1. **State the problem:** We have a rectangle with original length $L$ and width $W$. When the length is increased by 2 cm, the area becomes 36 cm². When the width is increased by 3 cm, the area becomes 49 cm². We need to find the original width $W$. 2. **Write the equations from the problem:** - Increasing length by 2 cm: $$(L + 2) \times W = 36$$ - Increasing width by 3 cm: $$L \times (W + 3) = 49$$ 3. **Express $L$ from the first equation:** $$L + 2 = \frac{36}{W} \implies L = \frac{36}{W} - 2$$ 4. **Substitute $L$ into the second equation:** $$\left(\frac{36}{W} - 2\right) \times (W + 3) = 49$$ 5. **Expand and simplify:** $$\left(\frac{36}{W} - 2\right)(W + 3) = 49$$ $$\frac{36}{W} \times (W + 3) - 2(W + 3) = 49$$ $$36 \times \frac{W + 3}{W} - 2W - 6 = 49$$ 6. **Rewrite the fraction:** $$36 \times \left(1 + \frac{3}{W}\right) - 2W - 6 = 49$$ $$36 + \frac{108}{W} - 2W - 6 = 49$$ 7. **Combine like terms:** $$30 + \frac{108}{W} - 2W = 49$$ 8. **Isolate terms:** $$\frac{108}{W} - 2W = 19$$ 9. **Multiply both sides by $W$ to clear denominator:** $$\cancel{W} \times \frac{108}{\cancel{W}} - 2W \times W = 19W$$ $$108 - 2W^2 = 19W$$ 10. **Rearrange to standard quadratic form:** $$-2W^2 - 19W + 108 = 0$$ Multiply both sides by $-1$ to simplify: $$2W^2 + 19W - 108 = 0$$ 11. **Solve quadratic equation:** Use quadratic formula: $$W = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=19$, $c=-108$. Calculate discriminant: $$\Delta = 19^2 - 4 \times 2 \times (-108) = 361 + 864 = 1225$$ Square root: $$\sqrt{1225} = 35$$ Calculate roots: $$W = \frac{-19 \pm 35}{2 \times 2} = \frac{-19 \pm 35}{4}$$ Two solutions: - $$W = \frac{-19 + 35}{4} = \frac{16}{4} = 4$$ - $$W = \frac{-19 - 35}{4} = \frac{-54}{4} = -13.5$$ (not possible since width must be positive) 12. **Final answer:** The original width is $\boxed{4}$ cm. **Answer choice:** 4 cm