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\[6x/(x+3) + 2/(x^2-9) - 3/(x-3)\]
\[\frac{ 6x }{ (x+3) }+\frac{ 2 }{ (x ^{2}-9) }-\frac{ 3 }{ (x-3) }\]the x^2-9 is a difference of 2sqrt.. can be written (x-3)(x+3)
Yeah, I got the bottom, but can't figure out the top.
we find the common denominator...then ...
\[\frac{ 6x(x-3)+2-3(x+3) }{ (x+3)(x-3) }\]
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we can distribute the 6x to (x-3)... and the 3 to(x+3)
\[\frac{ 6x ^{2}-18x+2-3x-9 }{ (x+3)(x-3) }\]now simplify..
Ah, I see now. Thanks!
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