(x+1)/(3) + (y)/(2) = 0

Multiply the entire equation by the least common multiple of 3 and 2, which is 6.

$$6 \left( \frac{x+1}{3} \right) + 6 \left( \frac{y}{2} \right) = 6(0)$$ $$2(x+1) + 3y = 0$$ $$2x + 2 + 3y = 0$$ $$2x + 3y = -2 \quad \text{(Equation 1')}$$

Multiply the entire equation by the least common multiple of 2 and 3, which is 6.

$$6 \left( \frac{x}{2} \right) - 6 \left( \frac{y-1}{3} \right) = 6(1)$$ $$3x - 2(y-1) = 6$$ $$3x - 2y + 2 = 6$$ $$3x - 2y = 6 - 2$$ $$3x - 2y = 4 \quad \text{(Equation 2')}$$

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:

$$(4x + 6y) + (9x - 6y) = -4 + 12$$ $$13x = 8$$ $$x = \frac{8}{13}$$

$$2x + 3y = -2$$ $$2 \left( \frac{8}{13} \right) + 3y = -2$$ $$\frac{16}{13} + 3y = -2$$ $$3y = -2 - \frac{16}{13}$$ $$3y = -\frac{26}{13} - \frac{16}{13}$$ $$3y = -\frac{42}{13}$$ $$y = -\frac{42}{13 \times 3}$$ $$y = -\frac{14}{13}$$

The solution to the system of equations is $x = \frac{8}{13}$ and $y = -\frac{14}{13}$.

$$x = \frac{8}{13}, y = -\frac{14}{13}$$