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

The solution to 1/(x+1)>(2x)/3 is x<-(1+sqrt(7))/2 \lor-1<x<(sqrt(7)-1)/2

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

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

x < - \frac{1 + 7}{2} or - 1 < x < \frac{7 - 1}{2}

Interval Notation

(- \infty, - \frac{1 + 7}{2}) \cup (- 1, \frac{7 - 1}{2})

Decimal Notation

x < - 1.82287 \dots or - 1 < x < 0.82287 \dots

Solution steps

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

Rewrite in standard form

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

Factor

\frac{3 - 2 x ( x + 1 )}{x + 1} : \frac{- ( x - \frac{- 1 + 7}{2} ) ( x + \frac{1 + 7}{2} )}{x + 1}

\frac{- ( x - \frac{- 1 + 7}{2} ) ( x + \frac{1 + 7}{2} )}{x + 1} > 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x - \frac{- 1 + 7}{2} ) ( x + \frac{1 + 7}{2} ) ) ( - 1 )}{x + 1} < 0 \cdot (- 1)

Simplify

\frac{( x - \frac{- 1 + 7}{2} ) ( x + \frac{1 + 7}{2} )}{x + 1} < 0

Identify the intervals

x < - \frac{1 + 7}{2} or - 1 < x < \frac{7 - 1}{2}

5 x^{2} - 12 x - 9

x^{2} + 29 x + 100

\frac{5 ^{9}}{5 ^{5}}

3 x - 4 \leq 23

\frac{( 1 + \frac{y}{x} )}{( x + y )}

What is 1/(x+1)>(2x)/3 ?
The solution to 1/(x+1)>(2x)/3 is x<-(1+sqrt(7))/2 \lor-1<x<(sqrt(7)-1)/2