Solve x+3/x+8<2

$$\frac { x + 3 } { x + 8 } < 2$$

Solve for x

$$x\in \left(-\infty,-13\right)\cup \left(-8,\infty\right)$$

Solution Steps

Denominator x+8 cannot be zero since division by zero is not defined. There are two cases.

$$x+8>0 x+8<0$$

Consider the case when x+8 is positive. Move 8 to the right hand side.

$$x>-8$$

The initial inequality does not change the direction when multiplied by x+8 for x+8>0.

$$x+3<2\left(x+8\right)$$

Multiply out the right hand side.

$$x+3<2x+16$$

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

$$x-2x<-3+16$$

Combine like terms.

$$-x<13$$

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

$$x>-13$$

Consider condition x>-8 specified above.

$$x>-8$$

Now consider the case when x+8 is negative. Move 8 to the right hand side.

$$x<-8$$

The initial inequality changes the direction when multiplied by x+8 for x+8<0.

$$x+3>2\left(x+8\right)$$

Multiply out the right hand side.

$$x+3>2x+16$$

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

$$x-2x>-3+16$$

Combine like terms.

$$-x>13$$

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

$$x<-13$$

Consider condition x<-8 specified above. The result remains the same.

$$x<-13$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,-13\right)\cup \left(-8,\infty\right)$$