1. **State the problem:** Solve the equation $$\frac{x - 2}{x^2 + x - 1} = \frac{1}{x - 3}$$ for $x$. 2. **Recall the formula and rules:** To solve an equation with fractions, we can cross-multiply to eliminate denominators, but we must check for values that make denominators zero (excluded values). 3. **Identify excluded values:** - Denominator $x^2 + x - 1 = 0$ - Denominator $x - 3 = 0$ gives $x = 3$ excluded. 4. **Cross-multiply:** $$ (x - 2)(x - 3) = 1 \cdot (x^2 + x - 1) $$ 5. **Expand left side:** $$ (x - 2)(x - 3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6 $$ 6. **Write the equation:** $$ x^2 - 5x + 6 = x^2 + x - 1 $$ 7. **Subtract $x^2$ from both sides:** $$ \cancel{x^2} - 5x + 6 = \cancel{x^2} + x - 1 $$ $$ -5x + 6 = x - 1 $$ 8. **Bring all terms to one side:** $$ -5x + 6 - x + 1 = 0 $$ $$ -6x + 7 = 0 $$ 9. **Solve for $x$:** $$ -6x = -7 $$ $$ x = \frac{-7}{-6} = \frac{7}{6} $$ 10. **Check excluded values:** - Check if $x=\frac{7}{6}$ makes any denominator zero: - $x^2 + x - 1 = \left(\frac{7}{6}\right)^2 + \frac{7}{6} - 1 = \frac{49}{36} + \frac{7}{6} - 1 = \frac{49}{36} + \frac{42}{36} - \frac{36}{36} = \frac{55}{36} \neq 0$ - $x - 3 = \frac{7}{6} - 3 = \frac{7}{6} - \frac{18}{6} = -\frac{11}{6} \neq 0$ So $x=\frac{7}{6}$ is valid. 11. **Answer:** The solution is $$x = \frac{7}{6}$$. Among the options, this corresponds to option c).