# Thread: Help with solving inequalities

1. Originally Posted by Jhevon
i always use \geq, that works as well.
Gotcha! Okay, I've been copying/pasting everybody's code and modifying. Thanks for the link to the tutorial - you're good!

2. This note is addressed to lilrhino
As I wrote to you, it is all about the factors.
Consider $(x+a)$. That factor is positive for $x>-a$ and is negative for $x<-a$. For $(x+b)$ the factor is positive for $x>-b$ and is negative for $x<-b$.
Now consider $(x+a)(x+b)$, that product is positive if both factors have the same ‘sign’; that product is negative if the factors have different ‘signs’.

This idea can be extended to any number of factors or quotients.
If you want to understand how inequalities work, then you need to understand the fundamental way factors contribute to the ‘signs’ of products and quotients.

3. Originally Posted by Plato
This note is addressed to lilrhino
As I wrote to you, it is all about the factors.
Consider $(x+a)$. That factor is positive for $x>-a$ and is negative for $x<-a$. For $(x+b)$ the factor is positive for $x>-b$ and is negative for $x<-b$.
Now consider $(x+a)(x+b)$, that product is positive if both factors have the same ‘sign’; that product is negative if the factors have different ‘signs’.

This idea can be extended to any number of factors or quotients.
If you want to understand how inequalities work, then you need to understand the fundamental way factors contribute to the ‘signs’ of products and quotients.
Plato you are correct. Anyway, I was reading through the text and since our sign is < we only care about numbers that are negative is that correct?

4. Originally Posted by lilrhino
Plato you are correct. Anyway, I was reading through the text and since our sign is < we only care about numbers that are negative is that correct?
we only care about factors that are negative

5. Originally Posted by Plato
we only care about factors that are negative
Got it! I'm getting there...

6. This is an inequality that's just one variable (not sure how to proceed):

$\frac{x+1}{2x-3}$ ≥ 0
In my 2nd question, I came up with 3 intervals to test:

$(-\infty{, -1})$
$(-1, \frac{3}{2})$
$(\frac{3}{2}, \infty)$

But after testing (-4, 0, 3) I found that the factors were only negative in:

$(-\infty{, -1})$
$(-1, \frac{3}{2})$

So, is this my correct solution set?

7. Originally Posted by lilrhino
In my 2nd question, I came up with 3 intervals to test:

$(-\infty{, -1})$
$(-1, \frac{3}{2})$
$(\frac{3}{2}, \infty)$

But after testing (-4, 0, 3) I found that the factors were only negative in:

$(-\infty{, -1})$
$(-1, \frac{3}{2})$

So, is this my correct solution set?
Your intervals were correct, but your interpretation was wrong. All the factors do is tell you where to split the intervals.

Look at the actual problem:
$\frac{x + 1}{2x - 3} \geq 0$

We wish to test if $\frac{x + 1}{2x - 3}$ is positive, not just the factors.

So test the expression on each interval:
$( -\infty, -1] \to \frac{x + 1}{2x - 3} \geq 0$ (Check!)
$\left ( -1, \frac{3}{2} \right ) \to \frac{x + 1}{2x - 3} < 0$ (No!)
$\left ( \frac{3}{2}, \infty \right ) \to \frac{x + 1}{2x - 3} > 0$ (Check!)

So the solution set is $( -\infty, -1] \cup \left ( \frac{3}{2}, \infty \right )$. (Note: we need to include x = -1, but can't include x = 3/2. Do you see why?)

-Dan

8. Originally Posted by topsquark
Your intervals were correct, but your interpretation was wrong. All the factors do is tell you where to split the intervals.

Look at the actual problem:
$\frac{x + 1}{2x - 3} \geq 0$

We wish to test if $\frac{x + 1}{2x - 3}$ is positive, not just the factors.

So test the expression on each interval:
$( -\infty, -1] \to \frac{x + 1}{2x - 3} \geq 0$ (Check!)
$\left ( -1, \frac{3}{2} \right ) \to \frac{x + 1}{2x - 3} < 0$ (No!)
$\left ( \frac{3}{2}, \infty \right ) \to \frac{x + 1}{2x - 3} > 0$ (Check!)

So the solution set is $( -\infty, -1] \cup \left ( \frac{3}{2}, \infty \right )$. (Note: we need to include x = -1, but can't include x = 3/2. Do you see why?)

-Dan
Is it because it will cause the denominator to equal 0?

9. Originally Posted by lilrhino
Is it because it will cause the denominator to equal 0?
Yup!

-Dan

10. Originally Posted by topsquark
Your intervals were correct, but your interpretation was wrong. All the factors do is tell you where to split the intervals.

Look at the actual problem:
$\frac{x + 1}{2x - 3} \geq 0$

We wish to test if $\frac{x + 1}{2x - 3}$ is positive, not just the factors.

So test the expression on each interval:
$( -\infty, -1] \to \frac{x + 1}{2x - 3} \geq 0$ (Check!)
$\left ( -1, \frac{3}{2} \right ) \to \frac{x + 1}{2x - 3} < 0$ (No!)
$\left ( \frac{3}{2}, \infty \right ) \to \frac{x + 1}{2x - 3} > 0$ (Check!)

So the solution set is $( -\infty, -1] \cup \left ( \frac{3}{2}, \infty \right )$. (Note: we need to include x = -1, but can't include x = 3/2. Do you see why?)

-Dan
As I did with Plato's solution, I'd like to illustrate that Dan's solution is correct with a graph. See below

11. Originally Posted by topsquark
Yup!

-Dan
Thanks Dan, I appreciate your help and the explanation !

12. Originally Posted by Jhevon
As I did with Plato's solution, I'd like to illustrate that Dan's solution is correct with a graph. See below
Thanks as always Jhevon. Do you mind sharing how you're generating the graphs .

13. Originally Posted by lilrhino
Thanks as always Jhevon. Do you mind sharing how you're generating the graphs .
i used a program called graph which TPH introduced to me

14. Originally Posted by lilrhino
$x^2 + 4x - 32 < 0$
Here is another way.

$x^2 + 4x - 32 = x^2 + 4x +4 - 36 = (x+2)^2 - 36$

Thus,
$(x+2)^2 - 36 \geq 0 \Rightarrow (x+2)^2 \geq 36$

Thus,
$x+2 \geq 6 \mbox{ or }x+2 \leq -6$

The rest is trivial.

15. Originally Posted by lilrhino
$\frac{x+1}{2x-3} \geq 0$
Here is another way.

Either the numerator is non-negative and the denominator is postive.

Or,

EIther the denoimator is non-positive and the denoimator is negative.

Thus,
$x+1 \geq 0 \mbox{ and } 2x - 3>0$
Or,
$x+1\leq 0 \mbox{ and }2x- 3<0$

The rest is trivial.

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