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

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

Interval Notation

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

Decimal Notation

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

Solution steps

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

Rewrite in standard form

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

Divide both sides by

9

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

Simplify

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

Identify the intervals

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

72 + 4 - 14 d = 36

- 2 \leq \frac{\frac{1}{x + 4} - \frac{1}{4}}{x} \leq x + 3

x (3 x - 4) \leq 7

1 - e^{- 2}

9 + \frac{14 - 2}{3 ^{2} + 3} - 3

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