What is |((x-3))/((x+1))|<= 2 ?

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

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|((x-3))/((x+1))|<= 2

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

x \leq - 5 or x \geq \frac{1}{3}

Solution steps

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

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

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

\frac{x - 3}{x + 1} \geq - 2 and \frac{x - 3}{x + 1} \leq 2

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

Merge Overlapping Intervals

x \leq - 5 or x \geq \frac{1}{3}

x^{2} + 3 x + x^{2} - 2 \geq 0

∣ x - 2 ∣ < \frac{1}{6}

x^{2} - 1 < ∣ x ∣

∣ 5 x - 10 ∣ > 0

∣ 2 x - 4 ∣ < ∣ x - 1 ∣

What is |((x-3))/((x+1))|<= 2 ?
The solution to |((x-3))/((x+1))|<= 2 is x<=-5\lor x>= 1/3