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Well on #32, I am having trouble multiplying the denominators to make them the same

"m over m^2 - 4 + 2 over 3m + 6"
\[
\frac{m}{m^2-4}+\frac{2}{3m+6}
\]

why

Just use this: \[
\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}
\]

I think the + is separating two fractions.

yes

yea

can you do it with me and then I'll do 33.) my self

what do you mean?

Equation editor

can we get back to the problem

the example he gave isn't going nowhere for me

let \(b=m^2-4\) and \(d=3m+6\)

\[
\frac{m}{m^2-4}+\frac{2}{3m+6}=\frac{(m)(3m+6)+(2)(m^2-4)}{(m^2-4)(3m+6)}
\]

Yes? Kind of how did you get that

I need full out steps, like getting the denominators equal, etc.

Yes. mathslover will be able to show you.

yes

Can you factor \(m^2-4\) @Firejay5 ?

yes

Can you tell me what would be the factorized form for \(x^2-4\) ?

(x - 2) & (x + 2)

Now take 3 common from 3m + 6. What you get?

I did the other factorization as well

3(m + 2)

Now take LCM .

m + 2

Check it again

and m - 2

LCM of \(\cfrac{1}{a+b} + \cfrac{1}{2(a-b)} \) (denominators) is (a+b)(a-b)2

Similarly do that there. we have \((x+2)(x-2) \quad\textbf{and}\quad 3(m+2)\) as the denominators

it would be 3(m - 2) (m + 2)

Yeah . and sorry there for using x on the place of m

So I have : 3(m-2)(m+2) as LCM
can you solve the fraction :

so what would it be entered in equation editor

I didn't get you,, please explain!

type the problem in fraction form of what we have left

hmn well it would be hard to say that I have :
\(\cfrac{\textbf{yet to come}}{3(m+2)(m-2)}\)

we have to solve numerator now

can u do that?

wouldn't the numerator m + 2

No! See : \(\cfrac{1}{(a+b)} + \cfrac{2}{(a-b)} = \cfrac{1(a-b) + 2(a+b)}{(a-b)(a+b)}\)

how does it work, you should + across

|dw:1364264397027:dw|

doesn't make sense

getting it?

Wow, finding the LCM is makes it complicated.

????

Does that mean you don't get it @Firejay5 ?

kind of

are the denominators the same

No.

|dw:1364264678904:dw|

8/15

But how?

whatever you multiply to the bottom, you must do to the top

work the problem a whole different way

|dw:1364264834069:dw|

2(a - c) or 2(a + c)

The LCM would be : 2(a+c)(a-c)

let's work the problem a different way and not factorize, and get the denominators the same

Yes , that would be easy for u.

Well which denominator you want to be changed

You're way is not what we were taught in class and is confusing me. 3m + 6

Sorry, ok let us change 3(m+2)

Would you try that for once?

what

to change the denominator ?

I have been trying

where did you get m - 2 over 3

don't jump ahead, it feels like you are just picking random numbers and variable from the problem

Are you all clear now, Firejay?

kind of would you explain it the same way

Do you want to talk about the original problem or what?

It feels like I am being taught a whole different then I learned it in class

different way

I just want to make sure we are talking about the right problem. Is this it:

|dw:1364265923023:dw|

yes! :D

|dw:1364265995425:dw|

@mathslover you did good though! :D

|dw:1364266190884:dw|

Are you with me so far?

|dw:1364266301603:dw|

And finally:
|dw:1364266416514:dw|

does it matter which order the m - 2 or m + 2 goes

uh???

I put 5m - 4 over 3(m + 2)(m - 2)

It doesn't matter.

How would you type 33.) y over y + 3 (-) 6y over y^2 - 9 in Wolframalpha

\[\frac{y}{y+3}-\frac{6y}{y^2-9}\]

yea

For wolf: y/(y+3)-(6y)/(y^2-9)

which of the alternative forms are the answer

|dw:1364267512777:dw|

That's the abbreviated version.

is it the first, second, or third one

|dw:1364267994242:dw|