5) \( \begin{aligned} x+y & =\frac{1}{2} \\ x-3 y & =\frac{1}{6}\end{aligned} \)

To solve the system of equations given, we can use substitution or elimination. Let's use substitution to find \( x \) and \( y \). From the first equation, we can express \( x \) in terms of \( y \): \[ x = \frac{1}{2} - y \] Now we can substitute this expression for \( x \) into the second equation: \[ \left(\frac{1}{2} - y\right) - 3y = \frac{1}{6} \] Simplifying, we get: \[ \frac{1}{2} - 4y = \frac{1}{6} \] To eliminate the fraction, let's find a common denominator (which is 6) for the constant terms: \[ \frac{3}{6} - 4y = \frac{1}{6} \] Subtracting \( \frac{1}{6} \) from both sides gives: \[ \frac{3}{6} - \frac{1}{6} = 4y \] \[ \frac{2}{6} = 4y \] Simplifying the left side gives: \[ \frac{1}{3} = 4y \] Now, we can solve for \( y \): \[ y = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12} \] Now that we have \( y \), we can substitute back to find \( x \): \[ x = \frac{1}{2} - \frac{1}{12} \] Finding a common denominator (which is 12) and calculating: \[ \frac{6}{12} - \frac{1}{12} = \frac{5}{12} \] Thus, the solution we find is: \[ x = \frac{5}{12}, \quad y = \frac{1}{12} \] So the final answer is: \( x = \frac{5}{12}, y = \frac{1}{12} \)