document.write( "Question 7128: ((2x-14)/2x)*((6x^2)/X^2-49))=\r
\n" ); document.write( "\n" ); document.write( "I have broken the question down to:
\n" ); document.write( "((2(x-7)/2x))*((3x*3x)/(x(x-49))
\n" ); document.write( "and this is where I am lost.....
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Algebra.Com's Answer #3931 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Let's see:\r
\n" ); document.write( "\n" ); document.write( "Simplify: \"%28%282x-14%29%2F%282x%29%29%2A%28%286x%5E2%29%2F%28x%5E2-49%29%29\"\r
\n" ); document.write( "\n" ); document.write( "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\" \r
\n" ); document.write( "\n" ); document.write( "\"%282%28x-7%29%2F%282x%29%29%2A%28%286x%5E2%29%2F%28x%2B7%29%28x-7%29%29\"\r
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "\"%281%2F1%29%2A%286x%2F%28x%2B7%29%29\" = \"6x%2F%28x%2B7%29\" and this is as far as you can go.
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