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

- anonymous

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

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- anonymous

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

- Nurali

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

- Nurali

Hope this will help

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## More answers

- anonymous

Ok c: Thanks

- Nurali

My Pleasure.

- anonymous

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

- anonymous

@Nurali

- UsukiDoll

post an example

- UsukiDoll

I think the other user is afk or something

- anonymous

ill draw a picture of it

- anonymous

|dw:1433843733453:dw|

- UsukiDoll

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

- anonymous

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

- UsukiDoll

yeah

- anonymous

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

- UsukiDoll

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

- anonymous

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

- UsukiDoll

sure

- anonymous

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*

- UsukiDoll

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

- anonymous

so like cross multiplying?

- UsukiDoll

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?

- anonymous

It looks kinda confusing sine the problem big :x

- anonymous

on cliff notes

- UsukiDoll

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

- anonymous

ok that makes alot more since lol

- UsukiDoll

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

- anonymous

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

- UsukiDoll

yes...

- UsukiDoll

\[\frac{(3x+2)(x+3)-2x^2}{2x(x+3)}\] 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

- UsukiDoll

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

- UsukiDoll

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

- anonymous

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

- UsukiDoll

ok

- anonymous

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

- UsukiDoll

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

- anonymous

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

- UsukiDoll

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

- UsukiDoll

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

- UsukiDoll

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

- anonymous

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

- anonymous

would we cancel our the 6 from 3?

- UsukiDoll

no

- UsukiDoll

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

- anonymous

maybe thats the restrictions they talk about?

- anonymous

i think

- UsukiDoll

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

- UsukiDoll

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

- anonymous

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

- anonymous

ok so the next question wants me to do this |dw:1433845443996:dw|

- UsukiDoll

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

- anonymous

its x-1 cx

- UsukiDoll

we can cancel out the x+3... it's in the numerator and denominator
\[\frac{x^2+11x-6}{2x(x+3)} \times \frac{x+3}{x-1}\]

- anonymous

oh true! so they would disapear?

- UsukiDoll

YUP! :D

- anonymous

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

- anonymous

*divided by 2x not 2

- anonymous

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

- UsukiDoll

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

- UsukiDoll

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

- anonymous

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

- anonymous

what happened :o

- UsukiDoll

what is the next question?

- anonymous

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 expressionâ€™s degree)

- UsukiDoll

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

- anonymous

yea

- UsukiDoll

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

- UsukiDoll

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

- anonymous

ohh and it turns into a mutplication by flipping?

- UsukiDoll

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

- anonymous

Oh ok, that makes it easier cx

- UsukiDoll

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

- UsukiDoll

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

- anonymous

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

- anonymous

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

- anonymous

what the flip does it mean by that lol :x

- UsukiDoll

what is that?!

- UsukiDoll

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

- anonymous

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

- UsukiDoll

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

- UsukiDoll

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 ..

- anonymous

maybe they mean degree by like * x, x^2, x^3

- UsukiDoll

yeah I'm thinking that we compare the last answer we have to the first question? hmmmmmmmm?! the degrees are the same if we look at it that way?!

- anonymous

haha yea, ill just say, i shouldnt get that many points taken off since were doing the rest right

- anonymous

say that*

- anonymous

Ok, so thats our final answer for Q 4, now we just have one left!

- UsukiDoll

o-o I thought that was the final one?!

- anonymous

no this one is the final one lol,

- anonymous

i had 5 Q's

- anonymous

4* i mean

- UsukiDoll

????????????

- anonymous

its the very last one, please helpp :3

- UsukiDoll

ok

- anonymous

|dw:1433847285555:dw|

- anonymous

its that one^

- UsukiDoll

ok first combine the second and third fraction

- anonymous

i meant subract 6

- anonymous

not + and ok cx

- anonymous

2x/2x+2 ?

- UsukiDoll

|dw:1433847473742:dw|

- UsukiDoll

x + x = 2x .. the denominator stays the same.....
OMG EVEN I DID +6 #FAIL

- anonymous

I|dw:1433847507052:dw|

- anonymous

lol its, its late at night lol xD

- UsukiDoll

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

- UsukiDoll

now use the least common denominator which is 12x(x+1)

- anonymous

so we would do that whole timezing the first by the equation and the second by the second one?

- UsukiDoll

but only multiply what is misssing for the first fraction we have 12x.. x+1 is missing so multiply x+1/x+1

- UsukiDoll

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

- UsukiDoll

OOPS

- anonymous

wait, so why did we mutiply 2x instead of 12?

- UsukiDoll

cuz I typoed

- UsukiDoll

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

- anonymous

ok lol, and for example on the first bottom ones we would do (12x)(x+1) ?

- UsukiDoll

yeah

- UsukiDoll

numerator on the left is going to be a bit tough though

- anonymous

ok :x ill do the denomnaters and show u

- anonymous

|dw:1433847901459:dw|

- UsukiDoll

|dw:1433847957533:dw|

- UsukiDoll

\[(x^2+11x-6)(x+1) = x^3+11x^2-6x+x^2+11x-6 = x^3+12x^2+5x-6\]

- UsukiDoll

\[x^3+12x^2+5x-6+24x^2 = x^3+36x^2+5x-6\]

- UsukiDoll

\[\frac{x^3+36x^2+5x-6}{12x(x+1)}\]

- anonymous

is that the end point? xO

- anonymous

or do we do stuff with the denomatr to the numerator now?

- UsukiDoll

hmmm... I think that's it?!

- UsukiDoll

we can't use factor by grouping with the numerator

- UsukiDoll

not a perfect cube either

- anonymous

yea, it should be lol i cant belive 4 problems took an hour :X, and thank you soOO much! i've asked this Q many times on here and noone has helped me like you have! and yea it doesnt look like a perfect cube

- UsukiDoll

I think it was due to my latex writing too.. that takes a bit of time and thinking xD

- anonymous

i dont think so cx these types of Q's look like theyd take forever lol, and we were moving pretty fast cx

- anonymous

again thank you so muccccch! very appreciated that you helped me cx

- UsukiDoll

:D

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