Firejay5
  • Firejay5
Simplify Each Expression. Show work and explain it. 32. m over m^2 - 4 + 2 over 3m + 6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Firejay5
  • Firejay5
Well on #32, I am having trouble multiplying the denominators to make them the same
anonymous
  • anonymous
"m over m^2 - 4 + 2 over 3m + 6" \[ \frac{m}{m^2-4}+\frac{2}{3m+6} \]
Firejay5
  • Firejay5
why

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anonymous
  • anonymous
Just use this: \[ \frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd} \]
anonymous
  • anonymous
I think the + is separating two fractions.
Firejay5
  • Firejay5
yes
Firejay5
  • Firejay5
yea
Firejay5
  • Firejay5
can you do it with me and then I'll do 33.) my self
Firejay5
  • Firejay5
what do you mean?
Firejay5
  • Firejay5
Equation editor
Firejay5
  • Firejay5
can we get back to the problem
Firejay5
  • Firejay5
or not @Hero
Firejay5
  • Firejay5
the example he gave isn't going nowhere for me
anonymous
  • anonymous
let \(b=m^2-4\) and \(d=3m+6\)
anonymous
  • anonymous
\[ \frac{m}{m^2-4}+\frac{2}{3m+6}=\frac{(m)(3m+6)+(2)(m^2-4)}{(m^2-4)(3m+6)} \]
anonymous
  • anonymous
@Firejay5 Do you see the pattern?
Firejay5
  • Firejay5
Yes? Kind of how did you get that
Firejay5
  • Firejay5
I need full out steps, like getting the denominators equal, etc.
Firejay5
  • Firejay5
@Mertsj could you help with that
mathslover
  • mathslover
hmn . \[\frac{m}{m^2-4}+\frac{2}{3m+6}\] Well I think that it would be Ok to factorize both denominators
Mertsj
  • Mertsj
Yes. mathslover will be able to show you.
Firejay5
  • Firejay5
@Mertsj you sure
mathslover
  • mathslover
\(\color{blue}{m^2-4}\) is in the form of \(a^2-b^2\) \(a^2-b^2 = (a+b)(a-b)\) Use the above stated identity to factorize \(m^2-4 \) .
Mertsj
  • Mertsj
yes
mathslover
  • mathslover
Can you factor \(m^2-4\) @Firejay5 ?
Firejay5
  • Firejay5
yes
mathslover
  • mathslover
Can you tell me what would be the factorized form for \(x^2-4\) ?
Firejay5
  • Firejay5
(x - 2) & (x + 2)
mathslover
  • mathslover
Great! So \(x^2-4 = \color{blue}{(x+2)(x-2)}\) Now. we have the complete question as : \(\cfrac{m}{\color{red}{(m+2)(m-2)}}+\cfrac{2}{3m+6}\)
mathslover
  • mathslover
Now take 3 common from 3m + 6. What you get?
Firejay5
  • Firejay5
I did the other factorization as well
Firejay5
  • Firejay5
3(m + 2)
mathslover
  • mathslover
Great! \(\cfrac{m}{\color{red}{(m+2)(m-2)}}+\cfrac{2}{\color{orange}{3(m+2)}}\). Is what we have now.
mathslover
  • mathslover
Now take LCM .
Firejay5
  • Firejay5
m + 2
mathslover
  • mathslover
Check it again
Firejay5
  • Firejay5
and m - 2
mathslover
  • mathslover
LCM of \(\cfrac{1}{a+b} + \cfrac{1}{2(a-b)} \) (denominators) is (a+b)(a-b)2
mathslover
  • mathslover
Similarly do that there. we have \((x+2)(x-2) \quad\textbf{and}\quad 3(m+2)\) as the denominators
Firejay5
  • Firejay5
it would be 3(m - 2) (m + 2)
mathslover
  • mathslover
Yeah . and sorry there for using x on the place of m
mathslover
  • mathslover
So I have : 3(m-2)(m+2) as LCM can you solve the fraction :
Firejay5
  • Firejay5
so what would it be entered in equation editor
mathslover
  • mathslover
I didn't get you,, please explain!
Firejay5
  • Firejay5
type the problem in fraction form of what we have left
mathslover
  • mathslover
hmn well it would be hard to say that I have : \(\cfrac{\textbf{yet to come}}{3(m+2)(m-2)}\)
mathslover
  • mathslover
we have to solve numerator now
mathslover
  • mathslover
can u do that?
Firejay5
  • Firejay5
wouldn't the numerator m + 2
mathslover
  • mathslover
No! See : \(\cfrac{1}{(a+b)} + \cfrac{2}{(a-b)} = \cfrac{1(a-b) + 2(a+b)}{(a-b)(a+b)}\)
Firejay5
  • Firejay5
how does it work, you should + across
mathslover
  • mathslover
|dw:1364264397027:dw|
Firejay5
  • Firejay5
doesn't make sense
mathslover
  • mathslover
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
mathslover
  • mathslover
getting it?
anonymous
  • anonymous
Wow, finding the LCM is makes it complicated.
Firejay5
  • Firejay5
????
mathslover
  • mathslover
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) \(\cfrac{\cancel{3(m+2)} (m-2)}{\cancel{3(m+2)} }\)
mathslover
  • mathslover
Does that mean you don't get it @Firejay5 ?
Firejay5
  • Firejay5
kind of
Firejay5
  • Firejay5
are the denominators the same
mathslover
  • mathslover
No.
mathslover
  • mathslover
|dw:1364264678904:dw|
Firejay5
  • Firejay5
8/15
mathslover
  • mathslover
But how?
Firejay5
  • Firejay5
whatever you multiply to the bottom, you must do to the top
Firejay5
  • Firejay5
work the problem a whole different way
mathslover
  • mathslover
|dw:1364264834069:dw|
Firejay5
  • Firejay5
2(a - c) or 2(a + c)
mathslover
  • mathslover
The LCM would be : 2(a+c)(a-c)
Firejay5
  • Firejay5
let's work the problem a different way and not factorize, and get the denominators the same
mathslover
  • mathslover
Yes , that would be easy for u.
mathslover
  • mathslover
Well which denominator you want to be changed
Firejay5
  • Firejay5
You're way is not what we were taught in class and is confusing me. 3m + 6
mathslover
  • mathslover
Sorry, ok let us change 3(m+2)
mathslover
  • mathslover
Would you try that for once?
Firejay5
  • Firejay5
what
mathslover
  • mathslover
to change the denominator ?
Firejay5
  • Firejay5
I have been trying
mathslover
  • mathslover
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 \(\cfrac{(m-2)}{3}\)
Firejay5
  • Firejay5
where did you get m - 2 over 3
Firejay5
  • Firejay5
don't jump ahead, it feels like you are just picking random numbers and variable from the problem
Mertsj
  • Mertsj
Are you all clear now, Firejay?
Firejay5
  • Firejay5
kind of would you explain it the same way
Mertsj
  • Mertsj
Do you want to talk about the original problem or what?
Firejay5
  • Firejay5
It feels like I am being taught a whole different then I learned it in class
Firejay5
  • Firejay5
different way
Mertsj
  • Mertsj
I just want to make sure we are talking about the right problem. Is this it:
Mertsj
  • Mertsj
|dw:1364265923023:dw|
Firejay5
  • Firejay5
yes! :D
Mertsj
  • Mertsj
|dw:1364265995425:dw|
Mertsj
  • Mertsj
When you add fractions you need to have the denominators the same. So we have to build the common denominator.
Firejay5
  • Firejay5
@mathslover you did good though! :D
Mertsj
  • Mertsj
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)
Mertsj
  • Mertsj
|dw:1364266190884:dw|
Mertsj
  • Mertsj
Are you with me so far?
Mertsj
  • Mertsj
|dw:1364266301603:dw|
Mertsj
  • Mertsj
And finally: |dw:1364266416514:dw|
Firejay5
  • Firejay5
does it matter which order the m - 2 or m + 2 goes
Firejay5
  • Firejay5
uh???
Firejay5
  • Firejay5
I put 5m - 4 over 3(m + 2)(m - 2)
Mertsj
  • Mertsj
It doesn't matter.
Firejay5
  • Firejay5
How would you type 33.) y over y + 3 (-) 6y over y^2 - 9 in Wolframalpha
Mertsj
  • Mertsj
\[\frac{y}{y+3}-\frac{6y}{y^2-9}\]
Firejay5
  • Firejay5
yea
Mertsj
  • Mertsj
For wolf: y/(y+3)-(6y)/(y^2-9)
Firejay5
  • Firejay5
which of the alternative forms are the answer
Mertsj
  • Mertsj
|dw:1364267512777:dw|
Mertsj
  • Mertsj
That's the abbreviated version.
Firejay5
  • Firejay5
is it the first, second, or third one
Mertsj
  • Mertsj
|dw:1364267994242:dw|

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