What is (\frac{x+3)/(x+1)-2}{x-1} ?

The solution to (\frac{x+3)/(x+1)-2}{x-1} is -1/(x+1)

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simplify (\frac{x+3)/(x+1)-2}{x-1}

simplify

\frac{\frac{x + 3}{x + 1} - 2}{x - 1}

- \frac{1}{x + 1}

Solution steps

\frac{\frac{x + 3}{x + 1} - 2}{x - 1}

Join

\frac{x + 3}{x + 1} - 2 : \frac{- x + 1}{x + 1}

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

Apply the fraction rule:

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

= \frac{- x + 1}{( x + 1 ) ( x - 1 )}

Cancel

\frac{- x + 1}{( x + 1 ) ( x - 1 )} : - \frac{1}{x + 1}

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

0.2 (0.05 + 0.1)

(3 x + 1) (3 x + 8)

5 - (t - 1)^{2}

2^{- 3} + (- 3)^{2}

\frac{5}{8} \div \frac{2}{3}

What is (\frac{x+3)/(x+1)-2}{x-1} ?
The solution to (\frac{x+3)/(x+1)-2}{x-1} is -1/(x+1)