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

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

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Answer 1 · 2017-07-21T18:00:00

$x = - \frac{1}{10} \pm \frac{\sqrt{10}}{5}$ $x=-1/10pmsqrt10/5$

Explanation:

First our restrictions :

$x \neq 0 and x \neq 1 and x \neq - 1$ $x!=0andx!=1andx!=-1$

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

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

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

$(x + 1) (x + 1) x + 2 (x - 1) (x + 1) - x (x - 1) x = 0 \Leftrightarrow$ $(x+1)(x+1)x+2(x-1)(x+1)-x(x-1)x=0iff$

${(x + 1)}^{2} x + 2 (x^{2} - 1) - x^{2} (x - 1) = 0 \Leftrightarrow$ $(x+1)^2x+2(x^2-1)-x^2(x-1)=0iff$

$(x^{2} + 2 x + 1) x + 2 x^{2} - 2 - x^{3} + x^{2} = 0 \Leftrightarrow$ $(x^2+2x+1)x+2x^2-2-x^3+x^2=0 iff$

$x^{3} + 2 x^{2} + x + 2 x^{2} - 2 - x^{3} + x^{2} = 0 \Leftrightarrow$ $x^3+2x^2+x+2x^2-2-x^3+x^2=0iff$

$5 x^{2} + x - 2 = 0 \Leftrightarrow$ $5x^2+x-2=0 iff$

$x = \frac{- 1 \pm \sqrt{40}}{10} = - \frac{1}{10} \pm \frac{2}{10} \sqrt{10} = - \frac{1}{10} \pm \frac{\sqrt{10}}{5}$ $x=(-1pmsqrt(40))/10=-1/10pm2/10sqrt10=-1/10pmsqrt10/5$

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

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

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