Question 132040
Given:
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{{{(2x+1)/(x^2-16) + (x^2 - 6x - 7)/(x^2 - 11x + 28)}}}
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Factor the denominators of both fractions and the numerator of the second fractions. They 
factor as follows:
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{{{x^2 - 16 = (x -4)*(x + 4)}}}
{{{x^2 - 11x + 28 = (x - 7)*(x - 4)}}}
{{{x^2 - 6x - 7 = (x - 7)*(x + 1)}}}
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Substitute these factors at the appropriate places in the original equation and you get:
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{{{(2x + 1)/((x-4)*(x+4)) + ((x-7)*(x+1))/((x-7)*(x-4)))}}}
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Notice that in the second fraction the term (x -7) appears in both the numerator and the
denominator. Therefore, it can be canceled out as follows:
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{{{(2x + 1)/((x-4)*(x+4)) + (cross(x-7)*(x+1))/(cross(x-7)*(x-4))}}}
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and the problem is then reduced to:
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{{{(2x + 1)/((x-4)*(x+4)) + (x+1)/(x-4)}}}
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The second fraction can be put over a common denominator of {{{(x-4)*(x+4)}}} by multiplying 
it by {{{(x+4)/(x+4)}}}. Notice that since the numerator of this multiplier is the same as 
the denominator, the multiplier is equivalent to 1 so it doesn't really change the second fraction. 
The multiplication is as follows:
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{{{(2x + 1)/((x-4)*(x+4)) + ((x+1)/(x-4))*((x+4)/(x+4))}}}
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and this becomes:
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{{{(2x + 1)/((x-4)*(x+4)) + ((x+1)*(x+4))/((x-4)*(x+4))}}}
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Notice now that the two fractions have the common denominator of {{{(x-4)*(x+4)}}}
so the numerators can be combined over this common denominator to get:
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{{{(2x + 1+ (x+1)*(x+4))/((x-4)*(x+4))}}}
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Simplify the numerator by multiplying out the two factors to make the expression become:
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{{{(2x + 1+ x^2 + 5x + 4)/((x-4)*(x+4))}}}
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In the numerator combine the 2x and the 5x to get +7x and combine the 1 and the 4 to get +5.
Substitute these results and you get the answer of:
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{{{(x^2+7x + 5)/((x-4)*(x+4))}}}
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This is the answer. The numerator does not factor. 
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Or you can, if you so desire, multiply out the denominator to get the answer of:
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{{{(x^2+7x + 5)/(x^2 - 16)}}}
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Hope this helps you to see your way through the problem.
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