What is (-3x)/(x^2-9)>0 ?

The solution to (-3x)/(x^2-9)>0 is x<-3\lor 0<x<3

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(-3x)/(x^2-9)>0

\frac{- 3 x}{x ^{2} - 9} > 0

x < - 3 or 0 < x < 3

Interval Notation

(- \infty, - 3) \cup (0, 3)

Solution steps

\frac{- 3 x}{x ^{2} - 9} > 0

Rewrite in standard form

\frac{x}{x ^{2} - 9} < 0

Factor

\frac{x}{x ^{2} - 9} : \frac{x}{( x + 3 ) ( x - 3 )}

\frac{x}{( x + 3 ) ( x - 3 )} < 0

Identify the intervals

x < - 3 or 0 < x < 3

\frac{x - 8}{x + 2} \geq 0

\frac{4}{2 20}

x + 2 \geq 4

45 n^{2} + 35 n + 140

\frac{5}{11}