$\left\{\begin{array}{l}\frac{x+y}{2}+\frac{y}{3}=6 \\ \frac{x-y}{2}-\frac{y}{2}=-4\end{array}\right.$

Step 1 ：We are given the system of equations: \(\left\{\begin{array}{l}\frac{x+y}{2}+\frac{y}{3}=6 \\ \frac{x-y}{2}-\frac{y}{2}=-4\end{array}\right.\)

Step 2 ：We can solve this system by either substitution or elimination method. We will use the elimination method.

Step 3 ：First, we multiply the first equation by 2 and the second equation by 3 to make the coefficients of y the same in both equations. This gives us the new system of equations: \(\left\{\begin{array}{l}x + \frac{5y}{3}=12 \\ \frac{3x}{2} - 3y=-12\end{array}\right.\)

Step 4 ：Next, we subtract the second equation from the first to eliminate y. This gives us the equation: \(-\frac{x}{2} + \frac{14y}{3}=24\)

Step 5 ：Solving this equation for x, we get: \(x = \frac{28y}{3} - \frac{48}{11}\)

Step 6 ：Substituting this value of x into the first equation, we can solve for y, giving us: \(y = \frac{60}{11}\)

Step 7 ：Finally, substituting this value of y into the equation for x, we get the solution to the system of equations: \(\boxed{x = \frac{28}{3} - \frac{48}{11}, y = \frac{60}{11}}\)