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?

Question

6571 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-24T13:42:53

See answer below

Given that

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

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

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

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

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

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

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

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

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