution. $$\begin{aligned}x = \sqrt{2},\,y = - 3 \sqrt{2}\end{aligned}$$