Solve x/x-3<4

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

Solve for x

$$x\in \left(-\infty,3\right)\cup \left(4,\infty\right)$$

Solution Steps

Denominator x-3 cannot be zero since division by zero is not defined. There are two cases.

$$x-3>0 x-3<0$$

Consider the case when x-3 is positive. Move -3 to the right hand side.

$$x>3$$

The initial inequality does not change the direction when multiplied by x-3 for x-3>0.

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

Multiply out the right hand side.

$$x<4x-12$$

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

$$x-4x<-12$$

Combine like terms.

$$-3x<-12$$

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

$$x>4$$

Consider condition x>3 specified above. The result remains the same.

$$x>4$$

Now consider the case when x-3 is negative. Move -3 to the right hand side.

$$x<3$$

The initial inequality changes the direction when multiplied by x-3 for x-3<0.

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

Multiply out the right hand side.

$$x>4x-12$$

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

$$x-4x>-12$$

Combine like terms.

$$-3x>-12$$

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

$$x<4$$

Consider condition x<3 specified above.

$$x<3$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,3\right)\cup \left(4,\infty\right)$$