1. **State the problem:** Solve the rational equation $$\frac{7}{x + 2} + \frac{3x + 5}{x - 2} = \frac{2x + 7}{x - 2}$$. 2. **Identify the denominators:** The denominators are $x+2$ and $x-2$. To eliminate the fractions, multiply both sides of the equation by the least common denominator (LCD), which is $(x+2)(x-2)$. 3. **Multiply both sides by the LCD:** $$ (x+2)(x-2) \times \left( \frac{7}{x + 2} + \frac{3x + 5}{x - 2} \right) = (x+2)(x-2) \times \frac{2x + 7}{x - 2} $$ 4. **Distribute and simplify:** $$ 7 \cancel{(x+2)} \frac{(x-2)}{\cancel{(x+2)}} + (3x + 5) \cancel{(x-2)} \frac{(x+2)}{\cancel{(x-2)}} = (2x + 7) \cancel{(x-2)} \frac{(x+2)}{\cancel{(x-2)}} $$ This simplifies to: $$ 7(x-2) + (3x + 5)(x+2) = (2x + 7)(x+2) $$ 5. **Expand each term:** $$ 7x - 14 + (3x^2 + 6x + 5x + 10) = 2x^2 + 4x + 7x + 14 $$ Simplify inside the parentheses: $$ 7x - 14 + 3x^2 + 11x + 10 = 2x^2 + 11x + 14 $$ 6. **Combine like terms on the left:** $$ 3x^2 + (7x + 11x) + (-14 + 10) = 2x^2 + 11x + 14 $$ $$ 3x^2 + 18x - 4 = 2x^2 + 11x + 14 $$ 7. **Bring all terms to one side to set equation to zero:** $$ 3x^2 + 18x - 4 - 2x^2 - 11x - 14 = 0 $$ Simplify: $$ (3x^2 - 2x^2) + (18x - 11x) + (-4 - 14) = 0 $$ $$ x^2 + 7x - 18 = 0 $$ 8. **Solve the quadratic equation:** Use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=1$, $b=7$, $c=-18$. Calculate the discriminant: $$ \Delta = 7^2 - 4 \times 1 \times (-18) = 49 + 72 = 121 $$ Calculate the roots: $$ x = \frac{-7 \pm \sqrt{121}}{2} = \frac{-7 \pm 11}{2} $$ 9. **Find the two solutions:** - For $+$: $$ x = \frac{-7 + 11}{2} = \frac{4}{2} = 2 $$ - For $-$: $$ x = \frac{-7 - 11}{2} = \frac{-18}{2} = -9 $$ 10. **Check for restrictions:** The original denominators are $x+2$ and $x-2$, so $x \neq -2$ and $x \neq 2$ to avoid division by zero. Since $x=2$ is not allowed, discard it. 11. **Final solution:** $$ \boxed{x = -9} $$