What is (\frac{x+1)/(x-1)+1}{(x+1)/(x-1)-1} ?

The solution to (\frac{x+1)/(x-1)+1}{(x+1)/(x-1)-1} is x

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simplify (\frac{x+1)/(x-1)+1}{(x+1)/(x-1)-1}

simplify

\frac{\frac{x + 1}{x - 1} + 1}{\frac{x + 1}{x - 1} - 1}

x

Solution steps

\frac{\frac{x + 1}{x - 1} + 1}{\frac{x + 1}{x - 1} - 1}

Join

\frac{x + 1}{x - 1} + 1 : \frac{2 x}{x - 1}

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

Join

\frac{x + 1}{x - 1} - 1 : \frac{2}{x - 1}

= \frac{\frac{2 x}{x - 1}}{\frac{2}{x - 1}}

Apply the fraction rule:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{2 x ( x - 1 )}{( x - 1 ) \cdot 2}

Cancel

\frac{2 x ( x - 1 )}{( x - 1 ) \cdot 2} : x

= x

\frac{72}{16}

- 0 + 2

(- 8 x^{2} + 3) + (- 9 x^{2} + 3 x - 3)

\frac{5 ^{2}}{5 ^{3}}

- 3 (10 b + 10) + 5 (b + 2)

What is (\frac{x+1)/(x-1)+1}{(x+1)/(x-1)-1} ?
The solution to (\frac{x+1)/(x-1)+1}{(x+1)/(x-1)-1} is x