Santos is traveling 333 miles away from Sam. They are traveling towards each other. If Sam travels 7mph faster than Santos and they meet after 9 hours, how fast was each traveling?
asked Jul 26 in Algebra 1 Answers by anonymous

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Let Santos’ speed be S then Sam’s is S+7.

If they meet x miles from where Sam started, they will meet after time t=x/(S+7).

Santos will meet Sam after the same time t=(333-x)/S.

We know t=9 hours.

x/(S+7)=(333-x)/S=9.

So xS=(333-x)(S+7), xS=333S+2331-xS-7x, x(2S+7)=333S+2331, x=(333S+2331)/(2S+7).

So we can substitute for x: (333S+2331)/((2S+7)(S+7))=9.

333S+2331=9(2S+7)(S+7), divided by 9: 37S+259=2S²+16S+49, 2S²-16S-210=0, S²-8S-105=(S-15)(S+7).

Therefore S=15mph, Santos’ speed, Sam’s speed=22mph.

CHECK

x=9×22=198 miles, the distance from Sam’s starting position, because he travels at 22mph for 9 hours. So this is 333-198=135 miles from Santos’ starting position, and it would take 135/15=9 hours for them to meet.

 

answered Jul 26 by Rod Top Rated User (582,840 points)

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