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can anyone solve for v? -3v/v+3=-9/v^2+v-6
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-3v^3 -3v^2 + 6v = -9v - 27
is right hand side = -9 / (v^2+v-6) ?
what i mean is are all three terms in the bottom of fraction?
are u there nest?
\[\frac{-3v}{v+3}=\frac{-9}{v ^{2}+v-6}\]\[\frac{-3v}{v+3}=\frac{-9}{(v+3)(v-2)}\]\[-3v(v-2)=-3v ^{2}+6v\]\[-3v ^{2}+6v=-9\]\[-3v^2+6v+9=0\]\[-3(v ^{2}-2v-3)=0\]\[(v+1)(v-3)=0\]\[v=3 \]\[ v =-1\]
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