# Okla Homer Baggage

When Homer Smith and his wife Okla flew to Rome, they had together 94 pounds of baggage. Homer paid \$15.00 and Okla paid \$20.00 for the excess weight of their baggage. If Homer had made the trip by himself with the combined baggage of both of them, he would have had to pay \$135.00 for excess baggage. How many pounds of baggage can one person take along without charge?

The urge to go scrounging for equations is almost irresistible.
Let x be the number of pounds one can bring along for free, and
let y be the price, in dollars, per pound of excess baggage.

If Homer had gone alone, he would have had to pay \$100 for the baggage that Okla brought for free. So xy = 100. This equation doesn’t look too promising though. True, x has to be a positive integer, but y could be any number of decimals.

So we back off a bit and think some more. Taking an extreme view, suppose no baggage could be carried for free and Homer had to pay the excess fee for all 94 pounds. He’d pay the original \$15.00 and \$20.00, plus the \$100 for Okla’s “free” baggage (that’s \$135.00), plus another \$100 for his own “free” baggage. That would be \$235.00 in excess fees for 94 pounds. Aha!

We divide: \$235.00/94 = \$2.50. So a pound of overweight baggage costs \$2.50.

Now we can use our original equation: xy = 100 and 2.50x = 100, so x = 40. They could each take 40 pounds of baggage without charge.

Checking, Homer’s original fee of \$15.00 paid for 6 pounds of overweight; he brought 46 pounds of baggage. Okla’s \$20.00 paid for 8 pounds of overweight; she brought 48 pounds of baggage. And 46 + 48 = 94. If Homer had traveled alone, he would have had 54 pounds of excess baggage, and 54(\$2.50) = \$135.00.

(We observe that Okla and Homer must have made the trip to Rome a number of years ago, when the airlines’ rules were very different from what they are now.)