Solve x+2/x-3<2

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

Solve for x

$$x\in \left(-\infty,3\right)\cup \left(8,\infty\right)$$

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.

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

Multiply out the right hand side.

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

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

$$x-2x<-2-6$$

Combine like terms.

$$-x<-8$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x>8$$

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

$$x>8$$

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.

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

Multiply out the right hand side.

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

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

$$x-2x>-2-6$$

Combine like terms.

$$-x>-8$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x<8$$

Consider condition x<3 specified above.

$$x<3$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,3\right)\cup \left(8,\infty\right)$$