Question 95996
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58. {{{3/((x^2+4x+3))}}} - {{{1/((x^2-9))}}}
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Factor to:
{{{3/((x+1)(x+3))}}} - {{{1/((x+3)(x-3))}}}
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Put both fractions over a common denominator
{{{(3(x-3) - 1(x+1))/((x+1)(x+3)(x-3))}}}
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{{{((3x - 9 - x - 1))/((x+1)(x+3)(x-3))}}}
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{{{((2x - 10))/((x+1)(x+3)(x-3))}}} = {{{(2(x - 5))/((x+1)(x+3)(x-3))}}}; simplified expression
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62. 
{{{(4y)/((y^2+6y+5))}}}+{{{(2y)/((y^2-1))}}}
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Factor:
{{{(4y)/((y+1)(y+5))}}}+ {{{(2y)/((y+1)(y-1))}}}
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both fractions over a common denominator:
{{{(4y(y-1) + 2y(y+5))/((y+1)(y+5)(y-1))}}}
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{{{(4y^2 - 4y + 2y^2 + 10y)/((y+1)(y+5)(y-1))}}}
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{{{(6y^2 + 6y)/((y+1)(y+5)(y-1))}}}
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Factor out 6y
{{{(6y(y + 1))/((y+1)(y+5)(y-1))}}}
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Cancel out the (y+1)'s
{{{(6y)/((y+5)(y-1))}}}
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36. (5a-12)/(a^2-8a+15)-(3a-2)/(a^2-8a+15)...(I have 2/(a-3)as the answer,is it correct?) 
Hey! That is the right answer, way to go!
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24. x^2/(x-3)-9/(x-3)...(I have x+3 for the answer is it correct?)
Right again!  You just need a little help.