Question

Hi there, I know math has a lot of rules so I'm double checking what I want to do with this expression (equation?)...

Answer 1

(x - 2)(x + 2)
____________
(x + 2)(x - 6)
Can I cross out (x + 2) and (x + 2) even though they are both the same? and not (x + 2) and (2 + x)?

Answer 2

You're asking if you can do this correct?\[\frac{(x-2)\cancel{(x+2)}}{\cancel{(x+2)}(x-6)}\]

Answer 3

yes

Answer 4

You can cross them out or rather cancel them because they are the same value,

Answer 5

and also x + 2 = 2 + x

Answer 6

Awesome, thanks :) math always confuses me us so I wanted to double check :P

Answer 7

You can also cross it out if it's (x + 2) and (2 + x) because they're both positive.

Answer 8

You can do that, but be careful because the first way of writing it implies \(x\neq-2\). After canceling, this is not true

Answer 9

So after you have cancelled, you have actually changed the expression. To counteract this, merely state that \(x\neq-2\) next to the equation.

Answer 10

Right, got it. I've been doing these problems over and over since I keep messing up the negative signs with the plus signs :/ Thanks again guys!

Answer 11

I got the EXACT same thing as them only I got x CAN'T equal 2 and -6...why is it -2?

Answer 12

Suppose \(x=-2\). Then you have the equation \[\frac{(-2)^2-2}{((-2)+2)((-2)-6)}=\frac{4-2}{0(-8)}=\frac{2}{0}\]This is undefined since you are dividing by 0.

Answer 13

That should be \((-2)^2-4\) on top, but the reasoning still applies

Answer 14

I'm still kinda confused, I get the top part no sweat, but the bottom is a little iffy.

Answer 15

wouldn't that -2 be a plus 4 since it's really -2^2?

Answer 16

the first -2 I mean

Answer 17

on the bottom

Answer 18

On the bottom, you have \(((-2)+2)\cdot((-2)-6)=(2-2)\cdot(-2-6)=0\cdot-8=0\). So if you had \(x=-2\), your denominator, you would be dividing by 0, which is bad.

Also, in the denominator, you just have \((x+2)(x-6)\). No \(x^2\)'s in that form.

Answer 19

Oh, I thought you were doing this from the beginning expression, I get it now. I understand why it's -2, but how come I didn't get that? Isn't it supposed to be: x - 2 = 0 and you simplify it?

Answer 20

If you prefer, we can start before it's been factorized leaving us with \[\frac{(-2)^2-4}{(-2)^2-4(-2)-12}=\frac{0}{4+8-12}=\frac{0}{0}\]

Why would it be \(x-2=0\)? When solving problems like these that restrict the values of \(x\), you need to look at what's in the denominator, and there's an \(x+2\) there.

Answer 21

Well, in the explanation it has
x - 2
______
x - 6

they taught me to solve this by saying
x - 2 = 0
x - 6 = 0
and you have to get x by itself
x - 2 + 2 = 0 + 2
x = 2
x - 6 + 6 = 0 + 6
x = 6
isn't that right?

Answer 22

If you're solving \[\frac{x-2}{x-6}=0\]you would want to set \(x-2=0\) and solve. Then to find out what the restricted domain is, look at \(x-6=0\) and solve for \(x\) there. So when \(x=2\), the equation is equal to 0.

Answer 23

Right! So why does it say that the restricted domain is -2?

Answer 24

Because that's left over from before. This new equation you have should really be \[\frac{x-2}{x-6}\qquad x\neq-2\]When you factored the original equation, you were left with a \(x+2\) in the denominator. This means that \(x\neq-2\), and you must keep that restriction for the rest of the problem.

Answer 25

Just because you canceled the term, doesn't mean that the restrictions it previously imposed are gone. That would be changing the equation.

Answer 26

So..wouldn't that mean there would be three restrictions? Since you're including the one from before? Because in some of the other equations they include the remaining equation even though they canceled something out

Answer 27

What would be the third restriction? We have \(x\neq-2\), and \(x\neq6\). Where is the third one coming from?

Answer 28

Well, in the other equations they have (for example)
x + 2
_____
x - 9

and then they do the x + 2 = 0 stuff...after they do that, they take the restrictions from what we did previously(up there^ in my sentence) and then those are the two restrictions even though we canceled two equations out. But now, you're saying that even though (iun the expression we're doing) we're going to take the equation that we canceled out and replace it for the remaining equation. Does that make sense?

Answer 29

I think I see. From what I'm getting, we're just doing things in a different order. You're teacher/the website finds the zeroes of the numerator first, and then removes the restrictions. I removed the restrictions, and then found the zeroes of the numerator.

Answer 30

I think so, I have another expression that explains what I'm saying...it'll show the differences.
(I think)

Answer 31

They used the equatiosn from the final equations they got to get the restrictions, but in this one, they didn't! Instead they used the equation we crossed out to get the answer...How am I supposed to get the right answer if idk which they'll do!

Answer 32

They're doing the same thing in each problem. However, if you're worried about how to get the restrictions, here what to do.

Look at the denominator of the original equation. Ignore everything else (for now). Factor it, and solve for the zeroes. Those zeroes are the restrictions. Now that you have those, go back to the original equation, and simplify/solve/whatever they want you to do with it.

Answer 33

Ohh ok, I just want to make sure I'm doing this right. I did the same thing I did for the top expression I just sent for the second one but got it wrong. I couldn't figure out why :P

Answer 34

If you're having more problems with this stuff, feel free to ask me for help.

Answer 35

Thanks! I appreciate it :)