How do you solve $-3 + \frac{1}{x+1}=\frac{2}{x}$ by finding the least common multiple?

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Answer 1 · 2018-07-24T13:42:53

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Given that

$-3+1/{x+1}=2/x$

$-3+1/{x+1}-2/x=0$

${-3x(x+1)+x-2(x+1)}/{x(x+1)}=0$

${-3x^2-3x+x-2x-2}/{x(x+1)}=0$

${-3x^2-4x-2}/{x(x+1)}=0$

$-3x^2-4x-2=0\quad (\forall \ x\ne 0, x\ne -1)$

$3x^2+4x+2=0$

$B^2-4AC=4^2-4(3)(2)=-8<0$

The given equation has no real root .

The complex roots are given by quadratic formula as follows

$x=\frac{-4\pm\sqrt{(4)^2-4(3)(2)}}{2(3)}$

$=\frac{-4\pm2i\sqrt2}{6}$

$=\frac{-2\pmi\sqrt2}{3}$

How do you solve -3 + \frac{1}{x+1}=\frac{2}{x}-3 + \frac{1}{x+1}=\frac{2}{x} by finding the least common multiple?