Posted by **john** on Tuesday, February 17, 2009 at 11:16pm.

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**, Tuesday, February 17, 2009 at 11:35pm
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)(x-45) = 0

x = -10, which is silly or

x = 45 mph

check:

30/45 + 20/60 = 1

- math....please help! -
**Reiny**, Tuesday, February 17, 2009 at 11:36pm
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)(x-45) = 0

x = -10, which is silly or

x = 45 mph

- math....please help! -
**john**, Tuesday, February 17, 2009 at 11:41pm
thank you so much, I really appreciate your time.

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