Solve the rational equation $\frac{4}{3x^2-10x+3} + \frac{3}{3x^2+2x-1} = \frac{2}{x^2-2x-3}$ by applying the five-step strategy for equations with rational expressions.

Step 1 — Identify restricted values. Factor all three denominators: $3x^2-10x+3 = (3x-1)(x-3)$ $3x^2+2x-1 = (3x-1)(x+1)$ $x^2-2x-3 = (x-3)(x+1)$ Setting each unique factor to zero gives the restricted values: $x \neq -1$, $x \neq \frac{1}{3}$, and $x \neq 3$.

Step 2 — Find the LCD. The factored denominators share the factors $(3x-1)$, $(x-3)$, and $(x+1)$. The LCD is $(3x-1)(x+1)(x-3)$.

Step 3 — Clear the fractions. Multiply every term on both sides by the LCD $(3x-1)(x+1)(x-3)$ and cancel matching denominator factors: $(3x-1)(x+1)(x-3) \cdot \frac{4}{(3x-1)(x-3)} + (3x-1)(x+1)(x-3) \cdot \frac{3}{(3x-1)(x+1)} = (3x-1)(x+1)(x-3) \cdot \frac{2}{(x-3)(x+1)}$ Simplify each term to obtain a fraction-free equation: $4(x+1) + 3(x-3) = 2(3x-1)$

Step 4 — Solve the resulting equation. Distribute: $4x + 4 + 3x - 9 = 6x - 2$. Combine like terms: $7x - 5 = 6x - 2$. Subtract $6x$ from both sides: $x - 5 = -2$. Add $5$ to both sides: $x = 3$.

Step 5 — Check. The algebraic solution $x = 3$ exactly matches one of the restricted values identified in Step 1. Substituting $x = 3$ into the original equation would make two of the denominators equal to zero, which is mathematically undefined. Therefore, $x = 3$ is an extraneous solution and must be discarded.

Because the only candidate solution is extraneous, the original equation has no solution.