What is ((2x))/((2x^2+5x+2))> 1/((x+1)) ?

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

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((2x))/((2x^2+5x+2))> 1/((x+1))

\frac{( 2 x )}{( 2 x ^{2} + 5 x + 2 )} > \frac{1}{( x + 1 )}

- 2 < x < - 1 or - \frac{2}{3} < x < - \frac{1}{2}

Interval Notation

(- 2, - 1) \cup (- \frac{2}{3}, - \frac{1}{2})

Decimal Notation

- 2 < x < - 1 or - 0.66666 \dots < x < - 0.5

Solution steps

\frac{2 x}{2 x ^{2} + 5 x + 2} > \frac{1}{x + 1}

Rewrite in standard form

\frac{- 3 x - 2}{( x + 2 ) ( x + 1 ) ( 2 x + 1 )} > 0

Identify the intervals

- 2 < x < - 1 or - \frac{2}{3} < x < - \frac{1}{2}

\frac{1}{2 ( 3 - 8 x )} > 20 (\frac{1 - 1}{5 x})

\frac{1}{( x - 2 )} \leq 0

\frac{1}{( x ^{2} + x )} \leq \frac{1}{( 2 x ^{2} + 2 x + 3 )}

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

(x - 3) (x - 7) (x - 12) < 0

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