Solve x-2/2x+1<3

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

Solve for x

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

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.

$$x-2<3\left(2x+1\right)$$

Multiply out the right hand side.

$$x-2<6x+3$$

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

$$x-6x<2+3$$

Combine like terms.

$$-5x<5$$

Divide both sides by -5. Since -5 is negative, the inequality direction is changed.

$$x>-1$$

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

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

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.

$$x-2>3\left(2x+1\right)$$

Multiply out the right hand side.

$$x-2>6x+3$$

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

$$x-6x>2+3$$

Combine like terms.

$$-5x>5$$

Divide both sides by -5. Since -5 is negative, the inequality direction is changed.

$$x<-1$$

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

$$x<-1$$

The final solution is the union of the obtained solutions.

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