What is 1/((x+1))>= 2/((x+2)) ?

The solution to 1/((x+1))>= 2/((x+2)) is x<-2\lor-1<x<= 0

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1/((x+1))>= 2/((x+2))

\frac{1}{( x + 1 )} \geq \frac{2}{( x + 2 )}

x < - 2 or - 1 < x \leq 0

Interval Notation

(- \infty, - 2) \cup (- 1, 0]

Solution steps

\frac{1}{x + 1} \geq \frac{2}{x + 2}

Rewrite in standard form

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

Identify the intervals

x < - 2 or - 1 < x \leq 0

\frac{( 2 x - x ^{2} + 11 )}{( x ^{2} + 2 x + 3 )} \geq 1

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

\frac{( ( n + 5 ) ( n - 9 ) )}{( n - 12 )} \geq 0

(x - 1)^{3} \cdot (x + 1) \cdot (x + 2)^{6} < 0

\frac{( x + 4 )}{( x - 3 )} > 0

What is 1/((x+1))>= 2/((x+2)) ?
The solution to 1/((x+1))>= 2/((x+2)) is x<-2\lor-1<x<= 0