math....please help!
posted by john .
rachel allows herself 1 hour to reach a sales appt 50 miles away. after she has driven 30 miles, she realizes that she must increase her speed by 15mph in order to get there on time. what was her speed for the first 30 miles?

math....please help! 
Reiny
If you had worked out my solution from several pages back
http://www.jiskha.com/display.cgi?id=1234871808
you would have realized that it was a simple typo, but my factors and my solution was correct.
I will repeat the corrected version
let her speed for the first leg be x mph
let her speed for the second leg be x+15 mph
so her time for the first leg is 30/x
her time for the second leg is 20/(x+15)
but 30/x + 20/(x+15) = 1
multiplying both sides by x(x+15) and simplifying I got
x^2 + 35x  450 = 0
(x+10)(x45) = 0
x = 10, which is silly or
x = 45 mph
check:
30/45 + 20/60 = 1 
math....please help! 
Reiny
oops, copied it and forgot to change the typo, silly me.
let her speed for the first leg be x mph
let her speed for the second leg be x+15 mph
so her time for the first leg is 30/x
her time for the second leg is 20/(x+15)
but 30/x + 20/(x+15) = 1
multiplying both sides by x(x+15) and simplifying I got
x^2  35x  450 = 0
(x+10)(x45) = 0
x = 10, which is silly or
x = 45 mph 
math....please help! 
john
thank you so much, I really appreciate your time.
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