Math Problem Solver

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

\[a = 4 - \sqrt{10}, b = \sqrt{10} + 4\]\[a = \sqrt{10} + 4, b = 4 - \sqrt{10}\]

Solve \(a + b = 8\) for \(a\). \[a = 8 - b\]Substitute \(8 - b\) for \(a\) in \(a b = 6\) and simplify. $$\begin{aligned}a b &= 6 \\ \left(8 - b\right) b &= 6 \\ b^{2} - 8 b &= -6 \\b^{2} - 8 b + 6 &= 0 \\ b &= \frac{-(-8) \pm \sqrt{(-8)^{2} - 4(1)(6)}}{2(-8)} \\ b = 4 - \sqrt{10}&, b = \sqrt{10} + 4\end{aligned}$$Substitute \(4 - \sqrt{10}\) into \(a + b = 8\) to solve for \(a\). \[\begin{aligned}a - \sqrt{10} + 4 &= 8\\a + \left(4 - \sqrt{10}\right) &= 8\\a &= \sqrt{10} + 4\end{aligned}\]This yields the following solution. $$\begin{aligned}a = \sqrt{10} + 4,\,b = 4 - \sqrt{10}\end{aligned}$$Substitute \(\sqrt{10} + 4\) into \(a + b = 8\) to solve for \(a\). \[\begin{aligned}a + \sqrt{10} + 4 &= 8\\a + \left(\sqrt{10} + 4\right) &= 8\\a &= 4 - \sqrt{10}\end{aligned}\]This yields the following solution. $$\begin{aligned}a = 4 - \sqrt{10},\,b = \sqrt{10} + 4\end{aligned}$$