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Simplifying a fraction (block reduction, control)

  1. Dec 30, 2015 #1
    1. The problem statement, all variables and given/known data

    2*3*4*5/4+3*4*4*6

    2. Relevant equations

    2*3*4*5/4+3*4*4*6

    3. The attempt at a solution

    2*3*5/1+3*4*6

    Knowledge gap 1 - I do not understand why if top and bottom are divided through by 4 there should be a 4 left in the denominator and why the rest of the numerators and denomenators are not divided through by 4.

    I think I am just needing a prompt in the right direction here

    Any help would be greatley apprecited.

    I have attched photos to show where this comes from. Th blue pen working (337) is where I have changed the original problem (335) from letters to numbers to allow me to attempt to make more snse of it.

    IMAG0335.JPG IMAG0337.JPG
     
  2. jcsd
  3. Dec 30, 2015 #2

    Samy_A

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    How many 4's are there in 3*4*4*6?
     
  4. Dec 30, 2015 #3
    2 obviously
     
  5. Dec 30, 2015 #4
    I was thinking why two of the 4's come out on the bottom line and only one on the top. Is this because the two 4's on the bottom line are from different expresions?
     
  6. Dec 30, 2015 #5

    BvU

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    The 30/72 on the lower right is not correct. if you strike 4 it means you divide by four. The quotient is 1. So you get 30/(1+72) = 120/292 and all is right.
     
  7. Dec 30, 2015 #6

    Samy_A

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    So when you simplify your fraction by removing one 4 in each term, why should the second 4 in 3*4*4*6 just disappear?
     
  8. Dec 30, 2015 #7
    yes I have since noticed this, thanks. Appreciated
     
  9. Dec 30, 2015 #8

    BvU

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    Let $$A = {G_2\,G_3\,G_4\over 1+ G_3\,G_4\,H_3}$$ and we want to look at $$ {A\over 1 + {H_2\over G_4}A}\ ,$$ right ? So we have $$
    {{ G_2\,G_3\,G_4\over 1+ G_3\,G_4\,H_3} \over 1 + { G_2\,G_3\,G_4\over 1+ G_3\,G_4\,H_3}\ { H_2\over G_4}} \quad
    = \quad {{ G_2\,G_3\,G_4\over 1+ G_3\,G_4\,H_3} \over 1 + { G_2\,G_3\,G_4\, H_2 \over G_4+ G_3\,G_4\,G_4\,H_3}} \quad \equiv B
    $$ the block where your pen is pointing at (I call it ##B## for now). Then the dividing out of ##G_4##:

    $$ { G_2\,G_3\,G_4\, H_2 \over G_4+ G_3\,G_4\,G_4\,H_3}\quad = \quad { G_2\,G_3\,G_4\, H_2 \over G_4 ( 1 + G_3\,G_4\,H_3) } \quad = \quad
    {G_4\over G_4}\, { G_2\,G_3\, H_2 \over 1 + G_3\,G_4\,H_3 }
    \quad = \quad
    { G_2\,G_3\, H_2 \over 1 + G_3\,G_4\,H_3 }
    $$
    Then he multiplies B with 1 : $$ B\ { 1 + G_3\,G_4\,H_3 \over 1 + G_3\,G_4\,H_3 } \quad = \quad
    {G_2\,G_3\,G_4\over 1+ G_3\,G_4\,H_3 + G_2\,G_3\, H_2 } $$

    and this times ##G_1## yields the next rectangular block.

    etc.
    --
     
  10. Dec 30, 2015 #9
    Crystal clear, thanks very much BvU.
     

    Attached Files:

  11. Dec 30, 2015 #10
    --[/QUOTE]
    Please if you have the time BvU can you help me fill in the step in the attached.
    Sorry that should also read C(S)/R(S) =
    Thanks IMAG0341.jpg
     
  12. Dec 30, 2015 #11

    Mark44

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    The above does not agree with what you wrote in the image you attached. Because of the higher precedence of division over addition, the above is the same as ##2\cdot 3 \cdot 4 \cdot \frac 5 4 + 3 \cdot 4 \cdot 4 \cdot 6##

    To match what you wrote in the image, you need parentheses around the entire denominator, like so:
    2 * 3 * 4 * 5 / (4 + 3 * 4 * 4* 6)
     
  13. Dec 30, 2015 #12
    Ok fair enough.
     
  14. Dec 30, 2015 #13

    BvU

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    Sure: multiplying with 1 is allowed, so here goes:$$
    {C\over R} =
    { I\, S^2 \over I\, S^2\, } \ { {K_1\,K_2 \over I\, S^2} \over
    1 + {K_1\,K_2\,K_3 \over I \, S } + {K_1\,K_2\over I\, S^2} } =
    { { I\, S^2 }\ {K_1\,K_2 \over I\, S^2}
    \over {I\, S^2\, } \left ( 1 + {K_1\,K_2\, K_3 \over I\, S} + {K_1\,K_2\over I\, S^2}\right ) } =
    {K_1\,K_2\over I\, S^2 + {K_1\,K_2\,K_3\, S} + {K_1\,K_2} } $$
     
  15. Dec 31, 2015 #14
    Thats fantastic BvU, once again crystal clear,
     
  16. Dec 31, 2015 #15
    Just rusty on the syplifying. Thanks
     
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