Solve 1+1/2x+1>0

$$1 + \frac { 1 } { 2 x + 1 } > 0$$

Solve for x

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

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2x+1}{2x+1}.

$$\frac{2x+1}{2x+1}+\frac{1}{2x+1}>0$$

Since \frac{2x+1}{2x+1} and \frac{1}{2x+1} have the same denominator, add them by adding their numerators.

$$\frac{2x+1+1}{2x+1}>0$$

Combine like terms in 2x+1+1.

$$\frac{2x+2}{2x+1}>0$$

For the quotient to be positive, 2x+2 and 2x+1 have to be both negative or both positive. Consider the case when 2x+2 and 2x+1 are both negative.

$$2x+2<0 2x+1<0$$

The solution satisfying both inequalities is x<-1.

$$x<-1$$

Consider the case when 2x+2 and 2x+1 are both positive.

$$2x+1>0 2x+2>0$$

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

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

The final solution is the union of the obtained solutions.

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