Solve the following pair of linear equations by the elimination method and the substitution method:
$\frac{x}{2} + \frac{2y}{3} = -1$ and $x - \frac{y}{3} = 3$

(X=2, Y=-3) Given equations:
$(1)$ $\frac{x}{2} + \frac{2y}{3} = -1$
$(2)$ $x - \frac{y}{3} = 3$
Simplify equation $(1)$ by multiplying by $6$: $3x + 4y = -6$ $(3)$
Simplify equation $(2)$ by multiplying by $3$: $3x - y = 9$ $(4)$
Elimination Method:
Subtract $(4)$ from $(3)$: $(3x + 4y) - (3x - y) = -6 - 9$
$5y = -15 \implies y = -3$
Substitute $y = -3$ into $(4)$: $3x - (-3) = 9 \implies 3x + 3 = 9 \implies 3x = 6 \implies x = 2$
Substitution Method:
From $(4)$,$y = 3x - 9$
Substitute into $(3)$: $3x + 4(3x - 9) = -6$
$3x + 12x - 36 = -6 \implies 15x = 30 \implies x = 2$
Substitute $x = 2$ into $y = 3x - 9$: $y = 3(2) - 9 = 6 - 9 = -3$
Final solution: $x = 2, y = -3$.