Math Problem Solver

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

\[x = 3 + \sqrt{13}, y = -3 + \sqrt{13}\]\[x = 3 - \sqrt{13}, y = - \sqrt{13} - 3\]

Solve \(x - y = 6\) for \(x\). \[x = y + 6\]Substitute \(y + 6\) for \(x\) in \(x y = 4\) and simplify. $$\begin{aligned}x y &= 4 \\ \left(y + 6\right) y &= 4 \\ y^{2} + 6 y &= 4 \\y^{2} + 6 y - 4 &= 0 \\ y &= \frac{-(6) \pm \sqrt{(6)^{2} - 4(1)(-4)}}{2(6)} \\ y = -3 + \sqrt{13}&, y = - \sqrt{13} - 3\end{aligned}$$Substitute \(-3 + \sqrt{13}\) into \(x - y = 6\) to solve for \(x\). \[\begin{aligned}x - \sqrt{13} + 3 &= 6\\x + \left(3 - \sqrt{13}\right) &= 6\\x &= 3 + \sqrt{13}\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 3 + \sqrt{13},\,y = -3 + \sqrt{13}\end{aligned}$$Substitute \(- \sqrt{13} - 3\) into \(x - y = 6\) to solve for \(x\). \[\begin{aligned}x + 3 + \sqrt{13} &= 6\\x + \left(3 + \sqrt{13}\right) &= 6\\x &= 3 - \sqrt{13}\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 3 - \sqrt{13},\,y = - \sqrt{13} - 3\end{aligned}$$