Question

Can someone help me simplify this rational expression? (Will post it below) THANK YOU!

Answer 1

@zepdrix

Answer 2

Ooo what a mess >.<

Answer 3

I know :(

Answer 4

\[\Large\rm \frac{\frac{3}{x}+\color{orangered}{\frac{1}{x^2+x}}}{\frac{1}{x+1}-\frac{1}{x-1}}\]So if you look at this orange part first...
Hmm each term has something in common.
We can factor the denominator, yes?

Answer 5

Yes, the one in orange is (x+1)(x-1)

Answer 6

Noooo, no no no :O

Answer 7

\[\Large\rm (x+1)(x-1)=x^2-1\]

Answer 8

hahahah, i suck at this. hellpppp :(

Answer 9

They each have an x, yes?

Answer 10

yes

Answer 11

So factor an x out :d
\[\Large\rm x^2+x=x(x+1)\]

Answer 12

\[\Large\rm \frac{\frac{3}{x}+\color{orangered}{\frac{1}{x(x+1)}}}{\frac{1}{x+1}-\frac{1}{x-1}}\]Yah? :d

Answer 13

ohhhhh, right.. haha :P

Answer 14

I'm not sure how you have been taught to do these.....
but um....

I have learned .... in my years of doing math...
that ... fractions are pure evil.

So let's try something fancy.
Here is what I wanna do...

Answer 15

I wanna multiply the top and bottom by x(x+1)(x-1), just to get rid of every fraction.

Answer 16

What'dyou think? Too fancy?

Answer 17

That sounds like what my teacher was doing. So I think we can do it!

Answer 18

x, (x+1) and (x-1) are the factors showing up in our denominators.
\[\Large\rm \frac{\frac{3}{x}+\frac{1}{x(x+1)}}{\frac{1}{x+1}-\frac{1}{x-1}}\cdot\color{royalblue}{\left(\frac{x(x+1)(x-1)}{x(x+1)(x-1)}\right)}\]So by multiplying through by the LCM of our factors, we can get rid of these fractions.

Answer 19

I wanna make sure you understand what we're doing.
So...
When we distribute this x(x+1)(x-1) to the first term:\[\Large\rm \frac{3}{x}x(x+1)(x-1)\]What happens to all that good stuff, where does it go?
Numerator? Denominator of this fraction?
What do you think?

Answer 20

the x in the denominator and the x in the numerator cancel, and then the (x+1)(x-1) go on the numerator ? :) I hope haha

Answer 21

Mmm ok good.
So since we're left with nothing in the denominator, (Or a 1), we can simply write is as,\[\Large\rm 3(x+1)(x-1)\]ya? :d

Answer 22

yes! I got that :)

Answer 23

\[\large\rm \frac{3\cancel{x}(x+1)(x-1)}{\cancel{x}}=\frac{3(x+1)(x-1)}{1}=3(x+1)(x-1)\]

Answer 24

Ok cool, do it with the othersssss +_+
Figure it outtttt
You can do eeeet

Answer 25

the other one on the top is \[1x(x+1)\]

Answer 26

then on the bottom, \[1x(x-1) and 1(x-1)\]

Answer 27

wait wait wait :U
water you talking about?
1x(x+1)?

I think you need to cancel out your x(x+1) from those factors, yes?
leaving you with (x-1) for the second top term?

Answer 28

ohhh, yes i see that now :P sorry haha

Answer 29

\[\Large\rm \frac{3(x+1)(x-1)+(x-1)}{\frac{1}{x+1}-\frac{1}{x-1}}\cdot\left(\frac{1}{x(x+1)(x-1)}\right)\]And then what? You got some stuff in the bottom?

Answer 30

The second term in the bottom looks messed up :U

Answer 31

wait, can you explain why we are multiplying by (1/x(x+1)(x-1)) ?

Answer 32

We already distributed the numerator, it's taken care of.
So all that's left up there is a 1, we can't leave a 0 there.

Answer 33

\[\Large\rm \frac{\frac{3}{x}+\frac{1}{x(x+1)}}{\frac{1}{x+1}-\frac{1}{x-1}}\cdot\left(\frac{1\cdot x(x+1)(x-1)}{x(x+1)(x-1)}\right)\]You can think of it like this if you like, with a 1 on top.
After distributing the stuff to the numerator, and doing that fancy magic, we're left with:
\[\Large\rm \frac{3(x+1)(x-1)+(x-1)}{\frac{1}{x+1}-\frac{1}{x-1}}\cdot\left(\frac{1}{x(x+1)(x-1)}\right)\]

Answer 34

ohh okay.

Answer 35

So how do I get rid of the fraction on the bottom? the same way as we did with the numerator?

Answer 36

\[\Large\rm \frac{3(x+1)(x-1)+(x-1)}{\left(\frac{1}{x+1}-\frac{1}{x-1}\right)x(x+1)(x-1)}\cdot\left(\frac{1}{1}\right)\]Yah, distribute the stuff to each term.
Your first fraction looked ok.
You cancelled out the (x+1)'s.

Answer 37

\[\Large\rm \frac{3(x+1)(x-1)+(x-1)}{x(x-1)-\color{green}{\frac{1}{x-1}~x(x+1)(x-1)}}\]How bout the second term? :d

Answer 38

x(x+1)

Answer 39

\[\Large\rm \frac{3(x+1)(x-1)+(x-1)}{x(x-1)-x(x+1)}\]Yay good job \c:/

Answer 40

From there, you can uhhhh.. simplify if you want :d

Answer 41

oh jeez.. haha. greeeaattt. Can i cancel things out like the (x+1) and (x-1) or should I distribute everything, combine like terms and then cancel things out?

Answer 42

No canceling here :(
Gotta distribute everything, then combine like-terms,
then maybe if you're lucky, the leftover stuff will factor and you can try to factor maybe.

as a matter of factor, i factor that you can factor those factors.. after you factor them of course.

Ignore that last line ^ kinda lost my marbles there.

Answer 43

then maybe if you're lucky, the leftover stuff will factor and you can try to `cancel`* maybe.

Blah typo in there.

Answer 44

okay, im getting \[3x ^{2}+x-4 \div -2\]

Answer 45

ew ew ew D: you used the old school division operator! from kindergarden!! D:

Answer 46

sorry, all of that is over -2 not 4 divided by -2. I have a feeling that is not right! GAH!

Answer 47

I don't know how to use the line one! I'm not all fancy like you! :P

Answer 48

Mmmmm yah I think that's right.
I'm just eyeballing it, but it uhhhhhhh..... yah 3x^2+x-4 all over -2
wait wait wait wait.... .... !111!!!!1!one!

isn't it -2x on the bottom?

Answer 49

ahh, yes it is! Good catch! haha.
Wait! my friend just messaged me and the teacher gave her the answer: (3x+4)(x-1) all over -2x

so how do i get from 3x^2 +x -4 all over -2x to that ? D:

Answer 50

So we need to factor the numerator :U

Answer 51

friend to the rescue.

Answer 52

ohh haha. it didn't send. I was trying to send "Just factor it?" okayyy.

Answer 53

OH DANG. WE DID IT. hahah. Thank you so much!! You're wonderful!

Answer 54

yay team \c:/

now get that dog out of my face -_-
I'm allergic.

No I'm not, but he looks like a racist.
jk, im sure he's a sweetheart.

Answer 55

awhh. Ben is the sweetest thing you could ever meet. He is such a lover. He just cuddles and kisses you all day.

Answer 56

CATS RULE. <3
DOGS DROOL.

there it is.

Answer 57

I really am allergic to cats though D: