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

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Answer 1 · 2018-08-10T20:41:09

$x = - 1$ $x=-1$

Here,

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

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

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

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

$\Rightarrow (1 + \frac{1}{x}) [\frac{1}{2} - 1] = 0$ $=>(1+1/x)[1/2-1]=0$

$\Rightarrow (1 + \frac{1}{x}) [- \frac{1}{2}] = 0$ $=>(1+1/x)[-1/2]=0$

$=>1+1/x=0to[because-1/2!=0]$

$=>1/x=-1$

$=>x=-1$

Answer 2 · 2018-08-11T11:55:17

$x=-1$

$1/2*(1+1/x)-2=(1-x)/x$

$1/2*(1+1/x)-2=1/x-1$

$1/2*(1+1/x)=1/x+1$

$1+1/x=2/x+2$

$1/x=-1$

$x=1/(-1)=-1$

How do you solve \frac { 1} { 2} ( 1+ \frac { 1} { x } ) - 2= \frac { 1- x } { x }12(1+1x)−2=1−xx\frac { 1} { 2} ( 1+ \frac { 1} { x } ) - 2= \frac { 1- x } { x }?