1. **Problem 4:** Given the equation $$520 \div x = 40n$$, find the value of $$x$$ in terms of $$n$$. 2. **Step 1:** Write the equation clearly: $$\frac{520}{x} = 40n$$ 3. **Step 2:** To isolate $$x$$, multiply both sides by $$x$$: $$520 = 40n \times x$$ 4. **Step 3:** Now divide both sides by $$40n$$ to solve for $$x$$: $$x = \frac{520}{40n}$$ 5. **Step 4:** Simplify the fraction: $$x = \frac{520}{40n} = \frac{\cancel{40} \times 13}{\cancel{40} n} = \frac{13}{n}$$ 6. **Step 5:** So, $$x = \frac{13}{n}$$. Since the options are constants or expressions without $$n$$, none exactly match unless $$n=1$$, then $$x=13$$. --- 7. **Problem 5:** Given $$a - 16 = 8b + 6$$, find $$a + 3$$. 8. **Step 1:** Start with the equation: $$a - 16 = 8b + 6$$ 9. **Step 2:** Add 16 to both sides to isolate $$a$$: $$a = 8b + 6 + 16$$ 10. **Step 3:** Simplify the right side: $$a = 8b + 22$$ 11. **Step 4:** Now find $$a + 3$$: $$a + 3 = (8b + 22) + 3 = 8b + 25$$ 12. **Step 5:** So, $$a + 3 = 8b + 25$$. --- 13. **Problem 6:** Janice weighs $$x$$ pounds. Elaina weighs 23 pounds more than Janice. Find their combined weight in terms of $$x$$. 14. **Step 1:** Elaina's weight is: $$x + 23$$ 15. **Step 2:** Combined weight is Janice's weight plus Elaina's weight: $$x + (x + 23) = 2x + 23$$ 16. **Step 3:** So, the combined weight is $$2x + 23$$ pounds.