What is 2/(3x-1)< 2/(9x^2-1) ?

The solution to 2/(3x-1)< 2/(9x^2-1) is x<-1/3 \lor 0<x< 1/3

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2/(3x-1)< 2/(9x^2-1)

\frac{2}{3 x - 1} < \frac{2}{9 x ^{2} - 1}

x < - \frac{1}{3} or 0 < x < \frac{1}{3}

Interval Notation

(- \infty, - \frac{1}{3}) \cup (0, \frac{1}{3})

Decimal Notation

x < - 0.33333 \dots or 0 < x < 0.33333 \dots

Solution steps

\frac{2}{3 x - 1} < \frac{2}{9 x ^{2} - 1}

Rewrite in standard form

\frac{x}{( 3 x - 1 ) ( 3 x + 1 )} < 0

Identify the intervals

x < - \frac{1}{3} or 0 < x < \frac{1}{3}

3 f = 18

∣ 3 x - 6 ∣ \geq 21

75 x^{2} y

4 (x - 1 + i) = i (x + 5)

2 n^{2} + 5 n + 2

What is 2/(3x-1)< 2/(9x^2-1) ?
The solution to 2/(3x-1)< 2/(9x^2-1) is x<-1/3 \lor 0<x< 1/3