Solve the rational equation $\frac{15}{x^2+x-6} - \frac{3}{x-2} = \frac{2}{x+3}$.

Step 1 — Identify restricted values. Factor the quadratic denominator: $x^2+x-6 = (x-2)(x+3)$. The factors of all denominators are $(x-2)$ and $(x+3)$. Setting them to zero yields the restricted values: $x \neq 2$ and $x \neq -3$.

Step 2 — Find the LCD. The denominators are $(x-2)(x+3)$, $(x-2)$, and $(x+3)$. The LCD is $(x-2)(x+3)$.

Step 3 — Clear the fractions. Multiply every term on both sides by the LCD $(x-2)(x+3)$ and cancel the matching factors: $(x-2)(x+3) \cdot \frac{15}{(x-2)(x+3)} - (x-2)(x+3) \cdot \frac{3}{x-2} = (x-2)(x+3) \cdot \frac{2}{x+3}$ Simplify to obtain the fraction-free equation: $15 - 3(x+3) = 2(x-2)$

Step 4 — Solve the resulting equation. Distribute: $15 - 3x - 9 = 2x - 4$. Combine like terms: $6 - 3x = 2x - 4$. Add $3x$ to both sides: $6 = 5x - 4$. Add $4$ to both sides: $10 = 5x$. Divide by $5$: $x = 2$.

Step 5 — Check. The algebraic solution $x = 2$ matches a restricted value identified in Step 1. Because substituting $x = 2$ into the original equation would result in division by zero, it is an extraneous solution.

Since the only potential algebraic solution is extraneous, the equation has no solution.