Here's the solution to 11(a): 11. (a) We need to solve the following system of simultaneous equations: 1) $$ \frac{x+1}{3} + \frac{y}{2} = 0 $$ 2) $$ \frac{x}{2} - \frac{y-1}{3} = 1 $$ Step 1: Simplify the first equation. Multiply the entire equation by the least common multiple of 3 and 2, which is 6. $$ 6 \left( \frac{x+1}{3} \right) + 6 \left( \frac{y}{2} \right) = 6(0) $$ $$ 2(x+1) + 3y = 0 $$ $$ 2x + 2 + 3y = 0 $$ $$ 2x + 3y = -2 \quad \text{(Equation 1')} $$ Step 2: Simplify the second equation. Multiply the entire equation by the least common multiple of 2 and 3, which is 6. $$ 6 \left( \frac{x}{2} \right) - 6 \left( \frac{y-1}{3} \right) = 6(1) $$ $$ 3x - 2(y-1) = 6 $$ $$ 3x - 2y + 2 = 6 $$ $$ 3x - 2y = 6 - 2 $$ $$ 3x - 2y = 4 \quad \text{(Equation 2')} $$ Step 3: Solve the system of simplified equations using the elimination method. We have: 1') $2x + 3y = -2$ 2') $3x - 2y = 4$ To eliminate $y$, multiply Equation 1' by 2 and Equation 2' by 3: $$ 2(2x + 3y) = 2(-2) \implies 4x + 6y = -4 $$ $$ 3(3x - 2y) = 3(4) \implies 9x - 6y = 12 $$ Add the two new equations: $$ (4x + 6y) + (9x - 6y) = -4 + 12 $$ $$ 13x = 8 $$ $$ x = \frac{8}{13} $$ Step 4: Substitute the value of $x$ into Equation 1' to find $y$. $$ 2x + 3y = -2 $$ $$ 2 \left( \frac{8}{13} \right) + 3y = -2 $$ $$ \frac{16}{13} + 3y = -2 $$ $$ 3y = -2 - \frac{16}{13} $$ $$ 3y = -\frac{26}{13} - \frac{16}{13} $$ $$ 3y = -\frac{42}{13} $$ $$ y = -\frac{42}{13 \times 3} $$ $$ y = -\frac{14}{13} $$ The solution to the system of equations is $x = \frac{8}{13}$ and $y = -\frac{14}{13}$. $$ \boxed{x = \frac{8}{13}, y = -\frac{14}{13}} $$