Question

um
x+2/5(x-3) divide x-2/5(x-3)

Answer 1

Flip n multiply the second fraction. The 5(x-3) cancel.

x+2/(x-2)

Answer 2

can you cancel when there is no equal sign?

Answer 3

Yes, x/x = 1 (ie x/x simplifies to 1 even though it's not explicitly equal to anything

Answer 4

So what I'm saying is that you can simplify even when there is no equal sign.

Answer 5

Well you can leave it there if you want, it's the same equation, just not simplified. The top and bottom are being multiplied by the same number so it's unnecessary.

Answer 6

so what your saying is that ^ is = x+2/5(x-3) * 5(x-3)/x-2?

Answer 7

yeah

Answer 8

If you're dividing fractions you can flip the second one and multiply for ease of calculating :)

Answer 9

(x+2)/(x-2)

Answer 10

umm restriction is x cannot be 2 u have to state that else its wrong

Answer 11

er, multiply top by bottom?

Answer 12

wtf so confused lol

Answer 13

umm give me a sec

Answer 14

ah crap i dont know hw to use eqn thing but u flip when dividing
x+2 5(x-3)
------- x --------
5(x-3) x-2

Answer 15

do u see it

Answer 16

mhmm
but wouldn't that be (x+2)(x-2) instead of (x+2)/(x-2)?

Answer 17

how would u conclude that?

Answer 18

er... I don't know lol.

Answer 19

25(x-3)²/(x-2)(x+2)

Answer 20

lol no i already flipped it, just cancel the 5(x-3) out and restriction is x cannot be 3 or 2

Answer 21

x+2/5(x-3) * 5(x-3)/(x-2)

gives

(x+2)*5(x-3) / (5(x-3)*(x-2))

The 5(x-3)'s cancel, you're left with...
(x+2)/(x-2)

Answer 22

facepalm. what's this buisness with the restrictions?

Answer 23

It won't exist when the denominator = 0

Answer 24

that's to avoid division by zero

Answer 25

well u cannot divide by 0

Answer 26

so x can't = 2.. only?

Answer 27

and can't equal 3 as well

Answer 28

what? why? we're left with just (x+2)/(x-2)

Answer 29

Why 3?

Answer 30

because we need to ensure that the final expression is equal to the original expression

Answer 31

it's like saying f(x) = g(x), which would imply that f and g have the same domain

Answer 32

i sort of get it. (x-3) and all

Answer 33

mybrain won this round though.

Answer 34

mybrain 1 me 0

Answer 35

basically, to find the restrictions, look at the original expression, not the final one

Answer 36

Back to the basic example of x/x. Clearly x/x=1, but when x=0, x/x is undefined (it's actually an indeterminate expression, you'll learn about this later), but 1 is clearly defined. So something went terribly wrong here. So to avoid this, we just say that x can't equal zero.

Answer 37

hmm

Answer 38

thanks guys, I'll have my girlfriend's dad explain this to me in real life.

Answer 39

just remember that something like x=x is only true if EVERY value of x works (ie makes that equation true). If a very select few don't work, then you have to exclude them

Answer 40

But haven't we simplified so we can solve the equation when x = 3. I can tell you it will equal 5 but you're saying that's incorrect, because if we have the complete equation it would be indeterminable.

Also there is nothing that says i can't multiply (x-1)/(x-1) by the equation. It will be the same equation but now you also can't calculated it when x = 1.

Answer 41

So by that same logic, every equation is indeterminable at all values of x.

Answer 42

All I'm doing is looking at the original expression. Since there is an x-3 in the denominator, plugging in x=3 will result in a division by zero error. So this is why x cannot equal 3.

Answer 43

Or the orginal equation is absolute and simplifying it will result in a loss of information?

Answer 44

Yes that's one way of putting it. You "lose" the info that x=3 is not in the domain. So if that's a crucial or critical value, you've essentially lost it in translation.

Answer 45

But i could represent that equation as (x+n)/(x+n)*(x+2)/(x-2) with n = -infinity to +infinity and then it can not be calculated for any value of x. Which i know is not the case, right?

Answer 46

That may be true, but because you're introducing these discontinuities, I don't think they affect the original expression's domain. You could take your argument even further and take something really simple like x and blow it up to get

x*(x+m)/(x+m)*(x-n)/(x-n)*...

and you could go on forever introducing discontinuities, but clearly x can be any number.