How do you solve \frac { 1} { 2} ( 1+ \frac { 1} { x } ) - 2= \frac { 1- x } { x }?

Aug 10, 2018

$x = - 1$

Explanation:

Here,

$\frac{1}{2} \left(1 + \frac{1}{x}\right) - 2 = \frac{1}{x} - 1$

$\implies \frac{1}{2} \left(1 + \frac{1}{x}\right) - 2 - \frac{1}{x} + 1 = 0$

$\implies \frac{1}{2} \left(1 + \frac{1}{x}\right) - 1 - \frac{1}{x} = 0$

$\implies \frac{1}{2} \left(1 + \frac{1}{x}\right) - \left(1 + \frac{1}{x}\right) = 0$

$\implies \left(1 + \frac{1}{x}\right) \left[\frac{1}{2} - 1\right] = 0$

$\implies \left(1 + \frac{1}{x}\right) \left[- \frac{1}{2}\right] = 0$

$\implies 1 + \frac{1}{x} = 0 \to \left[\because - \frac{1}{2} \ne 0\right]$

$\implies \frac{1}{x} = - 1$

$\implies x = - 1$

Aug 11, 2018

$x = - 1$

Explanation:

$\frac{1}{2} \cdot \left(1 + \frac{1}{x}\right) - 2 = \frac{1 - x}{x}$

$\frac{1}{2} \cdot \left(1 + \frac{1}{x}\right) - 2 = \frac{1}{x} - 1$

$\frac{1}{2} \cdot \left(1 + \frac{1}{x}\right) = \frac{1}{x} + 1$

$1 + \frac{1}{x} = \frac{2}{x} + 2$

$\frac{1}{x} = - 1$

$x = \frac{1}{- 1} = - 1$