SOLUTION: mike and adam left a bus terminal at the same time and traveled in opposite direcrtions. mike's bus was in heavy traffic and had to travel 20 mph slower than adam's bus. after 3 ho

Question 7122: mike and adam left a bus terminal at the same time and traveled in opposite direcrtions. mike's bus was in heavy traffic and had to travel 20 mph slower than adam's bus. after 3 hours, their buses were 270 miles apart. how fast was each bus going?
Answer by prince_abubu(198) (Show Source): You can put this solution on YOUR website!
There are two ways of approaching this, and both will give you the same answers.

Let's take it as Mike's speed is slower than Adam's speed by 20 mph. So that translates to . (Since Adam's speed is faster, we'll need to cut down his speed by 20 mph to equal Mike's).

We know that we've got to add Mike and Adam's distances because in 3 hours, they should be 270 miles apart. Mike's distance would be his speed, the A - 20, times 3 hours. Adam's distance will be his speed, A, times 3 hours. SO, we've got

<----- used distrib property.

<----- 2 shots in 1 strike here - combined like terms and added 60 to both sides.

<----- So Adam's speed was 55 miles per hour. If Mike's is 20 mph slower than Adams, then Mike's speed would have been 35 miles per hour.
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What if we take it as Adam's speed being faster than Mike's by 20? If we think of it that way, the equation would look like (Mike's the slower one, so we need to crank up his speed by 20 mph to match Adam's).

Putting that in the total distance traveled by both guys, we have

. So Mike traveled at 35 mph. If Adam's was 20 mph faster, then Adam's is 55 mph. The answers in both thinking is the same.

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