Math Problem Solver

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

\[x = - \frac{38}{5}, y = \frac{16}{5}\]\[x = 8, y = -2\]

Solve \(x + 3 y = 2\) for \(x\). \[x = 2 - 3 y\]Substitute \(2 - 3 y\) for \(x\) in \(x^{2} + y^{2} = 68\) and simplify. $$\begin{aligned}x^{2} + y^{2} &= 68 \\ \left(2 - 3 y\right)^{2} + y^{2} &= 68 \\ 5 y^{2} - 6 y &= 32 \\5 y^{2} - 6 y - 32 &= 0 \\ \left(y + 2\right) \left(5 y - 16\right) &= 0 \\ y = -2&, y = \frac{16}{5}\end{aligned}$$Substitute \(-2\) into \(x + 3 y = 2\) to solve for \(x\). \[\begin{aligned}x - 6 &= 2\\x &= 8\end{aligned}\]This yields the following solution. $$\begin{aligned}x = 8,\,y = -2\end{aligned}$$Substitute \(\frac{16}{5}\) into \(x + 3 y = 2\) to solve for \(x\). \[\begin{aligned}x + \frac{48}{5} &= 2\\x &= - \frac{38}{5}\end{aligned}\]This yields the following solution. $$\begin{aligned}x = - \frac{38}{5},\,y = \frac{16}{5}\end{aligned}$$