What is 2/(3x+3)> 4/(3x-1) ?

The solution to 2/(3x+3)> 4/(3x-1) is x<-7/3 \lor-1<x< 1/3

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2/(3x+3)> 4/(3x-1)

\frac{2}{3 x + 3} > \frac{4}{3 x - 1}

x < - \frac{7}{3} or - 1 < x < \frac{1}{3}

Interval Notation

(- \infty, - \frac{7}{3}) \cup (- 1, \frac{1}{3})

Decimal Notation

x < - 2.33333 \dots or - 1 < x < 0.33333 \dots

Solution steps

\frac{2}{3 x + 3} > \frac{4}{3 x - 1}

Rewrite in standard form

\frac{- 6 x - 14}{( x + 1 ) ( 3 x - 1 )} > 0

Factor

\frac{- 6 x - 14}{( x + 1 ) ( 3 x - 1 )} : \frac{- 2 ( 3 x + 7 )}{( x + 1 ) ( 3 x - 1 )}

\frac{- 2 ( 3 x + 7 )}{( x + 1 ) ( 3 x - 1 )} > 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - 2 ( 3 x + 7 ) ) ( - 1 )}{( x + 1 ) ( 3 x - 1 )} < 0 \cdot (- 1)

Simplify

\frac{2 ( 3 x + 7 )}{( x + 1 ) ( 3 x - 1 )} < 0

Divide both sides by

2

\frac{\frac{2 ( 3 x + 7 )}{( x + 1 ) ( 3 x - 1 )}}{2} < \frac{0}{2}

Simplify

\frac{3 x + 7}{( x + 1 ) ( 3 x - 1 )} < 0

Identify the intervals

x < - \frac{7}{3} or - 1 < x < \frac{1}{3}

\frac{- 3 ( 4 - 2 x ) + 1}{2 ( x + 6 )} > 6 x - 8

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

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

(x + 3) (5 - 2 x) + (2 x + 1) (x - 2) < \frac{x}{2}

\frac{( x ^{2} - 5 x + 6 )}{( x ^{2} - 7 x + 12 )} \leq 0

What is 2/(3x+3)> 4/(3x-1) ?
The solution to 2/(3x+3)> 4/(3x-1) is x<-7/3 \lor-1<x< 1/3