Question

(x+3)/(x-4)>0 --I have no idea how to solve this...

Answer 1

Okay, here we need to know what are called the critical numbers (I think)
They're the numbers what would make the expression on the left equal to zero... or not exist...

In other words, zeros of the numerator and zeros of the denominator.

Answer 2

This one is fun!

Answer 3

negative * negative = positive
positive * positive = positive

Answer 4

We have
\[\frac{x+3}{x-4} >0\]
If you multiply both sides by x-4 we get
\[x+3 < 0\] which when we then solve for x.

Answer 5

Then we just repeat a similiar process by dividing by x+3.

Answer 6

Oh, that's what I did wrong, austinL...

Answer 7

Uhh.... I don't remember it being this simple...

Answer 8

One does not simply multiply both sides by x - 4

Answer 9

No, no, no, and no. You CANNOT multiply an inequality by "x-4". You MUST know if "x-4" is positive or negative in order to do that.

Answer 10

Because you then have to take cases... when x-4 is positive and when x-4 is negative, to decide whether or not the inequality gets flipped

Answer 11

As an answer I arrive at
\[ x < -3\]
and
\[x > 4\]

Answer 12

You can treat inequalties like equations. You just need to know how to properly adjust the sign.

Answer 13

Austin, you're right. I just don't know how to multiply. That's why I couldn't figure it out. :P

Answer 14

Additionally,
http://www.wolframalpha.com/input/?i=%28x%2B3%29%2F%28x-4%29%3E0+solve+for+x

Answer 15

Once I get x<-3 and x>4, I can make my solution sets, find my test numbers... etc. etc.

Answer 16

you can make 2 assumptions for an inequality:

(x-c) is positive when x>c
(x-c) is negative when x<c

but the you can still do the multiplication

Answer 17

I find @austinL work not reproducible and bordering on magic.

Let's just find @terenzreignz critical numbers.

x+3 > 0 ==> x > -3
x-4 > 0 ==> x > 4

So,
if x > 4, they are BOTH positive
if x < -3, they are BOTH negative

We should be about done.

Answer 18

*facepalm*

Answer 19

Lol, thank you guys!

Answer 20

I think this sets a rather dangerous precedent when you start dealing with inequalities involving more than just two factors of the form (x-c)

Answer 21

Nothing personal, but this part "You just need to know how to properly adjust the sign." simply has no rational mathematical definition. Obviously, it's author does know what's going on, but it is not sound practice until it is better defined.

Answer 22

in real analysis, you define the thrms relating to inequalities better; and introduce a 1st multiplication rule and a 2nd multiplication rule

Answer 23

I'm done, she has been helped. I was just assuming too much of what the asker had been taught I suppose.

Answer 24

@terenzreignz, thank you for trying to help, but whatever you're doing is way over my head. I'm just a lowly Honors Algebra II student, without a speck of knowledge to her name.

Answer 25

the case is undefined when x=4 since that zeros out the deniminator

x+3 > 0 ; x > 4
x > -3; the intersection of these is x>4

x+3 < 0 ; x < 4
x < -3;
the intersection of these is x < -3

Answer 26

From here
\[\frac{x+3}{x-4} >0\]

Multiplying both sides by x - 4
\[{x+3} <0\]
Why was the inequality flipped?

Answer 27

most likely the intermediary step is ignored since when x>4 the top is postive to begin with

Answer 28

the only thing to assess would be when x < 4

Answer 29

mathing proofs are filled with statements like: "its obvious that such and such is true so we will focus on ...."

Answer 30

thats the hardest part i get about reading proofs; is they simply assume im smart enough to deduce their "obviousicities" lol

Answer 31

As a professor of mathematics, it is a wonderful exercise for your students to SAY "it is obvious that", just to make your students think about why it is obvious.

As a student of mathematics, it is NEVER appropriate to use such language without explaining why. Perhaps, "It is obvious because of the mean value theorem" or "it is clear from the definition of continuity".

If a student submitted a simple "It is clear that" to me, I would mark it wrong and say "Clear to whom?"

Answer 32

Here's my favourite method :)
Find the critical numbers first, but you already know they are -3 and 4, right :)
Let's draw a table :

Answer 33

That line you see represents the number line, we mark out the critical numbers -3 and 4 (in order, of course :)

Answer 34

Let's look at x + 3.. its critical number is -3

Means it is negative for all values before that, and positive for all values after that...
(x - 3) is going to be negative for any x less than 3, and positive for any x > 3

Answer 35

Everything okay so far, @Liv_16 ? :)

Answer 36

Yes

Answer 37

What about (x -4 ) ?
Its critical number is 4, so any number before that, it's going to be negative, while any number AFTER that, it's going to be positive, so...