Math Problem Solver

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

\[x = - \frac{11}{7}, y = \frac{2}{7}\]\[x = 1, y = -1\]

Solve \(x^{2} - x y + y^{2} = 3\) for \(x\). \[x = \frac{y}{2} - \frac{\sqrt{12 - 3 y^{2}}}{2}, x = \frac{y}{2} + \frac{\sqrt{12 - 3 y^{2}}}{2}\]Substitute \(\frac{y}{2} - \frac{\sqrt{12 - 3 y^{2}}}{2}\) for \(x\) in \(x + 2 y + 1 = 0\) and simplify. $$\begin{aligned}x + 2 y + 1 &amp= 0 \\ \left(\frac{y}{2} - \frac{\sqrt{12 - 3 y^{2}}}{2}\right) + 2 y + 1 &= 0 \\ \frac{5 y}{2} - \frac{\sqrt{12 - 3 y^{2}}}{2} + 1 &= 0 \end{aligned}$$Substitute \(\frac{2}{7}\) into \(x^{2} - x y + y^{2} = 3\) to solve for \(x\). $$\begin{aligned}x^{2} - x \frac{2}{7} + \left(\frac{2}{7}\right)^{2} &= 3 \\ x^{2} - \frac{2 x}{7} + \frac{4}{49} &= 3 \\x^{2} - \frac{2 x}{7} - \frac{143}{49} &= 0 \\ \frac{\left(7 x - 13\right) \left(7 x + 11\right)}{49} &= 0 \\ x = - \frac{11}{7}&, x = \frac{13}{7}\end{aligned}$$This yields the following solution. $$\begin{aligned}x = - \frac{11}{7},\,y = \frac{2}{7}\end{aligned}$$Substitute \(\frac{y}{2} - \frac{\sqrt{12 - 3 y^{2}}}{2}\) for \(x\) in \(x + 2 y + 1 = 0\) and simplify. $$\begin{aligned}x + 2 y + 1 &amp= 0 \\ \left(\frac{y}{2} - \frac{\sqrt{12 - 3 y^{2}}}{2}\right) + 2 y + 1 &= 0 \\ \frac{5 y}{2} - \frac{\sqrt{12 - 3 y^{2}}}{2} + 1 &= 0 \end{aligned}$$Substitute \(\frac{y}{2} + \frac{\sqrt{12 - 3 y^{2}}}{2}\) for \(x\) in \(x + 2 y + 1 = 0\) and simplify. $$\begin{aligned}x + 2 y + 1 &amp= 0 \\ \left(\frac{y}{2} + \frac{\sqrt{12 - 3 y^{2}}}{2}\right) + 2 y + 1 &= 0 \\ \frac{5 y}{2} + \frac{\sqrt{12 - 3 y^{2}}}{2} + 1 &= 0 \end{aligned}$$Substitute \(-1\) into \(x^{2} - x y + y^{2} = 3\) to solve for \(x\). $$\begin{aligned}x^{2} - x \left(-1\right) + \left(-1\right)^{2} &= 3 \\ x^{2} + x + 1 &= 3 \\x^{2} + x - 2 &= 0 \\ \left(x - 1\right) \left(x + 2\right) &= 0 \\ x = -2&, x = 1\end{aligned}$$This yields the following solution. $$\begin{aligned}x = 1,\,y = -1\end{aligned}$$