What is 3x-1< 2/((x+2)) ?

The solution to 3x-1< 2/((x+2)) is x<(-5-sqrt(73))/6 \lor-2<x<(sqrt(73)-5)/6

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3x-1< 2/((x+2))

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

x < \frac{- 5 - 73}{6} or - 2 < x < \frac{73 - 5}{6}

Interval Notation

(- \infty, \frac{- 5 - 73}{6}) \cup (- 2, \frac{73 - 5}{6})

Decimal Notation

x < - 2.25733 \dots or - 2 < x < 0.59066 \dots

Solution steps

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

Rewrite in standard form

\frac{3 x ^{2} + 5 x - 4}{x + 2} < 0

Factor

3 x^{2} + 5 x - 4 : (x - \frac{- 5 + 73}{6}) (x - \frac{- 5 - 73}{6})

\frac{( x - \frac{- 5 + 73}{6} ) ( x - \frac{- 5 - 73}{6} )}{x + 2} < 0

Identify the intervals

x < \frac{- 5 - 73}{6} or - 2 < x < \frac{73 - 5}{6}

lo g_{10} (0.5) (\frac{x}{( x + 1 )}) < - 1

\frac{x}{( 4 x + 9 ) ( 3 - 7 x )} > 0

- 2 x^{2} - 2 x - 1 < 0

\frac{( 4 x + 3 )}{( 2 x - 5 )} < 6

\frac{( x - 4 )}{( 2 x - 3 )} < 3

What is 3x-1< 2/((x+2)) ?
The solution to 3x-1< 2/((x+2)) is x<(-5-sqrt(73))/6 \lor-2<x<(sqrt(73)-5)/6