Question 131900
Given to simplify:
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{{{(2x+6)/(x^2 - 9) + 1/(x+3)}}}
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The factor form you are looking for is:
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{{{a^2 - b^2 = (a-b)*(a+b)}}}
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The way to look at this is if you have the difference between two perfect squares, it factors into
the product of the sum and the difference of the two square roots.  Notice that {{{x^2}}} is a 
perfect square and so is {{{9}}}. So {{{x^2 - 9}}} factors into {{{(x-3)*(x+3)}}}. This makes the 
problem become:
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{{{(2x+6)/((x-3)*(x+3)) + 1/(x+3)}}}
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You can now put the second fraction over a common denominator of {{{(x-3)*(x+3)}}} by 
multiplying the second fraction by {{{(x-3)/(x-3)}}} to make the problem:
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{{{(2x+6)/((x-3)*(x+3)) + (1/(x+3))*((x-3)/(x-3))}}}
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and this makes the problem become:
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{{{(2x+6)/((x-3)*(x+3)) + (x-3)/((x-3)*(x+3))}}}
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Since the two fractions now have a common denominator, you can combine their two numerators over
the common denominator and you have:
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{{{(2x + 6 + x - 3)/((x-3)*(x+3))}}}
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In the numerator combine the 2x and + x to get 3x and also combine the +6 and -3 to get +3.
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This makes the fraction become:
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{{{(3x + 3)/((x -3)*(x+3))}}}
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Notice that the numerator can be factored by taking out a 3 to get:
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{{{(3(x+1))/((x -3)*(x+3))}}}
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This looks like answer B on your answer list.
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Hope this helps you to understand the problem and how to work it.
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