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Write a single rational expression

Guest Nov 7, 2017
 #1
avatar+92608 
+2

1 / [ 2m - 4n ]  - 1 / [ 2m + 4n ]  - m / [ m^2 - 4n^2]    factor the denominators

 

1 / [ 2 (m - 2n) ]  -   1  / [ 2 (m + 2n) ]  -  m  / [ (m + 2n) (m - 2n) ]

 

The common denominator  is    2 (m + 2n) (m - 2n).....so we have

 

 (  [  (m + 2n) ]   - [m - 2n] - [2m]  )  / [  2 (m + 2n) (m - 2n) ]

 

(  4n - 2m ]  / [ 2 (m + 2n) (m - 2n) ]

 

[ 2 (2n - m) ]  /  [  2 (m + 2n) (m - 2n) ]

 

( 2n - m)  /  [ (m + 2n) (m - 2n) ]  

 

- ( m - 2n) / [ (m + 2n) (m - 2n) ] 

 

- 1  / [ m + 2n ]

 

 

 

cool cool cool

CPhill  Nov 7, 2017
 #2
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0

How come you put "2m" for the third fraction instead of 2?

Guest Nov 8, 2017
 #3
avatar+7339 
+1

Are you asking about this part:

 

 (  [  (m + 2n) ]   - [m - 2n] - [2m]  )  / [  2 (m + 2n) (m - 2n) ]          ???

 

Starting from the previous step...

 

\(\frac{1}{2(m-2n)}-\frac{1}{2(m+2n)}-\frac{m}{(m+2n)(m-2n)} \\~\\ =\,\frac{(m+2n)}{2(m+2n)(m-2n)}-\frac{1}{2(m+2n)}-\frac{m}{(m+2n)(m-2n)} \\~\\ =\,\frac{(m+2n)}{2(m+2n)(m-2n)}-\frac{(m-2n)}{2(m+2n)(m-2n)}-\frac{m}{(m+2n)(m-2n)} \)

                                                                                               Multiply the third fraction by  2/2  .

\(=\,\frac{(m+2n)}{2(m+2n)(m-2n)}-\frac{(m-2n)}{2(m+2n)(m-2n)}-\frac{2m}{2(m+2n)(m-2n)} \\~\\ =\, \frac{(m+2n)-(m-2n)-(2m)}{2(m+2n)(m-2n)}\)

hectictar  Nov 8, 2017
edited by hectictar  Nov 8, 2017

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