1. **Problem 8:** The cost of one eraser and three pencils is 17, and the cost of three erasers and four pencils is 31. Find the cost of 5 erasers. 2. Let the cost of one eraser be $e$ and one pencil be $p$. 3. We have the system of equations: $$e + 3p = 17$$ $$3e + 4p = 31$$ 4. From the first equation, express $e$ in terms of $p$: $$e = 17 - 3p$$ 5. Substitute $e$ in the second equation: $$3(17 - 3p) + 4p = 31$$ $$51 - 9p + 4p = 31$$ $$51 - 5p = 31$$ 6. Solve for $p$: $$-5p = 31 - 51$$ $$-5p = -20$$ $$p = 4$$ 7. Substitute $p = 4$ back into $e = 17 - 3p$: $$e = 17 - 3(4) = 17 - 12 = 5$$ 8. The cost of 5 erasers is: $$5e = 5 imes 5 = 25$$ 9. **Problem 9:** The area of a rectangle is 440 cm², and the length is 2 cm more than the width. Find the width. 10. Let the width be $w$ cm. 11. Then the length is: $$w + 2$$ 12. The area is length $\times$ width: $$w(w + 2) = 440$$ 13. Expand and form quadratic equation: $$w^2 + 2w - 440 = 0$$ 14. Solve the quadratic equation using the quadratic formula: $$w = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-440)}}{2 \times 1} = \frac{-2 \pm \sqrt{4 + 1760}}{2} = \frac{-2 \pm \sqrt{1764}}{2}$$ $$\sqrt{1764} = 42$$ 15. Thus, $$w = \frac{-2 + 42}{2} = 20 \quad \text{or} \quad w = \frac{-2 - 42}{2} = -22$$ 16. Width cannot be negative, so: $$w = 20$$ cm **Final Answers:** - The cost of 5 erasers is 25. - The width of the rectangle is 20 cm.