What is ((2x))/((x^2-9))<= 1/((x+2)) ?

The solution to ((2x))/((x^2-9))<= 1/((x+2)) is x<-3\lor-2<x<3

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((2x))/((x^2-9))<= 1/((x+2))

\frac{( 2 x )}{( x ^{2} - 9 )} \leq \frac{1}{( x + 2 )}

x < - 3 or - 2 < x < 3

Interval Notation

(- \infty, - 3) \cup (- 2, 3)

Solution steps

\frac{2 x}{x ^{2} - 9} \leq \frac{1}{x + 2}

Rewrite in standard form

\frac{x ^{2} + 4 x + 9}{( x + 3 ) ( x - 3 ) ( x + 2 )} \leq 0

Identify the intervals

x < - 3 or - 2 < x < 3

x^{4} - x^{2} - 12 \leq 0

\frac{( 3 x - 1 )}{( 2 x )} \geq 1

2 x \cdot (x - 2) < x \cdot (x + 4)

4 x - x^{2} \geq 0

(1 - x^{2}) > \frac{9}{25}

What is ((2x))/((x^2-9))<= 1/((x+2)) ?
The solution to ((2x))/((x^2-9))<= 1/((x+2)) is x<-3\lor-2<x<3