Question

Last Summer School Question Please HELP
WILL FAN AND GIVE MEDAL FOR ANSWER
Use complete sentences to describe the steps taken to simplify this problem. Make sure you include information about the new common denominator, the final simplified expression, and any restrictions.
(3/x)-(3/x+1)-(3/x+2)

Answer 1

@theEric Can you help me with this one?

Answer 2

\[\left(\frac{3}{x}\right)-\left(\frac{3}{x}+1\right)-\left(\frac{3}{x}+2\right)\]
Any idea where to start?

Answer 3

oh the 1 and 2 are with the x's under the bar..

Answer 4

(3/x)-(3/(x+1))-(3/(x+2))? Okay! :)

Answer 5

Yup! So I know they all need the same denominator

Answer 6

\[\left(\frac{3}{x}\right)-\left(\frac{3}{x+1}\right)-\left(\frac{3}{x+2}\right)\]

Answer 7

Yeah that's right so you would.... multiply the first one by (x+1) and (x+2)?

Answer 8

Yeah, that's the basic idea... I looked on the internet quick to try to find a shortcut, but found none so far. We can do it that way, but it'll be messy!

And we always multiply by \(1\), so we don't change the actual amount! (Anything times \(1\) is itself.)

And\[1=\frac{x}{x}=\frac{x+1}{x+1}=\frac{x+2}{x+2}\]:)

Answer 9

whoa okay... uh what next then?

Answer 10

So you know we're headed towards a common denominator of \((x)(x+1)(x+2)\), so you want to write that down.

\[\left(\frac{3}{x}\right)\left(\frac{x+1}{x+1}\right)\left(\frac{x+2}{x+2}\right)-\left(\frac{3}{x+1}\right)-\left(\frac{3}{x+2}\right)\]That's what you said! Now

Answer 11

Okay I got
\[\left( 3x ^{2}+9x+6 \div x(x+1)(x+2)\right)\]

Answer 12

Let me see what I get.

Answer 13

Hmm... We got a different answer, but I don't know which of us did the wrong thing. So! Did you get\[\frac{3(x+1)(x+2)-3(x)(x+2)-3(x)(x+1)}{x(x+1)(x+2)}\] along the way?

Answer 14

After going back and doing that to all of them yes I did.
for my final answer I got...\[-3x ^{2}/x(x+1)(x+2)\]

Answer 15

Alright! Now there's just one slight difference. I would like you to look back at one part in your work, which will be enlarged here:\[\small\frac{3\Large (x+1)(x+2)\small-3(x)(x+2)-3(x)(x+1)}{x(x+1)(x+2)}\]

Do you find that \((x+1)(x+2)=x^2+3x+2\)?

Answer 16

Yes I did!

Answer 17

Alright! :) How did you cancel that \(2\) out?

Answer 18

Because I didn't, in my work.

Answer 19

Oh what I did was multiply them all out... And I'm mistaken. I canceled it out after multiplying them all out and cancelling out the 6 with at 6x... Whoops

Answer 20

Haha, easy mistake! So, how dos that affect your final answer? I did cancel out all my terms with a plain \(x\) in them. I got \[\frac{3(2-x^2)}{x(x+1)(x+2)}=\frac{6-6x^2}{x(x+1)(x+2)}\]

Answer 21

Yes that is what I got after going back through!

Answer 22

Cool! Now, about the denominator. Should we FOIL that?

Answer 23

I don't know if that is making it simpler or not.

Answer 24

Uhhh... Yes?
Well I guess what I would do would be to foil out (x+1)(x+2) and then multiply x with the answer to that?

Answer 25

Okay!

Answer 26

So it would be \[3x ^{3}+3x ^{2}+3x\]

Answer 27

?

Answer 28

\((x+1)(x+2)=(x\times x)+(x\times 2)+(1\times x)+(1\times 2)\\=x^2+2x+x+2\\=x^2+3x+2\)

Answer 29

\((x)(x^2+3x+2)=(x\times x^2)+(x\times 3x)+(x\times 2)\\=x^3+3x^2+2x\)

Answer 30

Did I do it right? Or can you find your mistake?

Answer 31

Oh! Rookie mistake I meant 2!

Answer 32

2*1=2 wow can you tell I'm tired? haha

Answer 33

Haha, tiny mistakes...

Answer 34

Oh gosh.. You did say this is your last question, though! Until you get more familiar and relaxed with algebra, it can be exhausting in itself.

Answer 35

on it's own, I mean.

Answer 36

Yes one of them! I think there are two more after this one but they shouldn't be that big of a deal... Or trouble... If you know what I mean

Answer 37

So, now you need to look through your work and make sure you know how you got from each step to the next (and which steps you should erase because they had mistakes).

Answer 38

I know what you mean. OpenStudy will be here. And you seem to know what you're doing anyway.

Answer 39

By the way, before you simplified, the denominator was \(x(x+1)(x+2)\). That, the denominator, cannot be \(0\) or else the expression is undefined. I think that is what the problem wants to know when it asks for "restrictions".

Answer 40

So, for the restrictions, you have to say what \(x\) cannot be.

Answer 41

Sorry! my wifi went haywire! And I would say that x cannot equal -1 or -2 because then you will be dividing by 0.

Answer 42

Ah, that stinks! Right! There's one more thing \(x\) can be!

Answer 43

x can't be 0...?

Answer 44

Right. :)

Answer 45

Okay! Thank you so much!

Answer 46

You're welcome! Good luck with the sentences!

Answer 47

Thank you! Do you have time for one more question?

Answer 48

Sure! Post it! Then send me a link to or it use the "@" thing.