Solve x+5/2x+1<2

$$\frac { x + 5 } { 2 x + 1 } < 2$$

Solve for x

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

Solution Steps

Denominator 2x+1 cannot be zero since division by zero is not defined. There are two cases.

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

Consider the case when 2x+1 is positive. Move 1 to the right hand side.

$$2x>-1$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

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

The initial inequality does not change the direction when multiplied by 2x+1 for 2x+1>0.

$$x+5<2\left(2x+1\right)$$

Multiply out the right hand side.

$$x+5<4x+2$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$x-4x<-5+2$$

Combine like terms.

$$-3x<-3$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

$$x>1$$

Consider condition x>-\frac{1}{2} specified above. The result remains the same.

$$x>1$$

Now consider the case when 2x+1 is negative. Move 1 to the right hand side.

$$2x<-1$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

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

The initial inequality changes the direction when multiplied by 2x+1 for 2x+1<0.

$$x+5>2\left(2x+1\right)$$

Multiply out the right hand side.

$$x+5>4x+2$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$x-4x>-5+2$$

Combine like terms.

$$-3x>-3$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

$$x<1$$

Consider condition x<-\frac{1}{2} specified above.

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

The final solution is the union of the obtained solutions.

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