Ben Sweets Ca2591

1. **State the problem:** Ben and Steven have some sweets. We want to find how many sweets Ben has initially. 2. **Define variables:** Let $B$ = number of sweets Ben has initially. Let $S$ = number of sweets Steven has initially. 3. **Analyze the first scenario:** - Ben eats 4 sweets per day. - Steven eats 3 sweets per day. - When Steven finishes all his sweets, Ben has 156 sweets left. The number of days until Steven finishes his sweets is $$\frac{S}{3}$$. In that time, Ben eats $$4 \times \frac{S}{3} = \frac{4S}{3}$$ sweets. So, Ben's sweets left after that time is: $$B - \frac{4S}{3} = 156$$ 4. **Analyze the second scenario:** - Ben eats 1 sweet per day. - Steven eats 4 sweets per day. - When Steven finishes all his sweets, Ben has 338 sweets left. The number of days until Steven finishes his sweets is $$\frac{S}{4}$$. Ben eats $$1 \times \frac{S}{4} = \frac{S}{4}$$ sweets in that time. Ben's sweets left after that time is: $$B - \frac{S}{4} = 338$$ 5. **Write the system of equations:** $$\begin{cases} B - \frac{4S}{3} = 156 \\ B - \frac{S}{4} = 338 \end{cases}$$ 6. **Solve the system:** From the second equation: $$B = 338 + \frac{S}{4}$$ Substitute into the first equation: $$338 + \frac{S}{4} - \frac{4S}{3} = 156$$ Simplify the $S$ terms: $$\frac{S}{4} - \frac{4S}{3} = 156 - 338 = -182$$ Find common denominator 12: $$\frac{3S}{12} - \frac{16S}{12} = -182$$ $$\frac{3S - 16S}{12} = -182$$ $$\frac{-13S}{12} = -182$$ Multiply both sides by 12: $$-13S = -182 \times 12$$ $$-13S = -2184$$ Divide both sides by -13: $$S = \frac{-2184}{-13} = 168$$ 7. **Find $B$:** $$B = 338 + \frac{168}{4} = 338 + 42 = 380$$ **Answer:** Ben has $\boxed{380}$ sweets initially.