Solve 1-5/x+2>0

$$1- \frac{ 5 }{ x+2 } > 0$$

Solve for x

$$x\in \left(-\infty,-2\right)\cup \left(3,\infty\right)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+2}{x+2}.

$$\frac{x+2}{x+2}-\frac{5}{x+2}>0$$

Since \frac{x+2}{x+2} and \frac{5}{x+2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x+2-5}{x+2}>0$$

Combine like terms in x+2-5.

$$\frac{x-3}{x+2}>0$$

For the quotient to be positive, x-3 and x+2 have to be both negative or both positive. Consider the case when x-3 and x+2 are both negative.

$$x-3<0 x+2<0$$

The solution satisfying both inequalities is x<-2.

$$x<-2$$

Consider the case when x-3 and x+2 are both positive.

$$x+2>0 x-3>0$$

The solution satisfying both inequalities is x>3.

$$x>3$$

The final solution is the union of the obtained solutions.

$$x<-2\text{; }x>3$$