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subtract 4/x+1 - x+3/x
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\[\frac{4}{x+1}-\frac{x+3}{x}\]
that one?
yes
\[\frac{4x-(x+1)(x-3)}{(x+1)x}\]
then some algebra
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is it x^2 +3/ x^2 +x
numerator is \[4x-(x^2-2x-3)\] \[-x^2+6x+3\] denominator is \[x^2+x\]
for a 'final answer" of \[\frac{-x^2+6x+3}{x^2+x}\]
4x+1−x+3x
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