1. **Problem (a): Solve** $$\frac{4}{x - 1} = \frac{x}{2x^2 + 3x - 5}$$ 2. **Step 1: Factor the denominator on the right side** $$2x^2 + 3x - 5 = (2x - 2)(x + \frac{5}{2})$$ but better to factor as $$2x^2 + 3x - 5 = (2x - 2)(x + 2.5)$$ is not exact, so use quadratic formula or factor by grouping: Try factors of -10 (2*-5): 5 and -2 work: $$2x^2 + 5x - 2x - 5 = (2x^2 - 2x) + (5x - 5) = 2x(x - 1) + 5(x - 1) = (2x + 5)(x - 1)$$ So denominator is $$ (2x + 5)(x - 1) $$ 3. **Rewrite equation:** $$\frac{4}{x - 1} = \frac{x}{(2x + 5)(x - 1)}$$ 4. **Multiply both sides by the common denominator** $$ (x - 1)(2x + 5) $$ to clear fractions: $$4(2x + 5) = x$$ 5. **Simplify:** $$8x + 20 = x$$ 6. **Solve for x:** $$8x - x = -20$$ $$7x = -20$$ $$x = -\frac{20}{7}$$ 7. **Check for restrictions:** Denominators cannot be zero: $$x \neq 1$$ and $$2x + 5 \neq 0 \Rightarrow x \neq -\frac{5}{2}$$ Our solution $$x = -\frac{20}{7}$$ is valid. --- **Final answer for (a):** $$x = -\frac{20}{7}$$