1. **State the problem:** Solve the equation $$\frac{8}{x + 2} - \frac{5}{3} = \frac{x - 1}{3}$$ for $x$. 2. **Identify the formula and rules:** To solve rational equations, first find a common denominator to eliminate fractions by multiplying both sides. 3. **Find the least common denominator (LCD):** The denominators are $x+2$ and $3$. The LCD is $3(x+2)$. 4. **Multiply both sides by the LCD:** $$3(x+2) \times \left(\frac{8}{x + 2} - \frac{5}{3}\right) = 3(x+2) \times \frac{x - 1}{3}$$ 5. **Distribute and simplify:** $$3(x+2) \times \frac{8}{x + 2} - 3(x+2) \times \frac{5}{3} = 3(x+2) \times \frac{x - 1}{3}$$ $$3 \cancel{(x+2)} \times \frac{8}{\cancel{x + 2}} - \cancel{3} (x+2) \times \frac{5}{\cancel{3}} = \cancel{3} (x+2) \times \frac{x - 1}{\cancel{3}}$$ $$3 \times 8 - 5(x+2) = (x+2)(x-1)$$ 6. **Simplify each term:** $$24 - 5x - 10 = (x+2)(x-1)$$ $$14 - 5x = (x+2)(x-1)$$ 7. **Expand the right side:** $$14 - 5x = x^2 - x + 2x - 2$$ $$14 - 5x = x^2 + x - 2$$ 8. **Bring all terms to one side:** $$0 = x^2 + x - 2 - 14 + 5x$$ $$0 = x^2 + 6x - 16$$ 9. **Solve the quadratic equation:** Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=6$, $c=-16$. $$x = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times (-16)}}{2 \times 1} = \frac{-6 \pm \sqrt{36 + 64}}{2} = \frac{-6 \pm \sqrt{100}}{2}$$ $$x = \frac{-6 \pm 10}{2}$$ 10. **Calculate the two solutions:** $$x_1 = \frac{-6 + 10}{2} = \frac{4}{2} = 2$$ $$x_2 = \frac{-6 - 10}{2} = \frac{-16}{2} = -8$$ 11. **Check for restrictions:** The original denominator $x+2$ cannot be zero, so $x \neq -2$. Both $x=2$ and $x=-8$ are valid. **Final answer:** $$x = 2 \text{ or } x = -8$$