Solve x/x+3>0.9

$$\frac{ x }{ x+3 } > 0.9$$

Solve for x

$$x\in \left(-\infty,-3\right)\cup \left(27,\infty\right)$$

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.

$$x>0.9\left(x+3\right)$$

Multiply out the right hand side.

$$x>0.9x+2.7$$

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

$$x-0.9x>2.7$$

Combine like terms.

$$0.1x>2.7$$

Divide both sides by 0.1. Since 0.1 is positive, the inequality direction remains the same.

$$x>27$$

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

$$x>27$$

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.

$$x<0.9\left(x+3\right)$$

Multiply out the right hand side.

$$x<0.9x+2.7$$

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

$$x-0.9x<2.7$$

Combine like terms.

$$0.1x<2.7$$

Divide both sides by 0.1. Since 0.1 is positive, the inequality direction remains the same.

$$x<27$$

Consider condition x<-3 specified above.

$$x<-3$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,-3\right)\cup \left(27,\infty\right)$$