Math Problem Solver

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

\[x = -1, y = 1\]\[x = \frac{7}{5}, y = \frac{1}{5}\]

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