SOLUTION: A man drives 600 miles to a convention. on the return trip he increases his speed by 10 mph and saves 2 hours driving time. how fast did he drive going to and from the convention

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Question 96803: A man drives 600 miles to a convention. on the return trip he increases his speed by 10 mph and saves 2 hours driving time. how fast did he drive going to and from the convention?
Not from a textbook..a handed out assignment.
Found 2 solutions by checkley71, scott8148:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
(600/x)-2=600/(x+10)
(600-2x)/x=600/(x+10) cross multiply
(600-2x)(x+10)=600x
600x-2x^2-20x+6000=600x
2x^2+20x-6000=0
x^2+10x-3000=0
(x+60)(x-50)=0
x-50=0
x=50 for the speed going to the convention
50+10=60 return trip speed.
proof
(600/50)-2=600/60
12-2=10
10=10

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
time=distance/rate ... let x=(speed to the convention)

600/x=(600/(x+10))+2 ... (600-2x)/x=600/(x+10) ... cross multiplying gives 6000+580x-2x^2=600x

combining and rearranging terms gives -2x^2-20x+6000=0 ... x^2+10x-3000=0

factoring gives (x+60)(x-50)=0 ... so x=-60 and x=50 ... negative value not realistic

50 mph to and 60 mph from