1. **State the problem:** Solve the equation $$\frac{x^2}{3x - 1} + 2 = \frac{2(x - 3)}{3x - 1}$$ for $x$. 2. **Identify the common denominator:** The denominators on both sides are $3x - 1$. To eliminate the fractions, multiply both sides of the equation by $3x - 1$ (noting $3x - 1 \neq 0$). 3. **Multiply both sides:** $$\left(\frac{x^2}{3x - 1} + 2\right)(3x - 1) = \frac{2(x - 3)}{3x - 1} (3x - 1)$$ Simplifies to: $$x^2 + 2(3x - 1) = 2(x - 3)$$ 4. **Expand terms:** $$x^2 + 6x - 2 = 2x - 6$$ 5. **Bring all terms to one side:** $$x^2 + 6x - 2 - 2x + 6 = 0$$ Simplify: $$x^2 + 4x + 4 = 0$$ 6. **Factor the quadratic:** $$x^2 + 4x + 4 = (x + 2)^2$$ 7. **Solve for $x$:** $$(x + 2)^2 = 0 \implies x + 2 = 0 \implies x = -2$$ 8. **Check for extraneous solutions:** Check denominator $3x - 1$ at $x = -2$: $$3(-2) - 1 = -6 - 1 = -7 \neq 0$$ So $x = -2$ is valid. **Final answer:** $$x = -2$$