What is 5/((3x+1))-(20)/((9x^2-1))< 2/((3x-1)) ?

The solution to 5/((3x+1))-(20)/((9x^2-1))< 2/((3x-1)) is x<-1/3 \lor 1/3 <x<3

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5/((3x+1))-(20)/((9x^2-1))< 2/((3x-1))

\frac{5}{( 3 x + 1 )} - \frac{20}{( 9 x ^{2} - 1 )} < \frac{2}{( 3 x - 1 )}

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

Interval Notation

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

Decimal Notation

x < - 0.33333 \dots or 0.33333 \dots < x < 3

Solution steps

\frac{5}{3 x + 1} - \frac{20}{9 x ^{2} - 1} < \frac{2}{3 x - 1}

Rewrite in standard form

\frac{9 x - 27}{( 3 x + 1 ) ( 3 x - 1 )} < 0

Factor

\frac{9 x - 27}{( 3 x + 1 ) ( 3 x - 1 )} : \frac{9 ( x - 3 )}{( 3 x + 1 ) ( 3 x - 1 )}

\frac{9 ( x - 3 )}{( 3 x + 1 ) ( 3 x - 1 )} < 0

Divide both sides by

9

\frac{\frac{9 ( x - 3 )}{( 3 x + 1 ) ( 3 x - 1 )}}{9} < \frac{0}{9}

Simplify

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

Identify the intervals

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

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

\frac{( 5 x - 1 )}{( x ^{2} + 3 )} < 1

9 x^{2} - 8 x - 1 > 0

\frac{1}{3 ( 9 x + 27 )} > x + 33

\frac{\frac{x}{l o g _{10} ( 1 )}}{2} (2 x + 3) > 0

What is 5/((3x+1))-(20)/((9x^2-1))< 2/((3x-1)) ?
The solution to 5/((3x+1))-(20)/((9x^2-1))< 2/((3x-1)) is x<-1/3 \lor 1/3 <x<3