1. **State the problem:** Solve the equation $$\frac{(x+3)(x-2)}{(x+3)(x-1)} = \frac{x-1}{x-2}$$ for $x$. 2. **Identify restrictions:** The denominators cannot be zero, so: - $x+3 \neq 0 \Rightarrow x \neq -3$ - $x-1 \neq 0 \Rightarrow x \neq 1$ - $x-2 \neq 0 \Rightarrow x \neq 2$ 3. **Simplify the equation:** Since $x \neq -3$, we can cancel $(x+3)$ from numerator and denominator on the left side: $$\frac{\cancel{(x+3)}(x-2)}{\cancel{(x+3)}(x-1)} = \frac{x-1}{x-2}$$ which simplifies to: $$\frac{x-2}{x-1} = \frac{x-1}{x-2}$$ 4. **Cross multiply to solve:** $$ (x-2)(x-2) = (x-1)(x-1) $$ 5. **Expand both sides:** $$ (x-2)^2 = (x-1)^2 $$ $$ x^2 - 4x + 4 = x^2 - 2x + 1 $$ 6. **Subtract $x^2$ from both sides:** $$ \cancel{x^2} - 4x + 4 = \cancel{x^2} - 2x + 1 $$ which simplifies to: $$ -4x + 4 = -2x + 1 $$ 7. **Bring all terms to one side:** $$ -4x + 4 + 2x - 1 = 0 $$ $$ -2x + 3 = 0 $$ 8. **Solve for $x$:** $$ -2x = -3 $$ $$ x = \frac{-3}{-2} = \frac{3}{2} $$ 9. **Check restrictions:** $x = \frac{3}{2}$ is not equal to $-3$, $1$, or $2$, so it is valid. **Final answer:** $$ x = \frac{3}{2} $$