Let S and L be the cost in dollars of a small and large box respectively.

3S+14L=203 and 11S+L=220, so L=220-11S. (Note that L=11(20-S), so S cannot exceed 20, and if L>S it cannot exceed $18.33. For example, if S=$18, then L=$22 and if S=$19, L=$11, cheaper than the small box!)

Therefore, by substitution: 3S+14(220-11S)=203.

3S+3080-154S=203; 2877=151S, S=2877/151=$19.05 approx. L=220-11*19.05=$10.42 approx.

S+L=$29.47 approx. However, this seems wrong because the cost of a small box is bigger than the cost of a large box, so S<$18.33 is violated. This suggest an error in the question if the calculations are correct.

If the question had, for example, contained $2023 instead of $203 the solution would be:

3S+14L=2023, 11S+L=220, so L=220-11S.

Substituting: 3S+14(220-11S)=2023, 3S+3080-154S=2023, 151S=1057, S=$7 and L=$143.

S+L=$150, cost of one small and one large box.

**A MORE LIKELY ANSWER AND CORRECTION**

**If the question had read “11 small boxes and 11 large boxes” then the answer is:**

**11S+11L=220, so S+L=20 and S=20-L.**

**Substituting for S: 3(20-L)+14L=203, 60-3L+14L=203, 11L=143 and L=$13 and S=$7. The small box of oranges costs $7 and the large one costs $13.**