# Inequality without factor table

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1. May 8, 2016

### Rectifier

I am trying to solve this inequality without using a factor table.

The problem

$$\frac{x+4}{x-1} > 0$$

The attempt at a solution
As I can see $x \neq 1$. I want to muliply both sides of the expression with x-1 to get rid of it, from the fraction. But before that, I have to consider two cases where x-1 is bigger and smaller than 0 because the sign gets inverted when multiplying (or dividing) by a negative number.

Case 1:
$x-1 > 0 \\ x >1$ gives
$$\frac{x+4}{x-1} > 0 \\ \\ x+4 > 0 \\ x > -4$$

Case 2:
$x-1 < 0 \\ x < 1$ gives
$$\frac{x+4}{x-1} > 0 \\ \\ x+4 < 0 \\ x < -4$$

This is the place where I get stuck. I am not sure how to take all this information and produce an answer.

2. May 8, 2016

### Math_QED

Well, you can't just multiply both sides with x - 1 since it contains information to solve the inequality. From what you have now, x can be any real number except 4 or 1. What if x = 0? I don't know whether you know a sign table. You can use this to solve this exercise.

3. May 8, 2016

### PeroK

Let's take case 1. $x > 1$ What can you say about the expression $\frac{x+4}{x-1}$ in this case?

Last edited by a moderator: May 8, 2016
4. May 8, 2016

### Rectifier

It is always positive for x > 1.

5. May 8, 2016

### PeroK

Hmm! Really? What about $x = 2$?

6. May 8, 2016

### Rectifier

You were fast! Edited the answer just before you commented :D. It is positive. I cannot tell that from my expression that I get in case 1. Not sure how to interpret x > -4

7. May 8, 2016

### PeroK

So, that's half your answer: If $x > 1$ the expression is always positive.

What about when $x < 1$?

8. May 8, 2016

### Rectifier

Well, that is the tricky part! :D x = 0 gives a negative number. x = - 5 gives a positive number. Not sure how to use my case studies to come to an answer though.

9. May 8, 2016

### PeroK

Is there another critical value for $x$? Perhaps something you've already calculated?

10. May 8, 2016

### Rectifier

I have a feeling that it is the interval that I calculated in case 2 but I am not sure why.

11. May 8, 2016

### PeroK

What you did in the OP was correct, but it created a logically slightly difficult solution that you weren't able to interpret. I suggest you solve the problem and I can try to explain what happened in your OP, if you like.

You actually should have used 3 cases here. The next case is $-4 \le x < 1$. Let's call this case 2.

12. May 8, 2016

### Rectifier

In the third case gives a negative outcome.

I feel like its much easier to solve these with a sign table :D

13. May 8, 2016

### PeroK

Not quite. When $x = -4$ the expression is zero. But, there are no solutions for $-4 \le x < 1$.

What about $x< -4$?

14. May 8, 2016

### Rectifier

Gives a positive outcome.

15. May 8, 2016

### PeroK

So, what's the overall solution?

16. May 8, 2016

### Rectifier

x>1 and x<-4

17. May 8, 2016

### PeroK

What happened in your original post?

You started by assuming that $x > 1$ and looked for solutions. You found solutions when $x > -4$. This looked strange but it was actually quite logical. What you had was:

$x > 1$ and $x > -4$

Which, logically, is simply $x > 1$.

There's something else that might have happened in a different problem. You might have got something like:

$x > 1$ and $x < -2$

Again, that is not a problem, but would tell you that there are no solutions for $x > 1$.

18. May 8, 2016

### Rectifier

Oh okay, thank you for the explanation! But why do I choose x>1 and not x > -4? I suspect that the answer is trivial.

19. May 8, 2016

### PeroK

Which numbers satisfy both equations?

20. May 8, 2016

### Rectifier

x > 1. So I should basically take the one with the smallest common solution set?