1. **State the problem:** We are given a rectangle with perimeter 40 cm and a relationship between length $l$ and width $w$: twice the length is five centimeters more than three times the width. 2. **Write the equations:** The perimeter $P$ of a rectangle is given by: $$P = 2l + 2w$$ Given $P = 40$, we have: $$2l + 2w = 40$$ The relationship between length and width is: $$2l = 3w + 5$$ 3. **Simplify the perimeter equation:** Divide both sides by 2: $$\cancel{2}l + \cancel{2}w = \cancel{2}20$$ $$l + w = 20$$ 4. **Use substitution:** From the second equation: $$2l = 3w + 5 \implies l = \frac{3w + 5}{2}$$ 5. **Substitute $l$ into $l + w = 20$:** $$\frac{3w + 5}{2} + w = 20$$ Multiply both sides by 2 to clear the denominator: $$(3w + 5) + 2w = 40$$ $$3w + 5 + 2w = 40$$ $$5w + 5 = 40$$ 6. **Solve for $w$:** $$5w = 40 - 5$$ $$5w = 35$$ $$w = \frac{35}{5} = 7$$ 7. **Find $l$ using $l + w = 20$:** $$l + 7 = 20$$ $$l = 20 - 7 = 13$$ **Final answer:** Length $l = 13$ cm, Width $w = 7$ cm.