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(8/x-2) + (2/x+3)
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\[\frac{8}{x-2}+\frac{2}{x+3}\] like that ?
Yes they want the sum of the equation
you need to find a common denominator. Do you know how to do it?
umm nooo....sorry
to add fractions you need \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\]
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put \(a=8,b=x-2,c=2,d=x+3\) and get \[\frac{8(x+3)+2(x-2)}{(x-2)(x-3)}\]
then you can clean up the numerator with some algebra
(8x+24) + (2x+4) for the top, yes?
Help please
no it would be \[8x+24+2x-4\]
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then combine like terms to get \[10x+20\]
so the bottom would be some x^2 -6
i was trying to put the bottom together like i did with the top, there are 2 x's in the bottom equation.
Sorry! You have done everything corret!
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