Question

Simplify Each Expression. Show work and explain it.

32. m over m^2 - 4 + 2 over 3m + 6

Answer 1

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

Answer 2

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

Answer 3

why

Answer 4

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

Answer 5

I think the + is separating two fractions.

Answer 6

yes

Answer 7

yea

Answer 8

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

Answer 9

what do you mean?

Answer 10

Equation editor

Answer 11

can we get back to the problem

Answer 12

or not @Hero

Answer 13

the example he gave isn't going nowhere for me

Answer 14

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

Answer 15

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

Answer 16

@Firejay5 Do you see the pattern?

Answer 17

Yes? Kind of how did you get that

Answer 18

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

Answer 19

@Mertsj could you help with that

Answer 20

hmn .
\[\frac{m}{m^2-4}+\frac{2}{3m+6}\]

Well I think that it would be Ok to factorize both denominators

Answer 21

Yes. mathslover will be able to show you.

Answer 22

@Mertsj you sure

Answer 23

\(\color{blue}{m^2-4}\) is in the form of \(a^2-b^2\)

\(a^2-b^2 = (a+b)(a-b)\)

Use the above stated identity to factorize \(m^2-4 \) .

Answer 24

yes

Answer 25

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

Answer 26

yes

Answer 27

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

Answer 28

(x - 2) & (x + 2)

Answer 29

Great! So \(x^2-4 = \color{blue}{(x+2)(x-2)}\)

Now. we have the complete question as :
\(\cfrac{m}{\color{red}{(m+2)(m-2)}}+\cfrac{2}{3m+6}\)

Answer 30

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

Answer 31

I did the other factorization as well

Answer 32

3(m + 2)

Answer 33

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

Answer 34

Now take LCM .

Answer 35

m + 2

Answer 36

Check it again

Answer 37

and m - 2

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

So I have : 3(m-2)(m+2) as LCM

can you solve the fraction :

Answer 43

so what would it be entered in equation editor

Answer 44

I didn't get you,, please explain!

Answer 45

type the problem in fraction form of what we have left

Answer 46

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

Answer 47

we have to solve numerator now

Answer 48

can u do that?

Answer 49

wouldn't the numerator m + 2

Answer 50

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

Answer 51

how does it work, you should + across