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solve the equation: 3/(x+1) = 1/(x^2-1)
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\[\frac{3}{x+1}=\frac{1}{x^2-1}\] Cross multiply
Notice x^2 - 1
\[3x^2-3=x+1\]
\[3x^2-x-4=0\]
\[(3x-4)(x+1)=0\]
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\[3x-4=0\]
\[x=\frac{4}{3}\]
\[x+1=0\] \[x=-1\]
But of course x cannot be -1 because then x^2-1 (the denominator) would be 0 so discard that and x = 4/3
Thank you. Can you help me with this one? 3/(x) = 12/x+7
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Same thing. Cross multiply. What do you get?
3x+21=12x
\[12x=3x+21\]
Yes. Not finish it up.
3x-12x+21 -9x=21
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That would be positive 9x
oh
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