SOLUTION: please help me on a sales trip gail drives 6oo miles to richmond averaging one speed. the return trip home i made at the same average speed ten mph slower. total time for the ro

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Question 62724This question is from textbook Elementary and intermediate algebra
: please help me
on a sales trip gail drives 6oo miles to richmond averaging one speed. the return trip home i made at the same average speed ten mph slower. total time for the round trip is 22 hours. find gails average speed for each part of the trip. this is what i've done:
s=600/(t1) and s-10=600/(t2)
(600/t1)-10=600/(t2)
(600/t2)-(10*t2)=(600*t1)
(600/t2)-(600*t1)-(10*t2)=0
(590*t2)-(600*t1)=0

where do i go from here or am i doing this right at all
thank you for your help
This question is from textbook Elementary and intermediate algebra

Answer by 303795(562) About Me  (Show Source):
You can put this solution on YOUR website!
%28600%2Ft1%29-10=600%2F%28t2%29
Correct but things go pear shaped in the next line
%28600%2Ft2%29-%2810%2At2%29=%28600%2At1%29
Should be multiply both sides by t2 to get
%28600t2%2Ft1%29-%2810%2At2%29=%28600%29
Multiply both sides by t1 to get
%28600t2%29-%2810%2At1t2%29=%28600t1%29
Don't forget that t1 + t2=22 hours so t2=22-t1
600%2822-t1%29-10%2At1%2822-t1%29=%28600t1%29
13200-600t1-220t1%2B10t1%5E2=600t1
10t1%5E2-820t1+%2B13200=600t1
10t1%5E2-1420t1+%2B13200=0
t1%5E2-142t1+%2B1320=0
%28t1-10%29%28t1-132%29=0
ie t1=10 hours or t1 = 132 hours
The second answer is not practical (return trip in -110 hours) so the first part of the trip took 10 hours(at an average speed of 60 mph) and the second part 12 hours(at an average speed of 50 mph).