Question 127381
Given:
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{{{4ab/(a^2-b^2) - (a-b)/(a+b)}}}
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Factor the denominator of the first term using the rule {{{x^2 - y^2 = (x - y)*(x + y)}}}
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After you apply this rule the equation becomes:
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{{{4ab/((a-b)*(a+b)) - (a-b)/(a+b)}}}
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Put the second term over the common denominator by multiplying the second term by {{{(a-b)/(a-b)}}}
to get:
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{{{4ab/((a-b)*(a+b)) - ((a-b)(a-b))/((a+b)*(a-b))}}}
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Since the two terms now have a common denominator, you can combine the two terms by algebraically
adding their numerators and placing the result over the common denominator. This results in:
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{{{(4ab - (a-b)*(a-b))/((a+b)*(a-b))}}}
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Multiply out the product in the numerator to get:
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{{{(4ab - (a^2-2ab+b^2))/((a+b)*(a-b))}}}
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Get rid of the parentheses in the numerator by changing the signs of all the terms inside
the parenthesis to get:
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{{{(4ab - a^2+2ab- b^2)/((a+b)*(a-b))}}}
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Combine the two terms in the numerator that contain "ab" and the problem becomes:
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{{{( - a^2+6ab- b^2)/((a+b)*(a-b))}}}
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You can mess around with this a little and get it into various forms such as:
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{{{( -(a^2-6ab+b^2))/((a+b)*(a-b))}}}
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or
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{{{2b/(a-b) + 4b/(a+b) - 1}}}
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but these are just various forms of the above answer.
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Hope this helps you to understand the problem a little better.
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