1. **State the problem:** Half of Robert's wire equals two-thirds of Maria's wire, and together their wires total 10 feet. We need to find how much longer Robert's wire is than Maria's. 2. **Define variables:** Let $R$ be the length of Robert's wire and $M$ be the length of Maria's wire. 3. **Write the equation from the problem:** Half of Robert's wire equals two-thirds of Maria's wire: $$\frac{1}{2}R = \frac{2}{3}M$$ 4. **Express $R$ in terms of $M$:** Multiply both sides by 2 to isolate $R$: $$R = 2 \times \frac{2}{3}M = \frac{4}{3}M$$ 5. **Use the total length equation:** The total length is 10 feet: $$R + M = 10$$ Substitute $R$: $$\frac{4}{3}M + M = 10$$ 6. **Combine like terms:** $$\frac{4}{3}M + \frac{3}{3}M = 10$$ $$\frac{7}{3}M = 10$$ 7. **Solve for $M$:** Multiply both sides by $\cancel{\frac{3}{7}}$: $$M = 10 \times \frac{3}{7} = \frac{30}{7} \approx 4.29$$ 8. **Find $R$:** $$R = \frac{4}{3} \times \frac{30}{7} = \frac{120}{21} = \frac{40}{7} \approx 5.71$$ 9. **Find how much longer Robert's wire is than Maria's:** $$R - M = \frac{40}{7} - \frac{30}{7} = \frac{10}{7} \approx 1.43$$ **Answer:** Robert's wire is $\frac{10}{7}$ feet (approximately 1.43 feet) longer than Maria's wire.