What is (|x+1|)/(x+1)> 1/(x-1) ?

The solution to (|x+1|)/(x+1)> 1/(x-1) is x>-1

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(|x+1|)/(x+1)> 1/(x-1)

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

x > - 1

Interval Notation

(- 1, \infty)

Solution steps

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

Move

\frac{1}{x - 1}

to the left side

\frac{∣ x + 1 ∣}{x + 1} - \frac{1}{x - 1} > 0

∣ x + 1 ∣ \geq 0

\frac{a}{b} > 0 and a \geq 0

then

a > 0 and b > 0

∣ x + 1 ∣ > 0 and (x + 1) > 0

∣ x + 1 ∣ > 0 : x < - 1 or x > - 1

x + 1 > 0 : x > - 1

Combine the intervals

(x < - 1 or x > - 1) and x > - 1

Merge Overlapping Intervals

x > - 1

\frac{1}{2} + (\frac{1}{3} + \frac{4}{5})^{2}

\frac{a ^{- 1} + b ^{- 1}}{a ^{- 2} - b ^{- 2}}

3 x + 2 + x = x + 5

ln (x) - 6 ln (y) + ln (z)

8 = \frac{z}{6}