What is |(x+1)/(x-1)|<= 4 ?

The solution to |(x+1)/(x-1)| = 5/3

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|(x+1)/(x-1)|<= 4

\frac{x + 1}{x - 1} \leq 4

x \leq \frac{3}{5} or x \geq \frac{5}{3}

Interval Notation

(- \infty, \frac{3}{5}] \cup [\frac{5}{3}, \infty)

Decimal Notation

x \leq 0.6 or x \geq 1.66666 \dots

Solution steps

\frac{x + 1}{x - 1} \leq 4

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 4 \leq \frac{x + 1}{x - 1} \leq 4

\frac{x + 1}{x - 1} \geq - 4 and \frac{x + 1}{x - 1} \leq 4

(x \leq \frac{3}{5} or x > 1) and (x < 1 or x \geq \frac{5}{3})

Merge Overlapping Intervals

x \leq \frac{3}{5} or x \geq \frac{5}{3}

x - 3 < 5

(\frac{1}{5})^{2} \cdot (- 3)^{3}

4 \cdot 10 \frac{x}{4} = 1

\frac{3 n}{n ^{2} + 4 n + 4} + \frac{7}{n ^{2} + 6 n + 8}

- \frac{1}{3} (- 3)

What is |(x+1)/(x-1)|<= 4 ?
The solution to |(x+1)/(x-1)|<= 4 is x<= 3/5 \lor x>= 5/3