5. Solve the following system: \( \left\{\begin{array}{l}3 x+4 y=13 \\ 2 x+y=7\end{array}\right. \)

To solve the system, we can use the substitution or elimination method. Let's use the substitution method here! From the second equation \(2x + y = 7\), we can express \(y\) in terms of \(x\): \[ y = 7 - 2x \] Now, substitute this expression for \(y\) into the first equation \(3x + 4y = 13\): \[ 3x + 4(7 - 2x) = 13 \] \[ 3x + 28 - 8x = 13 \] Combine like terms: \[ -5x + 28 = 13 \] Subtract 28 from both sides: \[ -5x = 13 - 28 \] \[ -5x = -15 \] Dividing by -5 gives us: \[ x = 3 \] Now substitute \(x = 3\) back into the equation for \(y\): \[ y = 7 - 2(3) = 7 - 6 = 1 \] Thus, the solution to the system is: \( (x, y) = (3, 1) \).