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

- Firejay5

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

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- Firejay5

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

- anonymous

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

- Firejay5

why

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## More answers

- anonymous

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

- anonymous

I think the + is separating two fractions.

- Firejay5

yes

- Firejay5

yea

- Firejay5

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

- Firejay5

what do you mean?

- Firejay5

Equation editor

- Firejay5

can we get back to the problem

- Firejay5

or not @Hero

- Firejay5

the example he gave isn't going nowhere for me

- anonymous

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

- anonymous

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

- anonymous

@Firejay5 Do you see the pattern?

- Firejay5

Yes? Kind of how did you get that

- Firejay5

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

- Firejay5

@Mertsj could you help with that

- mathslover

hmn .
\[\frac{m}{m^2-4}+\frac{2}{3m+6}\]
Well I think that it would be Ok to factorize both denominators

- Mertsj

Yes. mathslover will be able to show you.

- Firejay5

@Mertsj you sure

- 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

yes

- mathslover

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

- Firejay5

yes

- mathslover

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

- Firejay5

(x - 2) & (x + 2)

- 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

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

- Firejay5

I did the other factorization as well

- Firejay5

3(m + 2)

- mathslover

Great! \(\cfrac{m}{\color{red}{(m+2)(m-2)}}+\cfrac{2}{\color{orange}{3(m+2)}}\). Is what we have now.

- mathslover

Now take LCM .

- Firejay5

m + 2

- mathslover

Check it again

- Firejay5

and m - 2

- mathslover

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

- mathslover

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

- Firejay5

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

- mathslover

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

- mathslover

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

- Firejay5

so what would it be entered in equation editor

- mathslover

I didn't get you,, please explain!

- Firejay5

type the problem in fraction form of what we have left

- mathslover

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

- mathslover

we have to solve numerator now

- mathslover

can u do that?

- Firejay5

wouldn't the numerator m + 2

- mathslover

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

- Firejay5

how does it work, you should + across

- mathslover

|dw:1364264397027:dw|

- Firejay5

doesn't make sense

- 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

getting it?

- anonymous

Wow, finding the LCM is makes it complicated.

- Firejay5

????

- 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

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

- Firejay5

kind of

- Firejay5

are the denominators the same

- mathslover

No.

- mathslover

|dw:1364264678904:dw|

- Firejay5

8/15

- mathslover

But how?

- Firejay5

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

- Firejay5

work the problem a whole different way

- mathslover

|dw:1364264834069:dw|

- Firejay5

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

- mathslover

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

- Firejay5

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

- mathslover

Yes , that would be easy for u.

- mathslover

Well which denominator you want to be changed

- Firejay5

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

- mathslover

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

- mathslover

Would you try that for once?

- Firejay5

what

- mathslover

to change the denominator ?

- Firejay5

I have been trying

- 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

where did you get m - 2 over 3

- Firejay5

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

- Mertsj

Are you all clear now, Firejay?

- Firejay5

kind of would you explain it the same way

- Mertsj

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

- Firejay5

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

- Firejay5

different way

- Mertsj

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

- Mertsj

|dw:1364265923023:dw|

- Firejay5

yes! :D

- Mertsj

|dw:1364265995425:dw|

- Mertsj

When you add fractions you need to have the denominators the same. So we have to build the common denominator.

- Firejay5

@mathslover you did good though! :D

- 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

|dw:1364266190884:dw|

- Mertsj

Are you with me so far?

- Mertsj

|dw:1364266301603:dw|

- Mertsj

And finally:
|dw:1364266416514:dw|

- Firejay5

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

- Firejay5

uh???

- Firejay5

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

- Mertsj

It doesn't matter.

- Firejay5

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

- Mertsj

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

- Firejay5

yea

- Mertsj

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

- Firejay5

which of the alternative forms are the answer

- Mertsj

|dw:1364267512777:dw|

- Mertsj

That's the abbreviated version.

- Firejay5

is it the first, second, or third one

- Mertsj

|dw:1364267994242:dw|

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