Question

Given the fraction below. What are the exclusions? In your answer, list variables before integers and negatives before positives.

https://media.glynlyon.com/a_matalg02_2011/5/164.gif

x ≠ ?

@lgbasallote Can you help me please?

Answer 1

I think on the first fraction the denominator is 0 because x^2 is assumed to be 1 and 1-1 = 0

Answer 2

hmm

Answer 3

i did not get that logic sorry

Answer 4

\[x^2 - 1 = (x+1)(x-1)\] not 1 - 1

Answer 5

your denominators are
\[x^2 - 1 = (x+1)(x-1)\]
and
\[x^2 - y^2 = (x+y)(x-y)\]
so solve for x

Answer 6

lol sorry. I suck at math, so I'm like @.@

Answer 7

carry on

Answer 8

lol I can do this xD Give me a sec. I can't fail you xD

Answer 9

when solving for a in \[(a+b)(a-b)\]
you equate the factors to zero then solve for a

a + b = 0

therefore a = -b

when a - b =0
a = b

Answer 10

I don't know what to do :'( See, I ignore math for as long as possible because it intimidates me, then I go to do it, and I have no idea what I am doing and I get beyond frustrated and still hate it.

annnndd give me a sec I'll try to solve it with that new info

Answer 11

if even the dyslexic was able to learn calculus why can't you ;) it's just practice and experience

Answer 12

Okay so the first fraction would be -1, I think?

and yeah, you have a point.

Answer 13

yup \[x \ne 1 \; ; \; x \ne -1\]
those are the exclusions for the first denominator

Answer 14

:'D haha okay, now let me find the second one...

Answer 15

however the dyslexic was not able to learn multivariable calculus and took him years to get figures and graphs but whatever lol

Answer 16

lol xD I was fairly okay with graphs when I was in Alg I and geometry

How would I go about trying to solve the second fraction? Do I just assume x and y = 0?

Answer 17

will you elaborate?

Answer 18

Like, x^2 - y^2 how do I know how to find the drnominator for that one? Do I assume x and y is 0 or 1? maybe?

Answer 19

look again at what i did with (a+b)(a-b)

Answer 20

lol @rebeccaskell94 do you get it already? :)

Answer 21

Okay so I'm basically just repeating what you did, but with x and y

(x+y)(x-y)
x+y=0
Therefore x = -y
When x - y = 0
x=y


lol well um, erm...
I *think* maybe, possibly... x ≠ 0 and y ≠ 0


Idunnow D:

Answer 22

cut your post...

Okay so I'm basically just repeating what you did, but with x and y (x+y)(x-y) x+y=0 Therefore x = -y When x - y = 0 x=y

Answer 23

there that's correct

Answer 24

ohhhh

Answer 25

think about it

(x+y)(x-y)

if x = 0 then you'll have

(0 + y)(0 - y) = (y)(-y) = -y^2 <--this is not zero

if y = 0
(x +0)(x-0) = (x)(x) = x^2 <--this is not zero

if x = -y
(-y + y)(-y -y) = (0)(-2y) = 0

if x = y
(y+y)(y-y) = (2y)(0) = 0

Answer 26

remember your goal is to find the value of x that will make it zero

Answer 27

(x+y)(x-y)

when x + y = 0

x = -y

stop there.

when x - y = 0

x = y

stop there.

Answer 28

I thought I was just trying to figure out what x didn't equal? urg this is so confusing

Answer 29

when you already have x = something, you stop

Answer 30

lol yeah

Answer 31

\[x \ne 1, -1, y, -y\]

Answer 32

im just using the term equal because it's shorter

Answer 33

the difference is just a slash...

Answer 34

Ohhh okay, I see how you found that haha

Thank you so so so so much :)

Answer 35

<tips hat>