# SOLUTION: I have tried to solve this using several equations, none seem to work. The one that I think came the closest was (this did not come from a math textbook, it came from an online as

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 Word Problems: Travel and Distance Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Travel Word Problems Question 106224: I have tried to solve this using several equations, none seem to work. The one that I think came the closest was (this did not come from a math textbook, it came from an online assignment for intermediate algebra): when I solved it I got x = -50 and x = 40 Here is the word problem: Steve traveled 200 miles at a certain speed. Had he gone 10 mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle. Thanks for your help!! Answer by edjones(7569)   (Show Source): You can put this solution on YOUR website!d=st d=distance, s=speed, t=time d=200 miles, s=Steve's speed in mph, t= his time in hrs The distance of both trips is the same. So,: st=200 t=200/s (s+10)(t-1)=200 st+10t-s-10=200 s*200/s+10*200/s-s-10=200 plug 200/s for every t in the above equation. 200+2000/s-s-10=200 2000/s-s-10=0 -s^2-10s+2000=0 multiply both sides by s to eliminate the fraction. s^2+10s-2000=0 multiply both sides times -1 (s+50)(s-40)=0 s can't be a negative number. s=40 mph Check: at 40 mph it takes 5 hrs to go 200 mi at 50 mph it takes 1 hr less you had the right answer all the time and didn't know it! Ed