Solve x+27/x+6<9

$$\frac { x + 27 } { x + 6 } < 9$$

Solve for x

$$x\in \left(-\infty,-6\right)\cup \left(-\frac{27}{8},\infty\right)$$

Solution Steps

Denominator x+6 cannot be zero since division by zero is not defined. There are two cases.

$$x+6>0 x+6<0$$

Consider the case when x+6 is positive. Move 6 to the right hand side.

$$x>-6$$

The initial inequality does not change the direction when multiplied by x+6 for x+6>0.

$$x+27<9\left(x+6\right)$$

Multiply out the right hand side.

$$x+27<9x+54$$

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

$$x-9x<-27+54$$

Combine like terms.

$$-8x<27$$

Divide both sides by -8. Since -8 is negative, the inequality direction is changed.

$$x>-\frac{27}{8}$$

Consider condition x>-6 specified above. The result remains the same.

$$x>-\frac{27}{8}$$

Now consider the case when x+6 is negative. Move 6 to the right hand side.

$$x<-6$$

The initial inequality changes the direction when multiplied by x+6 for x+6<0.

$$x+27>9\left(x+6\right)$$

Multiply out the right hand side.

$$x+27>9x+54$$

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

$$x-9x>-27+54$$

Combine like terms.

$$-8x>27$$

Divide both sides by -8. Since -8 is negative, the inequality direction is changed.

$$x<-\frac{27}{8}$$

Consider condition x<-6 specified above.

$$x<-6$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,-6\right)\cup \left(-\frac{27}{8},\infty\right)$$