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

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(8007) (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

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