Solve 3x-1/8x-3>0

$$\frac { 3 x - 1 } { 8 x - 3 } > 0$$

Solve for x

$$x\in \left(-\infty,\frac{1}{3}\right)\cup \left(\frac{3}{8},\infty\right)$$

Solution Steps

For the quotient to be positive, 3x-1 and 8x-3 have to be both negative or both positive. Consider the case when 3x-1 and 8x-3 are both negative.

$$3x-1<0 8x-3<0$$

The solution satisfying both inequalities is x<\frac{1}{3}.

$$x<\frac{1}{3}$$

Consider the case when 3x-1 and 8x-3 are both positive.

$$8x-3>0 3x-1>0$$

The solution satisfying both inequalities is x>\frac{3}{8}.

$$x>\frac{3}{8}$$

The final solution is the union of the obtained solutions.

$$x<\frac{1}{3}\text{; }x>\frac{3}{8}$$