# Simple inequality

1. May 30, 2015

### Scheuerf

How would I solve the inequality (X-4)/X>0. I thought that inequalities were solved in the same way equations were, but when I solve that way I get X>4 which isn't the entire answer.

2. May 30, 2015

### RUber

For this one, it is best to look at critical points where either the top or bottom equal zero. From that, you should be able to quickly categorize the intervals where the expression is true.

3. May 30, 2015

### Staff: Mentor

No, they're not. For example, if you multiply both sides of an equation by, say, -1, you get a new equation that is equivalent to the one you started with.

If you multiply an inequality by -1, the inequality symbol changes direction.

4. May 30, 2015

### Svein

Well, what you do is:
x-4: Negative when x<4, positive when x>4
x: Negative when x<0, positive when x>0
Expression: Positive when x<0 (neg. and neg. makes pos.), negative when 0<x<4 and positive when x>4

5. May 30, 2015

### Staff: Mentor

Or, more simply, just multiply both sides of the inequality by x2, first making a note that x cannot be zero. For x ≠ 0, x2 > 0, so the direction of the inequality doesn't change.

6. May 31, 2015

### Scheuerf

When I solve that way I get x>0 and x>4. Am I doing something wrong?

7. May 31, 2015

### RUber

The correct answer was given in post 4.
$\frac{x-4}{x} >0$
Taking Mark44's recommendation, this could also be seen as:
$x^2\frac{x-4}{x} >0*x^2$
$x(x-4) >0$
Remember that (-)(-)=(+) and (+)(+)=(+), and (-)(+)=(-) just the same as (-)/(-)=(+) and (+)/(+)=(+), and (-)/(+)=(-).
So whether or not you multiply by $x^2$, you still need to find the signs of your terms (x-4) and (x) and the appropriate regions.
Build a simple table, the inequality will only hold true if both terms are negative or both are positive.