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Simplify Each Expression. Show work and explain it. 32. m over m^2 - 4 + 2 over 3m + 6

Mathematics
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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

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Other answers:

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
or not @Hero
the example he gave isn't going nowhere for me
let \(b=m^2-4\) and \(d=3m+6\)
\[ \frac{m}{m^2-4}+\frac{2}{3m+6}=\frac{(m)(3m+6)+(2)(m^2-4)}{(m^2-4)(3m+6)} \]
@Firejay5 Do you see the pattern?
Yes? Kind of how did you get that
I need full out steps, like getting the denominators equal, etc.
@Mertsj could you help with that
hmn . \[\frac{m}{m^2-4}+\frac{2}{3m+6}\] Well I think that it would be Ok to factorize both denominators
Yes. mathslover will be able to show you.
@Mertsj you sure
\(\color{blue}{m^2-4}\) is in the form of \(a^2-b^2\) \(a^2-b^2 = (a+b)(a-b)\) Use the above stated identity to factorize \(m^2-4 \) .
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)
Great! So \(x^2-4 = \color{blue}{(x+2)(x-2)}\) Now. we have the complete question as : \(\cfrac{m}{\color{red}{(m+2)(m-2)}}+\cfrac{2}{3m+6}\)
Now take 3 common from 3m + 6. What you get?
I did the other factorization as well
3(m + 2)
Great! \(\cfrac{m}{\color{red}{(m+2)(m-2)}}+\cfrac{2}{\color{orange}{3(m+2)}}\). Is what we have now.
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
See i have LCM as 3(m+2)(m-2) I divided LCM by the first fraction's denominator that is (m+2)(m-2) to get 3 . Now I multiplied this 3 by the first fractions' numerator that is m
getting it?
Wow, finding the LCM is makes it complicated.
????
Similarly for the second one.. I have LCM as 3(m+2)(m-2) I divided this LCM by the denominator of the second fraction 3(m+2) \(\cfrac{\cancel{3(m+2)} (m-2)}{\cancel{3(m+2)} }\)
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
ok! I have first denominator as (m+2)(m-2) and other one as 3(m+2) I want to make 3(m+2) = (m+2)(m-2) For that I think I should multiply 3(m+2) by \(\cfrac{(m-2)}{3}\)
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|
When you add fractions you need to have the denominators the same. So we have to build the common denominator.
@mathslover you did good though! :D
For the first denominator we need the factors m-2 and m+2. For the second denominator we need the factors 3 and m+2. But we already have m+2. So the 3 factors we need are 3(m-2)(m+2)
|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|

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