Math Problem Solver

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

\[x = 1, y = 0\]\[x = 2, y = - \frac{1}{2}\]

Solve \(x^{2} - x - 9 y = 2 y^{2} - 12 y\) for \(x\). \[x = \frac{1}{2} - \frac{\sqrt{8 y^{2} - 12 y + 1}}{2}, x = \frac{\sqrt{8 y^{2} - 12 y + 1}}{2} + \frac{1}{2}\]Substitute \(\frac{1}{2} - \frac{\sqrt{8 y^{2} - 12 y + 1}}{2}\) for \(x\) in \(x + 2 y - 1 = 0\) and simplify. $$\begin{aligned}x + 2 y - 1 &amp= 0 \\ \left(\frac{1}{2} - \frac{\sqrt{8 y^{2} - 12 y + 1}}{2}\right) + 2 y - 1 &= 0 \\ - 2 y + \frac{\sqrt{8 y^{2} - 12 y + 1}}{2} + \frac{1}{2} &= 0 \end{aligned}$$Substitute \(\frac{1}{2} - \frac{\sqrt{8 y^{2} - 12 y + 1}}{2}\) for \(x\) in \(x + 2 y - 1 = 0\) and simplify. $$\begin{aligned}x + 2 y - 1 &amp= 0 \\ \left(\frac{1}{2} - \frac{\sqrt{8 y^{2} - 12 y + 1}}{2}\right) + 2 y - 1 &= 0 \\ - 2 y + \frac{\sqrt{8 y^{2} - 12 y + 1}}{2} + \frac{1}{2} &= 0 \end{aligned}$$Substitute \(\frac{\sqrt{8 y^{2} - 12 y + 1}}{2} + \frac{1}{2}\) for \(x\) in \(x + 2 y - 1 = 0\) and simplify. $$\begin{aligned}x + 2 y - 1 &amp= 0 \\ \left(\frac{\sqrt{8 y^{2} - 12 y + 1}}{2} + \frac{1}{2}\right) + 2 y - 1 &= 0 \\ 2 y + \frac{\sqrt{8 y^{2} - 12 y + 1}}{2} - \frac{1}{2} &= 0 \end{aligned}$$Substitute \(- \frac{1}{2}\) into \(x^{2} - x - 9 y = 2 y^{2} - 12 y\) to solve for \(x\). $$\begin{aligned}x^{2} - x - - \frac{9}{2} &= 2 \left(- \frac{1}{2}\right)^{2} - -6 \\ x^{2} - x + \frac{9}{2} &= \frac{13}{2} \\x^{2} - x - 2 &= 0 \\ \left(x - 2\right) \left(x + 1\right) &= 0 \\ x = -1&, x = 2\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 2,\,y = - \frac{1}{2}\end{aligned}$$Substitute \(0\) into \(x^{2} - x - 9 y = 2 y^{2} - 12 y\) to solve for \(x\). $$\begin{aligned}x^{2} - x - 0 &= \left(-12\right) 0 + 2 \cdot 0^{2} \\ x^{2} - x &= 0 \\x^{2} - x &= 0 \\ x \left(x - 1\right) &= 0 \\ x = 0&, x = 1\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 1,\,y = 0\end{aligned}$$