Question 7128
Let's see:

Simplify: {{{((2x-14)/(2x))*((6x^2)/(x^2-49))}}}

Your first step was ok, factor out the 2 in the first fraction. The second part was not quite right.  In the second fraction, you can factor the binomial in the denominator: {{{(x^2-49) = (x+7)(x-7)}}} 

{{{(2(x-7)/(2x))*((6x^2)/(x+7)(x-7))}}}

In the first fraction, cancel the 2's, then cancel the (x-7)'s, finally, cancel an x from the denominator of the first fraction and the numerator of the second fraction.

{{{(1/1)*(6x/(x+7))}}} = {{{6x/(x+7)}}} and this is as far as you can go.