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- Oct 16th 2006, 08:38 PM #1

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- Oct 16th 2006, 08:42 PM #2

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- Oct 16th 2006, 08:51 PM #3

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- Oct 17th 2006, 06:26 AM #4
Note the format the Jameson used:

6/14 = 7/(x - 3)

First reduce the fraction on the LHS:

3/7 = 7/(x - 3)

Multiply both sides of the equation by 7(x - 3):

7(x - 3)*(3/7) = 7(x - 3)*[7/(x - 3)]

3(x - 3) = 7*7

3x - 9 = 49

3x = 58

x = 16

Since x is not equal to 3 (this is prohibited by the original expression) this solution is valid.

-Dan