Question

Eli flew home from vacation with a heavy bag. With the first airline he flew, Eli had to pay $29 to check his bag, plus $18 for every kilogram that his bag was over the weight limit. The next flight was with another airline that had the same weight limit. Eli had to pay $19 per kilogram that his bag was over the weight limit, in addition to the checked bag fee of $18. By coincidence, the fees ended up being the same with both airlines.

Answer 1

Let \( w \) be the weight limit (in kilograms) for the bag, and let \( x \) be the number of kilograms that Eli's bag is over the weight limit.

For the first airline, the total cost \( C_1 \) can be expressed as: \[ C_1 = 29 + 18x \]

For the second airline, the total cost \( C_2 \) can be expressed as: \[ C_2 = 18 + 19x \]

Since the total fees are equal for both airlines, we can set the two equations equal to each other: \[ 29 + 18x = 18 + 19x \]

Next, we'll solve this equation for \( x \).

1. Subtract \( 18x \) from both sides: \[ 29 = 18 + x \]

2. Subtract 18 from both sides: \[ 29 - 18 = x \] \[ 11 = x \]

Thus, Eli's bag is 11 kilograms over the weight limit.

Now we can find \( w \) by substituting \( x = 11 \) back into either of the cost equations. We'll use the equation for the first airline:

\[ C_1 = 29 + 18(11) \] \[ C_1 = 29 + 198 \] \[ C_1 = 227 \]

Now, let’s confirm this cost using the second airline’s equation:

\[ C_2 = 18 + 19(11) \] \[ C_2 = 18 + 209 \] \[ C_2 = 227 \]

Both costs are indeed equal, confirming our calculations are correct.

In conclusion, Eli's bag is 11 kilograms over the weight limit.