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

∣ 5 x ∣ < ∣ 10 - 5 x ∣

x^{2} - 3 x + ∣ x + 1 ∣ \geq 4

\frac{( x - 4 )}{( x + 2 )} \leq \frac{( x - 2 )}{( x - 1 )}

x^{2} + x - 6 < 6

∣ x - 1 ∣ \geq ∣ x - 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