1. **Stating the problem:** We need to solve the expression $\frac{3}{-4} - \left(-\frac{2}{5}\right)$ and the product $\frac{3}{-4} \cdot \left(-\frac{2}{5}\right)$. 2. **Recall the rules:** - Subtracting a negative is the same as adding a positive. - Multiplying two fractions: multiply numerators and denominators. - Simplify fractions by canceling common factors. 3. **Solve the subtraction:** $$\frac{3}{-4} - \left(-\frac{2}{5}\right) = \frac{3}{-4} + \frac{2}{5}$$ Find common denominator $20$: $$\frac{3}{-4} = \frac{3 \times 5}{-4 \times 5} = \frac{15}{-20} = -\frac{15}{20}$$ $$\frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20}$$ Add: $$-\frac{15}{20} + \frac{8}{20} = \frac{-15 + 8}{20} = \frac{-7}{20}$$ 4. **Solve the multiplication:** $$\frac{3}{-4} \cdot \left(-\frac{2}{5}\right) = \frac{3 \times (-2)}{-4 \times 5} = \frac{-6}{-20}$$ Cancel negatives: $$\frac{\cancel{-6}}{\cancel{-20}} = \frac{6}{20}$$ Simplify by dividing numerator and denominator by 2: $$\frac{\cancel{6}^3}{\cancel{20}^{10}} = \frac{3}{10}$$ **Final answers:** - $\frac{3}{-4} - \left(-\frac{2}{5}\right) = -\frac{7}{20}$ - $\frac{3}{-4} \cdot \left(-\frac{2}{5}\right) = \frac{3}{10}$