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

The solution to |((x^2-1))/((3-2x-x^2))| =-2

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

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

x \geq - 2

Interval Notation

[- 2, \infty)

Solution steps

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

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

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

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

x > - 3 and (x < - 3 or x \geq - 2)

Merge Overlapping Intervals

x \geq - 2

∣ 4 x - 5 ∣ > 8

\frac{( x + 2 )}{5} < 1

\frac{( x ^{2} - 5 x + 4 )}{( x ^{2} - 4 )} \geq 1

\frac{( x ^{2} + 12 )}{( 7 x )} \geq 1

\frac{( ∣ x ∣ - 4 )}{( ∣ x ∣ - 5 )} \leq 0

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