Firejay5
Simplify Each Expression. Show work and explain it.
32. m over m^2 - 4 + 2 over 3m + 6
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Firejay5
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Well on #32, I am having trouble multiplying the denominators to make them the same
wio
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"m over m^2 - 4 + 2 over 3m + 6"
\[
\frac{m}{m^2-4}+\frac{2}{3m+6}
\]
Firejay5
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why
wio
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Just use this: \[
\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}
\]
wio
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I think the + is separating two fractions.
Firejay5
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yes
Firejay5
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yea
Firejay5
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can you do it with me and then I'll do 33.) my self
Firejay5
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what do you mean?
Firejay5
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Equation editor
Firejay5
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can we get back to the problem
Firejay5
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or not @Hero
Firejay5
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the example he gave isn't going nowhere for me
wio
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let \(b=m^2-4\) and \(d=3m+6\)
wio
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\[
\frac{m}{m^2-4}+\frac{2}{3m+6}=\frac{(m)(3m+6)+(2)(m^2-4)}{(m^2-4)(3m+6)}
\]
wio
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@Firejay5 Do you see the pattern?
Firejay5
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Yes? Kind of how did you get that
Firejay5
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I need full out steps, like getting the denominators equal, etc.
Firejay5
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@Mertsj could you help with that
mathslover
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hmn .
\[\frac{m}{m^2-4}+\frac{2}{3m+6}\]
Well I think that it would be Ok to factorize both denominators
Mertsj
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Yes. mathslover will be able to show you.
Firejay5
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@Mertsj you sure
mathslover
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\(\color{blue}{m^2-4}\) 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
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yes
mathslover
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Can you factor \(m^2-4\) @Firejay5 ?
Firejay5
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yes
mathslover
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Can you tell me what would be the factorized form for \(x^2-4\) ?
Firejay5
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(x - 2) & (x + 2)
mathslover
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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
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Now take 3 common from 3m + 6. What you get?
Firejay5
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I did the other factorization as well
Firejay5
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3(m + 2)
mathslover
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Great! \(\cfrac{m}{\color{red}{(m+2)(m-2)}}+\cfrac{2}{\color{orange}{3(m+2)}}\). Is what we have now.
mathslover
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Now take LCM .
Firejay5
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m + 2
mathslover
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Check it again
Firejay5
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and m - 2
mathslover
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LCM of \(\cfrac{1}{a+b} + \cfrac{1}{2(a-b)} \) (denominators) is (a+b)(a-b)2
mathslover
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Similarly do that there. we have \((x+2)(x-2) \quad\textbf{and}\quad 3(m+2)\) as the denominators
Firejay5
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it would be 3(m - 2) (m + 2)
mathslover
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Yeah . and sorry there for using x on the place of m
mathslover
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So I have : 3(m-2)(m+2) as LCM
can you solve the fraction :
Firejay5
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so what would it be entered in equation editor
mathslover
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I didn't get you,, please explain!
Firejay5
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type the problem in fraction form of what we have left
mathslover
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hmn well it would be hard to say that I have :
\(\cfrac{\textbf{yet to come}}{3(m+2)(m-2)}\)
mathslover
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we have to solve numerator now
mathslover
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can u do that?
Firejay5
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wouldn't the numerator m + 2
mathslover
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No! See : \(\cfrac{1}{(a+b)} + \cfrac{2}{(a-b)} = \cfrac{1(a-b) + 2(a+b)}{(a-b)(a+b)}\)
Firejay5
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how does it work, you should + across
mathslover
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|dw:1364264397027:dw|
Firejay5
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doesn't make sense
mathslover
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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
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getting it?
wio
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Wow, finding the LCM is makes it complicated.
Firejay5
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????
mathslover
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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
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Does that mean you don't get it @Firejay5 ?
Firejay5
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kind of
Firejay5
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are the denominators the same
mathslover
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No.
mathslover
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|dw:1364264678904:dw|
Firejay5
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8/15
mathslover
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But how?
Firejay5
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whatever you multiply to the bottom, you must do to the top
Firejay5
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work the problem a whole different way
mathslover
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|dw:1364264834069:dw|
Firejay5
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2(a - c) or 2(a + c)
mathslover
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The LCM would be : 2(a+c)(a-c)
Firejay5
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let's work the problem a different way and not factorize, and get the denominators the same
mathslover
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Yes , that would be easy for u.
mathslover
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Well which denominator you want to be changed
Firejay5
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You're way is not what we were taught in class and is confusing me. 3m + 6
mathslover
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Sorry, ok let us change 3(m+2)
mathslover
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Would you try that for once?
Firejay5
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what
mathslover
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to change the denominator ?
Firejay5
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I have been trying
mathslover
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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
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where did you get m - 2 over 3
Firejay5
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don't jump ahead, it feels like you are just picking random numbers and variable from the problem
Mertsj
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Are you all clear now, Firejay?
Firejay5
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kind of would you explain it the same way
Mertsj
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Do you want to talk about the original problem or what?
Firejay5
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It feels like I am being taught a whole different then I learned it in class
Firejay5
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different way
Mertsj
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I just want to make sure we are talking about the right problem. Is this it:
Mertsj
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|dw:1364265923023:dw|
Firejay5
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yes! :D
Mertsj
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|dw:1364265995425:dw|
Mertsj
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When you add fractions you need to have the denominators the same. So we have to build the common denominator.
Firejay5
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@mathslover you did good though! :D
Mertsj
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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
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|dw:1364266190884:dw|
Mertsj
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Are you with me so far?
Mertsj
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|dw:1364266301603:dw|
Mertsj
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And finally:
|dw:1364266416514:dw|
Firejay5
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does it matter which order the m - 2 or m + 2 goes
Firejay5
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uh???
Firejay5
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I put 5m - 4 over 3(m + 2)(m - 2)
Mertsj
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It doesn't matter.
Firejay5
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How would you type 33.) y over y + 3 (-) 6y over y^2 - 9 in Wolframalpha
Mertsj
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\[\frac{y}{y+3}-\frac{6y}{y^2-9}\]
Firejay5
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yea
Mertsj
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For wolf: y/(y+3)-(6y)/(y^2-9)
Firejay5
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which of the alternative forms are the answer
Mertsj
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|dw:1364267512777:dw|
Mertsj
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That's the abbreviated version.
Firejay5
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is it the first, second, or third one
Mertsj
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|dw:1364267994242:dw|