Solve 1-2x/3+4x<0

$$\frac{ 1-2x }{ 3+4x } < 0$$

Solve for x

$$x\in \left(-\infty,-\frac{3}{4}\right)\cup \left(\frac{1}{2},\infty\right)$$

Solution Steps

For the quotient to be negative, 1-2x and 4x+3 have to be of the opposite signs. Consider the case when 1-2x is positive and 4x+3 is negative.

$$1-2x>0 4x+3<0$$

The solution satisfying both inequalities is x<-\frac{3}{4}.

$$x<-\frac{3}{4}$$

Consider the case when 4x+3 is positive and 1-2x is negative.

$$4x+3>0 1-2x<0$$

The solution satisfying both inequalities is x>\frac{1}{2}.

$$x>\frac{1}{2}$$

The final solution is the union of the obtained solutions.

$$x<-\frac{3}{4}\text{; }x>\frac{1}{2}$$