SOLUTION: ((2x-14)/2x)*((6x^2)/X^2-49))= I have broken the question down to: ((2(x-7)/2x))*((3x*3x)/(x(x-49)) and this is where I am lost.....

Algebra ->  Algebra  -> Numeric Fractions Calculators, Lesson and Practice  -> Lessons -> SOLUTION: ((2x-14)/2x)*((6x^2)/X^2-49))= I have broken the question down to: ((2(x-7)/2x))*((3x*3x)/(x(x-49)) and this is where I am lost.....      Log On

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Question 7128: ((2x-14)/2x)*((6x^2)/X^2-49))=
I have broken the question down to:
((2(x-7)/2x))*((3x*3x)/(x(x-49))
and this is where I am lost.....
Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
Let's see:
Simplify: %28%282x-14%29%2F%282x%29%29%2A%28%286x%5E2%29%2F%28x%5E2-49%29%29
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: %28x%5E2-49%29+=+%28x%2B7%29%28x-7%29
%282%28x-7%29%2F%282x%29%29%2A%28%286x%5E2%29%2F%28x%2B7%29%28x-7%29%29
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.
%281%2F1%29%2A%286x%2F%28x%2B7%29%29 = 6x%2F%28x%2B7%29 and this is as far as you can go.