Math Problem Solver

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

\[x = -2, y = 1\]\[x = 3, y = -1\]

Solve \(2 x + 5 y = 1\) for \(x\). \[x = \frac{1}{2} - \frac{5 y}{2}\]Substitute \(\frac{1}{2} - \frac{5 y}{2}\) for \(x\) in \(x^{2} + 5 x y - 4 y^{2} + 10 = 0\) and simplify. $$\begin{aligned}x^{2} + 5 x y - 4 y^{2} + 10 &= 0 \\ \left(\frac{1}{2} - \frac{5 y}{2}\right)^{2} + 5 \left(\frac{1}{2} - \frac{5 y}{2}\right) y - 4 y^{2} + 10 &= 0 \\ y^{2} &= 1 \\ \sqrt{y^{2}} &= \sqrt{1} \\ y = -1&, y = 1\end{aligned}$$Substitute \(-1\) into \(2 x + 5 y = 1\) to solve for \(x\). \[\begin{aligned}2 x - 5 &= 1\\2 x &= 6\\x &= 3\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 3,\,y = -1\end{aligned}$$Substitute \(1\) into \(2 x + 5 y = 1\) to solve for \(x\). \[\begin{aligned}2 x + 5 &= 1\\2 x &= -4\\x &= -2\end{aligned}\]This yields the following solution. $$\begin{aligned}x = -2,\,y = 1\end{aligned}$$