What is x/(x-1)-2/(x+1)< 8/(x^2-1) ?

The solution to x/(x-1)-2/(x+1)< 8/(x^2-1) is -2<x<-1\lor 1<x<3

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x/(x-1)-2/(x+1)< 8/(x^2-1)

\frac{x}{x - 1} - \frac{2}{x + 1} < \frac{8}{x ^{2} - 1}

- 2 < x < - 1 or 1 < x < 3

Interval Notation

(- 2, - 1) \cup (1, 3)

Solution steps

\frac{x}{x - 1} - \frac{2}{x + 1} < \frac{8}{x ^{2} - 1}

Rewrite in standard form

\frac{x ^{2} - x - 6}{( x - 1 ) ( x + 1 )} < 0

Factor

\frac{x ^{2} - x - 6}{( x - 1 ) ( x + 1 )} : \frac{( x + 2 ) ( x - 3 )}{( x - 1 ) ( x + 1 )}

\frac{( x + 2 ) ( x - 3 )}{( x - 1 ) ( x + 1 )} < 0

Identify the intervals

- 2 < x < - 1 or 1 < x < 3

7 (x - 3) + (x - 1) (x + 1) = x^{(2)} - 5 (x + 2)

\frac{x}{3} < 2

3 \cdot \frac{5}{2}

\frac{1}{\frac{7}{25}}

(x^{4})^{- \frac{1}{2}}

What is x/(x-1)-2/(x+1)< 8/(x^2-1) ?
The solution to x/(x-1)-2/(x+1)< 8/(x^2-1) is -2<x<-1\lor 1<x<3