Math Problem Solver

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

\[f = \sqrt{6} + 5, x = -2 + \sqrt{6}\]\[f = 5 - \sqrt{6}, x = - \sqrt{6} - 2\]

Solve \(f - x = 7\) for \(f\). \[f = x + 7\]Substitute \(x + 7\) for \(f\) in \(f x = 3 x + 2\) and simplify. $$\begin{aligned}f x &amp= 3 x + 2 \\ \left(x + 7\right) x &= 3 x + 2 \\ x^{2} + 4 x &= 2 \\x^{2} + 4 x - 2 &= 0 \\ x &= \frac{-(4) \pm \sqrt{(4)^{2} - 4(1)(-2)}}{2(4)} \\ x = -2 + \sqrt{6}&, x = - \sqrt{6} - 2\end{aligned}$$Substitute \(-2 + \sqrt{6}\) into \(f - x = 7\) to solve for \(f\). \[\begin{aligned}f - \sqrt{6} + 2 &= 7\\f + \left(2 - \sqrt{6}\right) &= 7\\f &= \sqrt{6} + 5\end{aligned}\]This yields the following solution. $$\begin{aligned}f = \sqrt{6} + 5,\,x = -2 + \sqrt{6}\end{aligned}$$Substitute \(- \sqrt{6} - 2\) into \(f - x = 7\) to solve for \(f\). \[\begin{aligned}f + 2 + \sqrt{6} &= 7\\f + \left(2 + \sqrt{6}\right) &= 7\\f &= 5 - \sqrt{6}\end{aligned}\]This yields the following solution. $$\begin{aligned}f = 5 - \sqrt{6},\,x = - \sqrt{6} - 2\end{aligned}$$