1. **State the problem:** We have two rectangles P and Q. - Width of P is $x - 3$ cm. - Width of Q is $2x - 5$ cm. - Area of P is 2 cm$^2$. - Area of Q is 3 cm$^2$. - Sum of lengths of P and Q is 3 cm. We need to find the lengths of both rectangles. 2. **Write formulas for area:** Area = width $\times$ length. For rectangle P: $$\text{Area}_P = (x - 3) \times L_P = 2$$ For rectangle Q: $$\text{Area}_Q = (2x - 5) \times L_Q = 3$$ 3. **Express lengths in terms of $x$:** $$L_P = \frac{2}{x - 3}$$ $$L_Q = \frac{3}{2x - 5}$$ 4. **Use the sum of lengths:** $$L_P + L_Q = 3$$ Substitute expressions: $$\frac{2}{x - 3} + \frac{3}{2x - 5} = 3$$ 5. **Solve the equation:** Multiply both sides by $(x - 3)(2x - 5)$ to clear denominators: $$2(2x - 5) + 3(x - 3) = 3(x - 3)(2x - 5)$$ Expand left side: $$4x - 10 + 3x - 9 = 3(x - 3)(2x - 5)$$ $$7x - 19 = 3(x - 3)(2x - 5)$$ Expand right side: $$(x - 3)(2x - 5) = 2x^2 - 5x - 6x + 15 = 2x^2 - 11x + 15$$ So: $$7x - 19 = 3(2x^2 - 11x + 15)$$ $$7x - 19 = 6x^2 - 33x + 45$$ Bring all terms to one side: $$0 = 6x^2 - 33x + 45 - 7x + 19$$ $$0 = 6x^2 - 40x + 64$$ 6. **Simplify the quadratic:** Divide entire equation by 2: $$0 = 3x^2 - 20x + 32$$ 7. **Solve quadratic equation:** Use quadratic formula: $$x = \frac{20 \pm \sqrt{(-20)^2 - 4 \times 3 \times 32}}{2 \times 3} = \frac{20 \pm \sqrt{400 - 384}}{6} = \frac{20 \pm \sqrt{16}}{6}$$ $$x = \frac{20 \pm 4}{6}$$ Two solutions: $$x_1 = \frac{20 + 4}{6} = \frac{24}{6} = 4$$ $$x_2 = \frac{20 - 4}{6} = \frac{16}{6} = \frac{8}{3} \approx 2.67$$ 8. **Check for valid widths (widths must be positive):** For $x=4$: $$x - 3 = 1 > 0$$ $$2x - 5 = 8 - 5 = 3 > 0$$ Valid. For $x=\frac{8}{3} \approx 2.67$: $$x - 3 = 2.67 - 3 = -0.33 < 0$$ Width of P is negative, invalid. 9. **Calculate lengths for $x=4$:** $$L_P = \frac{2}{4 - 3} = \frac{2}{1} = 2 \text{ cm}$$ $$L_Q = \frac{3}{2(4) - 5} = \frac{3}{8 - 5} = \frac{3}{3} = 1 \text{ cm}$$ 10. **Verify sum of lengths:** $$2 + 1 = 3 \text{ cm}$$ Correct. **Final answer:** Length of rectangle P is 2 cm. Length of rectangle Q is 1 cm.