OpenSAT

To solve for \(x + y\), we can add the two equations together. Adding the left-hand side of the first equation and the left-hand side of the second equation yields \(3x + 2y + (x - y) = 4x + y\). Adding the right-hand sides of the two equations yields \(10 + 1 = 11\). Thus, \(4x + y = 11\). We can rewrite the first equation as \(x = \frac{10 - 2y}{3}\). Substituting this into the equation \(4x + y = 11\) gives \(4(\frac{10 - 2y}{3}) + y = 11\). Simplifying, we have \(\frac{40 - 8y}{3} + y = 11\), or \(\frac{40 - 8y + 3y}{3} = 11\), or \(\frac{40 - 5y}{3} = 11\). Multiplying both sides by 3 gives \(40 - 5y = 33\). Subtracting 40 from both sides gives \(-5y = -7\). Dividing both sides by -5 gives \(y = \frac{7}{5}\). Substituting this value of y into the equation \(x - y = 1\) gives \(x - \frac{7}{5} = 1\), or \(x = \frac{12}{5}\). Therefore, \(x + y = \frac{12}{5} + \frac{7}{5} = \frac{19}{5}\), or 3.8. Of the choices given, 5 is closest to 3.8.