Solve x-2/x>1/x

$$\frac{ x-2 }{ x } > \frac{ 1 }{ x }$$

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(3,\infty\right)$$

Solution Steps

Subtract \frac{1}{x} from both sides.

$$\frac{x-2}{x}-\frac{1}{x}>0$$

Since \frac{x-2}{x} and \frac{1}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x-2-1}{x}>0$$

Combine like terms in x-2-1.

$$\frac{x-3}{x}>0$$

For the quotient to be positive, x-3 and x have to be both negative or both positive. Consider the case when x-3 and x are both negative.

$$x-3<0 x<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when x-3 and x are both positive.

$$x>0 x-3>0$$

The solution satisfying both inequalities is x>3.

$$x>3$$

The final solution is the union of the obtained solutions.

$$x<0\text{; }x>3$$