# algebra 2

Two families met at a par for a picnic at the end of the day, they part ways. Family A leaves 15 minutes after Family B and travels east at an average of 42 mph; while family B travels west at an average of 50mph. Both families have an approximately 160 miles to travel.
1.)find the time(in hours and minutes) that it takes them to be 120 miles apart
2.) when family B arrives at home how much further does Family A still have to travel?
I believe the answer to part 1 is 8 hours and 9 minutes but i have no idea for the second part on how to set up the equation. For the first part i made a table with R,T,D and for the second part i also made a table with R,T,D

1. Scott

they are moving apart at 92 mph (the sum of their speeds)

so it takes ... 120 mi / 92 mph ... for them to be 120 mi apart

B arrives home in ... 160 mi / 50 mph

use B's travel time (minus 15 min) to find how far A has traveled

so would the equation for b be
42(x-15)=160-50x

3. Scott

B's travel time is ... 160/50 = 3.2 hr

at this point, A has been traveling for
... 3.2 - .25 = 2.95 hr

A's remaining distance is
... 160 - (42 * 2.95)

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