Question

Does anyone know how to add and subtract rational expressions? help very last assignment D:

Answer 1

Its an assignment I have to do, but its all about adding and subtracting rational expression, only 5 problems

Answer 2

http://www.cliffsnotes.com/math/algebra/algebra-ii/rational-expressions/adding-and-subtracting-rational-expressions

Answer 3

Hope this will help

Answer 4

Ok c: Thanks

Answer 5

My Pleasure.

Answer 6

is it ok if i do one problem and show you to make sure im not doing anything wrong?

Answer 7

@Nurali

Answer 8

post an example

Answer 9

I think the other user is afk or something

Answer 10

ill draw a picture of it

Answer 11

hmmm... they share a common denominator so all we have to do is to combine like terms in the numerator

Answer 12

How would we do that? would we add 2x to x and 1 to the other 1?

Answer 13

yeah

Answer 14

would the numerator come out like this = 2x^2+2 or 3x+2

Answer 15

it would be 3x+2. because 2x+x = 3x
1 + 1 =2

Answer 16

Ok! thanks, can you help with four more adding/subracting q's?

Answer 17

sure

Answer 18

Thankyou so much! so the next question basically wants me to use the answer we just got (3x+2/2x) subtracted by x/x+3, but it also says to * identify any possible restrictions that exist with (or in) the resulting rational expression*

Answer 19

\[\frac{3x+2}{2x}-\frac{x}{x+3}\]

it said any possible restrictions... so I"m not sure what that means 100% maybe what value makes the result undefined ?>! Not sure.. any way our lcd = (2x)(x+3) so we multiply x+3 on the first fraction and multiply 2x on the second fraction

Answer 20

so like cross multiplying?

Answer 21

umm... similar but instead we need to find the least common denominator
see example 3 on the cliffnotes site... noticed how they took the lcd of that problem?

Answer 22

It looks kinda confusing sine the problem big :x

Answer 23

on cliff notes

Answer 24

\[\frac{3x+2}{2x} \times \frac{x+3}{x+3}-\frac{x}{x+3} \times \frac{2x}{2x}\]

Answer 25

ok that makes alot more since lol

Answer 26

example 3 on cliffnotes from Nurali was just x and y as the denominators. it was the later examples that I didn't choose because I knew it would be too much

Answer 27

would we just times the bottom demonitaers togethor? like 2x(x+3)

Answer 28

yes...

Answer 29

\[\frac{(3x+2)(x+3)-2x^2}{2x(x+3)}\] we have to remember that whenever we have a - sign on the right side of the equation, we need to distribute that sign all over

Answer 30

\[-x(2x) \rightarrow -2x^2\]

Answer 31

and we need to use FOIL method to expand the left side

Answer 32

I know how to do that cx ill show you after im done foiling it c:

Answer 33

ok

Answer 34

would we do the distributive property to the denomater? or does that stay the same

Answer 35

umm I think it depends on the numerator... because we wouldn't want to combine the denominator and end up that we can cancel something out...taking a step backwards

Answer 36

i foiled the top but didn't add up yet, is this correct so far? 4x+9x+2x+6

Answer 37

\[(3x+2)(x+3) -2x^2 = 3x^2+9x+2x+6 = 3x^2+11x+6 -2x^2 = x^2+11x+6\]

Answer 38

remember

\[(x)(x) \rightarrow (x^1)(x^1) \rightarrow x^{1+1} \rightarrow x^2\]

Answer 39

so... \[(3x)(x) \rightarrow 3x^{1+1} \rightarrow 3x^2\]

Answer 40

Ohhh ok, so now i have x^2+11x-6/2x(x+3)

Answer 41

would we cancel our the 6 from 3?

Answer 42

no

Answer 43

I don't think this can be factored... the only pairs for 6 is
6 x 1
1 x 6
2 x 3
3 x2

that's 6+1 =7
3+2 = 5 and subtracting will make it worse

Answer 44

maybe thats the restrictions they talk about?

Answer 45

i think

Answer 46

\[\frac{x^2+11x-6}{2x(x+3)}\] maybe I wish it was more specific.. there is also what values of x makes this undefined too

Answer 47

2x(x+3) = 0 split into 2 problems
2x=0
x+3=0
solving x ...
X = 0,-3

Answer 48

those are probably the restrictions Cx , i think it does mean undifined

Answer 49

ok so the next question wants me to do this

Answer 50

\[\frac{x^2+11x-6}{2x(x+3)} \times \frac{x+3}{x+1}\]

Answer 51

its x-1 cx

Answer 52

we can cancel out the x+3... it's in the numerator and denominator

\[\frac{x^2+11x-6}{2x(x+3)} \times \frac{x+3}{x-1}\]

Answer 53

oh true! so they would disapear?

Answer 54

YUP! :D

Answer 55

yes! the easiest step on here lol cx so then the problem would be x^2+11-6/2 times x-1

Answer 56

*divided by 2x not 2

Answer 57

how would that work? since the second one isnt a denominater nor a numerator?

Answer 58

\[\frac{x^2+11x-6}{2x} \times \frac{1}{x-1} \rightarrow \frac{x^2+11x-6}{2x(x-1)}\]

Answer 59

huh?! ok wait I just latex the fractions without x+3 because the numerator had one and the denominator had one .

Answer 60

ohhh ok i remebr learning that step now, so it would stay like that? :o

Answer 61

what happened :o

Answer 62

what is the next question?

Answer 63

you said they cancled eachother out cx, and now i have to divide that equation by 6/x-1, but it also says to (Discuss why the degree of the resulting denominator did not change from your expression’s degree)

Answer 64

\[\frac{x^2+11x-6}{2x(x-1)} \] divided by \[\frac{6}{x-1}\]

Answer 65

yea

Answer 66

if you flip the second fraction... surprise x-1 gets canceled out too.

Answer 67

\[\frac{x^2+11x-6}{2x(x-1)} \times \frac{x-1}{6}\]

Answer 68

ohh and it turns into a mutplication by flipping?

Answer 69

when dividing fractions... you flip the second fraction... in the end you're multiplying

Answer 70

Oh ok, that makes it easier cx

Answer 71

\[\frac{x^2+11x-6}{2x} \times \frac{1}{6} \rightarrow \frac{x^2+11x-6}{12x}\]

Answer 72

I think it's safe now to combine the denominator...

Answer 73

oh yeah i keep forget that the single number would be on bottom and the 1 on top, and yea it does xD

Answer 74

so then i would have to explain *Discuss why the degree of the resulting denominator did not change from your expression’s degree*

Answer 75

what the flip does it mean by that lol :x

Answer 76

what is that?!

Answer 77

12x that's first degree isn't it?! awww c*** I haven't dealt with these questions in a while. pure math doesn't have this

Answer 78

well we got the retrictions Q , and i just wanna finish this one assignemnt thingy lol, you look like your doing it right! cx its just these wierd Q's

Answer 79

how did it not change from the expression's degree? could it be referring to the original question?!

Answer 80

the first question you had ... contained 2x

our fourth question had the denominator 2x(6) -> 12x

I think they are the same first degree. That's just a guess ..