What is x/((x^2+3x+2))> 1/((x+1)) ?

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

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x/((x^2+3x+2))> 1/((x+1))

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

- 2 < x < - 1

Interval Notation

(- 2, - 1)

Solution steps

\frac{x}{x ^{2} + 3 x + 2} > \frac{1}{x + 1}

Rewrite in standard form

\frac{1}{( x + 1 ) ( x + 2 )} < 0

\frac{1}{a} < 0

then

a < 0

(x + 1) (x + 2) < 0

Identify the intervals

- 2 < x < - 1

2 x^{2} > 3 x

\frac{- 2 ( x - 6 ) + 1}{2 \cdot ( 2 - 3 x )} \geq \frac{- 3 + 1}{2 \cdot x}

- 8 x^{4} + 6 x^{2} + 5 \leq 0

\frac{( ( x + 7 ) ^{2} )}{( ( 3 + x ) ( 2 - x ) )} \geq 0

\frac{4 ( x ^{2} + 2 )}{( x ^{2} - 1 )} < - 2

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