Solve x/x+3>0.2

$$\frac { x } { x + 3 } > 0.2$$

Solve for x

$$x\in \left(-\infty,-3\right)\cup \left(0.75,\infty\right)$$

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.

$$x>0.2\left(x+3\right)$$

Multiply out the right hand side.

$$x>0.2x+0.6$$

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

$$x-0.2x>0.6$$

Combine like terms.

$$0.8x>0.6$$

Divide both sides by 0.8. Since 0.8 is positive, the inequality direction remains the same.

$$x>0.75$$

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

$$x>0.75$$

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.

$$x<0.2\left(x+3\right)$$

Multiply out the right hand side.

$$x<0.2x+0.6$$

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

$$x-0.2x<0.6$$

Combine like terms.

$$0.8x<0.6$$

Divide both sides by 0.8. Since 0.8 is positive, the inequality direction remains the same.

$$x<0.75$$

Consider condition x<-3 specified above.

$$x<-3$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,-3\right)\cup \left(0.75,\infty\right)$$