What is |(3x)/(x^2-4)|<= 1 ?

The solution to |(3x)/(x^2-4)| = 4

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|(3x)/(x^2-4)|<= 1

\frac{3 x}{x ^{2} - 4} \leq 1

x \leq - 4 or - 1 \leq x \leq 1 or x \geq 4

Interval Notation

(- \infty, - 4] \cup [- 1, 1] \cup [4, \infty)

Solution steps

\frac{3 x}{x ^{2} - 4} \leq 1

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 1 \leq \frac{3 x}{x ^{2} - 4} \leq 1

\frac{3 x}{x ^{2} - 4} \geq - 1 and \frac{3 x}{x ^{2} - 4} \leq 1

(x \leq - 4 or - 2 < x \leq 1 or x > 2) and (x < - 2 or - 1 \leq x < 2 or x \geq 4)

Merge Overlapping Intervals

x \leq - 4 or - 1 \leq x \leq 1 or x \geq 4

∣ 2 x - 5 ∣ \geq ∣ x + 2 ∣

∣ x ∣ \leq ∣ x + 1 ∣

\frac{( x ^{2} - ∣ x ∣ - 5 )}{( ∣ x ∣ - 7 )} \geq 2 x

∣ 3 x - 2 ∣ < x + 3

∣ x - 3 ∣ > x^{2} - 3

What is |(3x)/(x^2-4)|<= 1 ?
The solution to |(3x)/(x^2-4)|<= 1 is x<=-4\lor-1<= x<= 1\lor x>= 4