1. **State the problem:** Solve the equation $$\frac{6}{x+2} = \frac{5x-1}{2-3x}$$ for $x$. 2. **Understand the formula and rules:** To solve an equation with fractions, we can cross-multiply to eliminate the denominators, provided the denominators are not zero. 3. **Cross-multiply:** $$6(2-3x) = (5x-1)(x+2)$$ 4. **Expand both sides:** $$12 - 18x = 5x^2 + 10x - x - 2$$ Simplify the right side: $$12 - 18x = 5x^2 + 9x - 2$$ 5. **Bring all terms to one side to set the equation to zero:** $$0 = 5x^2 + 9x - 2 - 12 + 18x$$ Simplify: $$0 = 5x^2 + 27x - 14$$ 6. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=5$, $b=27$, and $c=-14$. Calculate the discriminant: $$\Delta = 27^2 - 4 \times 5 \times (-14) = 729 + 280 = 1009$$ 7. **Find the roots:** $$x = \frac{-27 \pm \sqrt{1009}}{10}$$ 8. **Check for restrictions:** Denominators cannot be zero: - $x + 2 \neq 0 \Rightarrow x \neq -2$ - $2 - 3x \neq 0 \Rightarrow x \neq \frac{2}{3}$ Neither root equals these values, so both are valid. **Final answer:** $$x = \frac{-27 + \sqrt{1009}}{10} \quad \text{or} \quad x = \frac{-27 - \sqrt{1009}}{10}$$