1. **Problem (a): Solve the equation** $$\frac{3}{1-x} - \frac{2}{3x-1} = 0$$. 2. **Step 1: Find a common denominator and combine the fractions.** The common denominator is $$(1-x)(3x-1)$$. 3. **Step 2: Write the equation as a single fraction equal to zero:** $$\frac{3(3x-1) - 2(1-x)}{(1-x)(3x-1)} = 0$$ 4. **Step 3: Set the numerator equal to zero (since denominator cannot be zero):** $$3(3x-1) - 2(1-x) = 0$$ 5. **Step 4: Expand the numerator:** $$9x - 3 - 2 + 2x = 0$$ $$9x + 2x - 3 - 2 = 0$$ $$11x - 5 = 0$$ 6. **Step 5: Solve for $x$:** $$11x = 5$$ $$x = \frac{5}{11}$$ 7. **Step 6: Check for restrictions:** Denominator terms cannot be zero: $$1 - x \neq 0 \Rightarrow x \neq 1$$ $$3x - 1 \neq 0 \Rightarrow x \neq \frac{1}{3}$$ Since $x=\frac{5}{11}$ is not restricted, it is a valid solution. --- 8. **Problem (b): Simplify** $$\frac{4x^2 - 25}{6x^2 - 13x - 5}$$. 9. **Step 1: Factor numerator and denominator.** Numerator is a difference of squares: $$4x^2 - 25 = (2x - 5)(2x + 5)$$ Denominator factors as: $$6x^2 - 13x - 5$$ Find factors of $6 \times (-5) = -30$ that sum to $-13$: $-15$ and $2$. Rewrite: $$6x^2 - 15x + 2x - 5$$ Group: $$(6x^2 - 15x) + (2x - 5)$$ Factor: $$3x(2x - 5) + 1(2x - 5)$$ $$= (3x + 1)(2x - 5)$$ 10. **Step 2: Write the fraction with factored terms:** $$\frac{(2x - 5)(2x + 5)}{(3x + 1)(2x - 5)}$$ 11. **Step 3: Cancel common factor $(2x - 5)$:** $$\frac{\cancel{(2x - 5)}(2x + 5)}{(3x + 1)\cancel{(2x - 5)}} = \frac{2x + 5}{3x + 1}$$ 12. **Final simplified expression:** $$\frac{2x + 5}{3x + 1}$$