Respuesta :
1. If the total baggage fees were equal, you would have to equate the two situations. Suppose x is the weight above the limit on the flight home, while y is the weight above the limit on the flight back to college. Then, the equation would be:
20 + 8x = 26 + 5y
Now, we have two unknowns, but only one independent equation. This is unsolvable unless we make an assumption. Suppose the excess weight for both trips are the same such that x=y, then,
20 + 8x = 26 + 5x
2. Then, we solve for x.
8x - 5x = 26 - 20
3x = 6
x = 6/3 = 2 lb
Thus, Elizabeth's bag was 2 lb over the weight limit.
3. The total baggage fee would then be:
20+8(2) = 26 + 5(2) = $36
20 + 8x = 26 + 5y
Now, we have two unknowns, but only one independent equation. This is unsolvable unless we make an assumption. Suppose the excess weight for both trips are the same such that x=y, then,
20 + 8x = 26 + 5x
2. Then, we solve for x.
8x - 5x = 26 - 20
3x = 6
x = 6/3 = 2 lb
Thus, Elizabeth's bag was 2 lb over the weight limit.
3. The total baggage fee would then be:
20+8(2) = 26 + 5(2) = $36