anonymous
  • anonymous
Does anyone know how to add and subtract rational expressions? help very last assignment D:
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
Its an assignment I have to do, but its all about adding and subtracting rational expression, only 5 problems
Nurali
  • Nurali
http://www.cliffsnotes.com/math/algebra/algebra-ii/rational-expressions/adding-and-subtracting-rational-expressions
Nurali
  • Nurali
Hope this will help

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anonymous
  • anonymous
Ok c: Thanks
Nurali
  • Nurali
My Pleasure.
anonymous
  • anonymous
is it ok if i do one problem and show you to make sure im not doing anything wrong?
anonymous
  • anonymous
@Nurali
UsukiDoll
  • UsukiDoll
post an example
UsukiDoll
  • UsukiDoll
I think the other user is afk or something
anonymous
  • anonymous
ill draw a picture of it
anonymous
  • anonymous
|dw:1433843733453:dw|
UsukiDoll
  • UsukiDoll
hmmm... they share a common denominator so all we have to do is to combine like terms in the numerator
anonymous
  • anonymous
How would we do that? would we add 2x to x and 1 to the other 1?
UsukiDoll
  • UsukiDoll
yeah
anonymous
  • anonymous
would the numerator come out like this = 2x^2+2 or 3x+2
UsukiDoll
  • UsukiDoll
it would be 3x+2. because 2x+x = 3x 1 + 1 =2
anonymous
  • anonymous
Ok! thanks, can you help with four more adding/subracting q's?
UsukiDoll
  • UsukiDoll
sure
anonymous
  • 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
  • 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
  • anonymous
so like cross multiplying?
UsukiDoll
  • 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
  • anonymous
It looks kinda confusing sine the problem big :x
anonymous
  • anonymous
on cliff notes
UsukiDoll
  • UsukiDoll
\[\frac{3x+2}{2x} \times \frac{x+3}{x+3}-\frac{x}{x+3} \times \frac{2x}{2x}\]
anonymous
  • anonymous
ok that makes alot more since lol
UsukiDoll
  • 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
  • anonymous
would we just times the bottom demonitaers togethor? like 2x(x+3)
UsukiDoll
  • UsukiDoll
yes...
UsukiDoll
  • 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
  • UsukiDoll
\[-x(2x) \rightarrow -2x^2\]
UsukiDoll
  • UsukiDoll
and we need to use FOIL method to expand the left side
anonymous
  • anonymous
I know how to do that cx ill show you after im done foiling it c:
UsukiDoll
  • UsukiDoll
ok
anonymous
  • anonymous
would we do the distributive property to the denomater? or does that stay the same
UsukiDoll
  • 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
  • anonymous
i foiled the top but didn't add up yet, is this correct so far? 4x+9x+2x+6
UsukiDoll
  • UsukiDoll
\[(3x+2)(x+3) -2x^2 = 3x^2+9x+2x+6 = 3x^2+11x+6 -2x^2 = x^2+11x+6\]
UsukiDoll
  • UsukiDoll
remember \[(x)(x) \rightarrow (x^1)(x^1) \rightarrow x^{1+1} \rightarrow x^2\]
UsukiDoll
  • UsukiDoll
so... \[(3x)(x) \rightarrow 3x^{1+1} \rightarrow 3x^2\]
anonymous
  • anonymous
Ohhh ok, so now i have x^2+11x-6/2x(x+3)
anonymous
  • anonymous
would we cancel our the 6 from 3?
UsukiDoll
  • UsukiDoll
no
UsukiDoll
  • 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
  • anonymous
maybe thats the restrictions they talk about?
anonymous
  • anonymous
i think
UsukiDoll
  • 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
  • UsukiDoll
2x(x+3) = 0 split into 2 problems 2x=0 x+3=0 solving x ... X = 0,-3
anonymous
  • anonymous
those are probably the restrictions Cx , i think it does mean undifined
anonymous
  • anonymous
ok so the next question wants me to do this |dw:1433845443996:dw|
UsukiDoll
  • UsukiDoll
\[\frac{x^2+11x-6}{2x(x+3)} \times \frac{x+3}{x+1}\]
anonymous
  • anonymous
its x-1 cx
UsukiDoll
  • 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
  • anonymous
oh true! so they would disapear?
UsukiDoll
  • UsukiDoll
YUP! :D
anonymous
  • anonymous
yes! the easiest step on here lol cx so then the problem would be x^2+11-6/2 times x-1
anonymous
  • anonymous
*divided by 2x not 2
anonymous
  • anonymous
how would that work? since the second one isnt a denominater nor a numerator?
UsukiDoll
  • UsukiDoll
\[\frac{x^2+11x-6}{2x} \times \frac{1}{x-1} \rightarrow \frac{x^2+11x-6}{2x(x-1)}\]
UsukiDoll
  • UsukiDoll
huh?! ok wait I just latex the fractions without x+3 because the numerator had one and the denominator had one .
anonymous
  • anonymous
ohhh ok i remebr learning that step now, so it would stay like that? :o
anonymous
  • anonymous
what happened :o
UsukiDoll
  • UsukiDoll
what is the next question?
anonymous
  • 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 expression’s degree)
UsukiDoll
  • UsukiDoll
\[\frac{x^2+11x-6}{2x(x-1)} \] divided by \[\frac{6}{x-1}\]
anonymous
  • anonymous
yea
UsukiDoll
  • UsukiDoll
if you flip the second fraction... surprise x-1 gets canceled out too.
UsukiDoll
  • UsukiDoll
\[\frac{x^2+11x-6}{2x(x-1)} \times \frac{x-1}{6}\]
anonymous
  • anonymous
ohh and it turns into a mutplication by flipping?
UsukiDoll
  • UsukiDoll
when dividing fractions... you flip the second fraction... in the end you're multiplying
anonymous
  • anonymous
Oh ok, that makes it easier cx
UsukiDoll
  • UsukiDoll
\[\frac{x^2+11x-6}{2x} \times \frac{1}{6} \rightarrow \frac{x^2+11x-6}{12x}\]
UsukiDoll
  • UsukiDoll
I think it's safe now to combine the denominator...
anonymous
  • 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
  • anonymous
so then i would have to explain *Discuss why the degree of the resulting denominator did not change from your expression’s degree*
anonymous
  • anonymous
what the flip does it mean by that lol :x
UsukiDoll
  • UsukiDoll
what is that?!
UsukiDoll
  • 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
  • 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
  • UsukiDoll
how did it not change from the expression's degree? could it be referring to the original question?!
UsukiDoll
  • 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
  • anonymous
maybe they mean degree by like * x, x^2, x^3
UsukiDoll
  • 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
  • anonymous
haha yea, ill just say, i shouldnt get that many points taken off since were doing the rest right
anonymous
  • anonymous
say that*
anonymous
  • anonymous
Ok, so thats our final answer for Q 4, now we just have one left!
UsukiDoll
  • UsukiDoll
o-o I thought that was the final one?!
anonymous
  • anonymous
no this one is the final one lol,
anonymous
  • anonymous
i had 5 Q's
anonymous
  • anonymous
4* i mean
UsukiDoll
  • UsukiDoll
????????????
anonymous
  • anonymous
its the very last one, please helpp :3
UsukiDoll
  • UsukiDoll
ok
anonymous
  • anonymous
|dw:1433847285555:dw|
anonymous
  • anonymous
its that one^
UsukiDoll
  • UsukiDoll
ok first combine the second and third fraction
anonymous
  • anonymous
i meant subract 6
anonymous
  • anonymous
not + and ok cx
anonymous
  • anonymous
2x/2x+2 ?
UsukiDoll
  • UsukiDoll
|dw:1433847473742:dw|
UsukiDoll
  • UsukiDoll
x + x = 2x .. the denominator stays the same..... OMG EVEN I DID +6 #FAIL
anonymous
  • anonymous
I|dw:1433847507052:dw|
anonymous
  • anonymous
lol its, its late at night lol xD
UsukiDoll
  • UsukiDoll
\[\frac{x^2+11x-6}{12x} + \frac{2x}{x+1}\]
UsukiDoll
  • UsukiDoll
now use the least common denominator which is 12x(x+1)
anonymous
  • anonymous
so we would do that whole timezing the first by the equation and the second by the second one?
UsukiDoll
  • 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
  • UsukiDoll
\[\frac{x^2+11x-6}{12x} \times \frac{x+1}{x+1} + \frac{2x}{x+1} \times\frac{2x}{2x}\]
UsukiDoll
  • UsukiDoll
OOPS
anonymous
  • anonymous
wait, so why did we mutiply 2x instead of 12?
UsukiDoll
  • UsukiDoll
cuz I typoed
UsukiDoll
  • UsukiDoll
\[\frac{x^2+11x-6}{12x} \times \frac{x+1}{x+1} + \frac{2x}{x+1} \times\frac{12x}{12x}\]
anonymous
  • anonymous
ok lol, and for example on the first bottom ones we would do (12x)(x+1) ?
UsukiDoll
  • UsukiDoll
yeah
UsukiDoll
  • UsukiDoll
numerator on the left is going to be a bit tough though
anonymous
  • anonymous
ok :x ill do the denomnaters and show u
anonymous
  • anonymous
|dw:1433847901459:dw|
UsukiDoll
  • UsukiDoll
|dw:1433847957533:dw|
UsukiDoll
  • UsukiDoll
\[(x^2+11x-6)(x+1) = x^3+11x^2-6x+x^2+11x-6 = x^3+12x^2+5x-6\]
UsukiDoll
  • UsukiDoll
\[x^3+12x^2+5x-6+24x^2 = x^3+36x^2+5x-6\]
UsukiDoll
  • UsukiDoll
\[\frac{x^3+36x^2+5x-6}{12x(x+1)}\]
anonymous
  • anonymous
is that the end point? xO
anonymous
  • anonymous
or do we do stuff with the denomatr to the numerator now?
UsukiDoll
  • UsukiDoll
hmmm... I think that's it?!
UsukiDoll
  • UsukiDoll
we can't use factor by grouping with the numerator
UsukiDoll
  • UsukiDoll
not a perfect cube either
anonymous
  • 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
  • UsukiDoll
I think it was due to my latex writing too.. that takes a bit of time and thinking xD
anonymous
  • 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
  • anonymous
again thank you so muccccch! very appreciated that you helped me cx
UsukiDoll
  • UsukiDoll
:D

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