Question 592166
Reduce to a single fraction in lowest terms:
{{{(2y)/((y^2-4))}}} + {{{1/((2-y))}}}
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Note that the 1st denominator is the "difference of squares" can be factored to
{{{(2y)/((y-2)(y+2))}}} + {{{1/((2-y))}}}
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We can make the 2nd denominator like the 1st by factoring out -1, so we have
{{{(2y)/((y-2)(y+2))}}} + {{{1/(-1(y-2))}}}
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Change the sign and we are rid of the -1
{{{(2y)/((y-2)(y+2))}}} - {{{1/((y-2))}}}
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we can see the common denominator will be (y-2)(y+2)
{{{(2y-(y+2))/((y-2)(y+2))}}} = {{{((2y-y-2))/((y-2)(y+2))}}} = {{{((y-2))/((y-2)(y+2))}}}
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Cancel (y-2), resulting in
{{{1/((y+2))}}}