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

x^{2} - 9 + x^{2} - 1 > 8

11 x - 9 < 3 x + 11 < 7 x + 13

- 3 < - 4 x + 1 < 13

1 \leq \frac{( 3 x ^{2} - 7 x + 8 )}{( x ^{2} + 1 )} \leq 2

2 < ∣ x ∣ \leq 5

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