OpenSAT

To solve for \(x + y\), we can use the elimination method. Multiplying the second equation by 2 gives \(2x - 2y = 2\). Adding this equation to the first equation gives \(5x = 12\), so \(x = \frac{12}{5}\). Substituting this value of x back into the second equation gives \(\frac{12}{5} - y = 1\), so \(y = \frac{7}{5}\). Therefore, \(x + y = \frac{12}{5} + \frac{7}{5} = \frac{19}{5} = 3 \frac{4}{5}\), but only 5 is a choice.