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March 4, 2015

March 4, 2015

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

- math....please help! -
**Reiny**, Tuesday, February 17, 2009 at 11:35pmIf 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 11:36pmoops, 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 11:41pmthank you so much, I really appreciate your time.

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