1. **State the problem:** Solve the equation $$1 + \frac{3}{x - 1} = \frac{2x + 7}{x^2 + x - 2}$$ for $x$. 2. **Identify the denominator factorization:** Note that $$x^2 + x - 2 = (x - 1)(x + 2)$$. 3. **Rewrite the equation with factored denominator:** $$1 + \frac{3}{x - 1} = \frac{2x + 7}{(x - 1)(x + 2)}$$ 4. **Multiply both sides by the common denominator $(x - 1)(x + 2)$ to clear fractions:** $$\left(1 + \frac{3}{x - 1}\right)(x - 1)(x + 2) = 2x + 7$$ 5. **Distribute:** $$1 \cdot (x - 1)(x + 2) + \frac{3}{x - 1} \cdot (x - 1)(x + 2) = 2x + 7$$ 6. **Simplify the second term by canceling $(x - 1)$:** $$ (x - 1)(x + 2) + 3(x + 2) = 2x + 7$$ 7. **Expand the terms:** $$ (x^2 + 2x - x - 2) + 3x + 6 = 2x + 7$$ $$ (x^2 + x - 2) + 3x + 6 = 2x + 7$$ 8. **Combine like terms on the left:** $$ x^2 + 4x + 4 = 2x + 7$$ 9. **Bring all terms to one side:** $$ x^2 + 4x + 4 - 2x - 7 = 0$$ $$ x^2 + 2x - 3 = 0$$ 10. **Factor the quadratic:** $$ (x + 3)(x - 1) = 0$$ 11. **Solve for $x$:** $$ x = -3 \quad \text{or} \quad x = 1$$ 12. **Check for restrictions:** The original denominators cannot be zero, so $x \neq 1$. 13. **Final solution:** $$ \boxed{x = -3}$$