1. **Problem (a):** Calculate the dimensions of the smaller rectangle given that the area of the larger rectangle is 4 times the area of the smaller rectangle. 2. **Given:** - Larger rectangle dimensions: height = 4 cm, width = 12 cm - Smaller rectangle dimensions: height = $x$ cm, width = $x + 2$ cm - Area of larger rectangle = 4 \times area of smaller rectangle 3. **Formula for area of rectangle:** $$\text{Area} = \text{height} \times \text{width}$$ 4. **Calculate areas:** - Area of larger rectangle = $4 \times 12 = 48$ cm$^2$ - Area of smaller rectangle = $x \times (x + 2) = x(x + 2)$ cm$^2$ 5. **Set up the equation:** $$48 = 4 \times x(x + 2)$$ 6. **Simplify the equation:** $$48 = 4x^2 + 8x$$ 7. **Divide both sides by 4:** $$\cancel{4} \times 12 = \cancel{4}x^2 + 2x$$ $$12 = x^2 + 2x$$ 8. **Rewrite as a quadratic equation:** $$x^2 + 2x - 12 = 0$$ 9. **Factor the quadratic:** $$ (x + 4)(x - 3) = 0$$ 10. **Solve for $x$:** $$x + 4 = 0 \Rightarrow x = -4 \quad \text{(not valid since length cannot be negative)}$$ $$x - 3 = 0 \Rightarrow x = 3$$ 11. **Dimensions of smaller rectangle:** - Height = $3$ cm - Width = $3 + 2 = 5$ cm --- 12. **Problem (b):** Solve for $x$ in the equation: $$\frac{2x - 3}{3} = \frac{x - 2}{5}$$ 13. **Cross multiply:** $$5(2x - 3) = 3(x - 2)$$ 14. **Expand both sides:** $$10x - 15 = 3x - 6$$ 15. **Bring all terms to one side:** $$10x - 3x = -6 + 15$$ $$7x = 9$$ 16. **Solve for $x$:** $$x = \frac{9}{7}$$ **Final answers:** - (a) Smaller rectangle dimensions: height = 3 cm, width = 5 cm - (b) $x = \frac{9}{7}$