Question

Please Help! I'm really stuck ;(
2 2 2
--- + --- - ---- simplify and explain please!
x x+1 x+2

Answer 1

Guys please help, it's the only question i'm having troubles with

Answer 2

\[\Large\rm \frac{2}{x}+\frac{2}{(x+1)}-\frac{2}{(x+2)}\]Hey kitty kat lady c:
So we have to combine these into a single fraction and simplify?

Answer 3

mhmm :P it's simple I just can't seem to figure out the Least common Denominator

Answer 4

If the first denominator was 2,
the second denominator would be 2+1, or 3, yah?
And the last denominator would be 4.

So the common denominator could be 2x3x4
or 2(2+1)(2+2)

Answer 5

Same idea here with x,
these are all different factors,
we assume they have nothing in common.

So our common denominator that we're looking for is
x(x+1)(x+2)

Answer 6

So in the first fraction, looks like we need an (x+1)(x+2) to get that common denominator, yes?\[\Large\rm \color{royalblue}{\frac{(x+1)(x+2)}{(x+1)(x+2)}}\cdot\frac{2}{x}+\frac{2}{(x+1)}-\frac{2}{(x+2)}\]

Answer 7

yea, trying to keep up but quick question, what are we multiplying all of the denominators by so that they are all the same ?

Answer 8

We're multiplying them all by different values.
They're each missing different factors.
Example:\[\Large\rm \frac{1}{2}+\frac{1}{3}\]Our common denominator is 2x3,
so our first fraction is missing a 3, while the second fraction is missing a 2.

So the multiplication is going to be different in each case.

Answer 9

oh ok, what are we trying to get each one to look like then in my case?

Answer 10

We're trying to get each one to have a bottom of \(\Large\rm x(x+1)(x+2)\)

Answer 11

oh i see

Answer 12

The first fraction already has \(\Large\rm x\), so it still needs \(\Large\rm (x+1)(x+2)\).
The second fraction has \(\Large\rm (x+1)\), so it still needs \(\Large\rm x(x+2)\).
The third fraction has \(\Large\rm (x+2)\), so it still needs \(\Large\rm x(x+1)\).

Answer 13

Then do we need to muliply the same way to the top?

Answer 14

\[\Large\rm \color{royalblue}{\frac{(x+1)(x+2)}{(x+1)(x+2)}}\cdot\frac{2}{x}+\frac{2}{(x+1)}-\frac{2}{(x+2)}\]
Doing the multiplication for the first one gives us:\[\Large\rm \frac{2(x+1)(x+2)}{x(x+1)(x+2)}+\frac{2}{(x+1)}-\frac{2}{(x+2)}\]That takes care of the first one. :)
Yes, same value multiplying top and bottom (This is the same as multiplying by 1).
If it's not equivalent to 1, then we're changing the value of the expression, which would be bad.

Answer 15

haha makes sense

Answer 16

what do we do from here

Answer 17

So multiply your second fraction (top and bottom) by the missing factors:\[\large\rm \frac{2(x+1)(x+2)}{x(x+1)(x+2)}+\color{orangered}{\frac{x(x+2)}{x(x+2)}}\cdot\frac{2}{(x+1)}-\frac{2}{(x+2)}\]

Answer 18

im stuck

Answer 19

Noooo you're not 0_O

....where?

Answer 20

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

Answer 21

so far....

Answer 22

idk

Answer 23

@zepdrix

Answer 24

Ok looks great so far!
Since they all have the SAME denominator, we can write it as one big fraction now:\[\Large\rm \frac{2(x+1)(x+2)+2x(x+2)-2x(x+1)}{x(x+1)(x+2)}\]

Answer 25

Hopefully you understand that the denominators were the same, yes?\[\Large\rm x(x+1)(x+2) = (x+1)(x+2)x\]

Answer 26

Multiplication is commutative, you can do it in any order.
So we can rearrange our factors to look the same.

Answer 27

yea gotcha!

Answer 28

Leave the bottom alone, he's good to go from here on out.
But now you've got the task of expanding out all of the numerator brackets and then combining like-terms.

That will be a little tedious :) Not too bad though.

Answer 29

is it.. 2(x^2+4x+2) please sayy yes :|

Answer 30

For the numerator?
Oooo good job kitty kat! \c:/ Meowwwww!

Answer 31

Thank so much! I got the answer now! ^_^ appreciate it

Answer 32

Also are the restrictions?

Answer 33

You can't allow your x to take on any value that would make the denominator zero.
That would cause division by 0! Bad bad bad.

Answer 34

Set your denominator equal to zero,
\[\Large\rm x(x+1)(x+2)=0\]And solve for x.
These are the BAD VALUES.
The restricted values.

You should end up with three of them.