Question

|2-x|/2-x = 0
How do I get the 2-x thats under the fraction over to the other side

Answer 1

just multiply it to both sides.

Answer 2

So would i multiply it by 2+x or 2-x?

Answer 3

well you want to cancel the 2-x so you multiply that to both sides.

Answer 4

because that will get you 2-x/2-x on the left side and thats just equal to 1

Answer 5

Then I would get |2-x|(2-x) = 0 wouldnt I?

Answer 6

You don't. What's under the fraction cannot equal zero (unless there's a limit around there). That means x can take any value from your domain ( I'll assume that would be R) except 2 (because division by zero has been outlawed centuries ago).

Because what's under the fraction can't equal zero, it means that what's above the fraction has to. However, in order for |x-2| to equal zero, x has to equal 2. But that goes against the previous paragraph. Which makes the exercise redundant.

Answer 7

He doesn't want what's under the fraction to be zero.

Answer 8

*The point of my previous post was that the exercise itself is redundant. Any attempt to solve it will fail because it's non-sense.

Answer 9

|2-x|/2-x = 0


ima multiply both sides by 2-x

(2-x)|2-x|/(2-x) = 0*(2-x)

Answer 10

in the left side you have 2-x/2-x

what happens when you have the same thing divided by itself for example

2/2 or 4/4

they equal 1

Answer 11

Ohh okay, i thought there was more steps than that, I was just overthinking it. Thank you!

Answer 12

yeah in the end you are left with

|2-x|=0

Answer 13

which pretty much yields you one solution for x

Answer 14

*facepalm* Which will get you to x=2 which will go against your inition |2-x| / (2-x) because you're dividing by zero. Which is non-sense, wrong and bad.

Answer 15

So i typed it into my hw and it tells me that the x-intercept does not exist

Answer 16

lolz
for any value that a sane person would substitute for x in the expression
\[\Large \frac{|2-x|}{2-x}\] the result would always be either 1 or -1, never anything else.
Imagine y = 2-x, so that
\[\Large \frac{|y|}y\] and it'd probably make more sense that y cannot be 0, and neither can 2-x

Answer 17

So are you saying |2-x|=0 at the end is not the answer

Answer 18

While it's tempting to just multiply both sides by 2-x at the beginning, don't forget that in every instance that you do that, say in algebra, or the like, you are assuming from the start that what you multiplied to both sides was NOT zero to begin with.

Because if you multiply zero to both sides, the resulting would ALWAYS be equal.

So, implicitly, what was done was this:
\[\Large \frac{|2-x|}{2-x}\times (2-x) = 0\times (2-x) \ \ assuming \ 2 -x \ne 0\]
\[\Large |2-x| = 0 \ \ \ \ a \ contradiction\]

Never mind these details for now, though. Just accept that there must NEVER be a zero in the denominator.

Answer 19

I understand that 2-x cannot equal zero and maybe I should have explained the question more. They give me |2-x| / 2-x and want me to graph it. So I need to do the piecewise function for {-x if x<0 , x>0} and thats why I was trying to get the problem into |absolute value|=c

Answer 20

Then I've already pretty much made it a dead giveaway :P
There are three cases then, one of which was discussed here in heated detail, and it's that if x = 2, the function does not exist, yes? So, let's put that it.
\[\Large f(x) = \begin{cases}???&x < 2\\ \emptyset & x = 2\\??? &x>2\end{cases}\]

So... I want you to do something for me (actually yourself, LOL)

Try plugging in random numbers that are less than 2, see what you get.

Answer 21

Where do I plug them in at? When the equation is still |2-x|/2-x ?

Answer 22

Yup.
Plug in certain number that are less than 2, say, 1, 0, -1, your choice, it can be as mind bogglingly 'complex' as \(\large -\pi\) for all I care XD

Answer 23

I get 1 each time

Answer 24

No you don't?

Answer 25

|2-1|/2-1 = 1/1 =1
|2-0|/2-0 = 2/2 = 1

Answer 26

Oops.
Yeah, you're right :3

okay, try plugging in numbers greater than 2 now

Answer 27

I get 1 again

Answer 28

Are you sure? :P

Answer 29

|2-3| / 2-3 = -1/-1 = 1

Answer 30

oh wait its a positive on top isnt it

Answer 31

There we go XD

Answer 32

so in fact, for all x greater than 2, the function is simply equal to...?

Answer 33

-1?

Answer 34

That's right.
So, in fact, for all x less than 2, the function is just equal to 1
for x = 2, the function has no value, so leave that as a hole of sorts
for x > 2, the function is equal to -1

Should be simple to graph now ^_^
\[\Large f(x) = \begin{cases}1&x < 2\\ \emptyset & x = 2\\-1&x>2\end{cases}\]

Answer 35

Thank you so much!! :)

Answer 36

No worries.