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

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

Solve for x

$$x\in \left(-\infty,-7\right)\cup \left(-2,\infty\right)$$

Solution Steps

Denominator 2x+4 cannot be zero since division by zero is not defined. There are two cases.

$$2x+4>0 2x+4<0$$

Consider the case when 2x+4 is positive. Move 4 to the right hand side.

$$2x>-4$$

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

$$x>-2$$

The initial inequality does not change the direction when multiplied by 2x+4 for 2x+4>0.

$$4x-2<3\left(2x+4\right)$$

Multiply out the right hand side.

$$4x-2<6x+12$$

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

$$4x-6x<2+12$$

Combine like terms.

$$-2x<14$$

Divide both sides by -2. Since -2 is negative, the inequality direction is changed.

$$x>-7$$

Consider condition x>-2 specified above.

$$x>-2$$

Now consider the case when 2x+4 is negative. Move 4 to the right hand side.

$$2x<-4$$

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

$$x<-2$$

The initial inequality changes the direction when multiplied by 2x+4 for 2x+4<0.

$$4x-2>3\left(2x+4\right)$$

Multiply out the right hand side.

$$4x-2>6x+12$$

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

$$4x-6x>2+12$$

Combine like terms.

$$-2x>14$$

Divide both sides by -2. Since -2 is negative, the inequality direction is changed.

$$x<-7$$

Consider condition x<-2 specified above. The result remains the same.

$$x<-7$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,-7\right)\cup \left(-2,\infty\right)$$