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Fractional equations

  1. May 8, 2007 #1
    1. The problem statement, all variables and given/known data
    (2x + 3¬5) -(7¬3) = (3x -1¬4)

    solve for x


    3. The attempt at a solution

    the lowest common multiple is 60 so:

    14(2x + 3¬5) - 20(7¬3) = 15(3x -1¬4)

    is

    (28x + 42¬70) - (140¬60) = (60x - 15¬60)

    60x - 28x = 32x

    time to combine like terms,

    now comes the confusion, the denominators are different but the book (cd-roms, this wasn't in the book) says nothing about this, does using the LCM multiple cancel that difference out?
     
    Last edited: May 8, 2007
  2. jcsd
  3. May 8, 2007 #2

    symbolipoint

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    You are not using the properties of numbers; you found common denominator of 60, so you could multiply left and right-hand members by 60; and THEN ... combine like-terms.

    Start with
    [tex]
    60(2x + \frac{3}{5}) - 60(\frac{7}{3}) = 60(3x - \frac{1}{4})
    [/tex]

    I don't know why that other junk is in there with my message, but what I try to say should still be comprehendable.
     
    Last edited by a moderator: May 9, 2007
  4. May 8, 2007 #3
    i'm afraid i still don't understand, i know i have to combine like terms last,

    a) 14(2x + 3¬5) - 20(7¬3) = 15(3x -1¬4)

    b) 60(2x + 3¬5) - 60(7¬3) = 60(3x -1¬4)

    what i'm trying to say is, should the denominators be the same instead of the same multiple? if B is correct do i touch the denominators? i understand fractions just not how this is done. :frown: explain it to me sloooow like :uhh:
     
    Last edited: May 8, 2007
  5. May 8, 2007 #4
    i'm going to give an example of what i'm talking about;

    if the LCM is 60 then which one is right:

    60(3¬5) = 180 for the numerator

    14(3¬5) = 42 for the numerator

    now should the denominator by multiplied by 60 (haha don't laugh) or the multiplier? just telling me which one is right will set me one the right course.
     
  6. May 9, 2007 #5

    symbolipoint

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    light_bulb,

    I have not given-up on your question yet; but, do you know how to multiply a fraction by a whole number? Do you know enough arithmetic to try simplifying before performing any low-level multiplication?
    ... give me a chance and I'll try to explain the first steps in the original question (unless you can clear your difficulty with it first).
     
  7. May 9, 2007 #6
    sure, multiply the whole number by the denominator. next q, find the the LCM, the multiply the numerator by the the LCM multilpier for each denominator.
     
  8. May 9, 2007 #7
    a(2¬3)=(5¬9) a=5¬6

    i understand you flip the first fraction then multiply it by the reciprocal to get the answer, my misunderstanding comes when i have two constants. eg.

    (2x + 3¬5) -(7¬3) = (3x -1¬4)

    is x = 3¬28 ?

    i'll come back to this
     
    Last edited: May 9, 2007
  9. May 9, 2007 #8
    another example of where i'm getting stuck:

    (2¬3)x - (5¬6) = (3¬2)

    i get to this point:

    4x = 14

    now it shows that i'm supposed to have:

    (4x¬4) = (14¬4)

    but how did it get those denominators? they skipped over that part! i'm going to assume that that the multiplier of the first fraction is being used on the reciprocal (4x¬4) <-- where did that denominator come from :grumpy:
     
    Last edited: May 9, 2007
  10. May 9, 2007 #9
    pic for clarification. tia
     

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    Last edited: May 9, 2007
  11. May 9, 2007 #10

    Integral

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    Light_bulb<
    Your notation is not clear to me. Please read this thread on how to format equations.

    It seems you are trying to clear the denomiatiors of this:

    [tex] \frac 2 3 x - \frac 5 6 = \frac 3 2 [/tex]

    multibly both sides by 6

    [tex] 6 (\frac 2 3 x - \frac 5 6 ) = 6 ( \frac 3 2 ) [/tex]

    distribute

    [tex]6 \frac 2 3 x - 6 \frac 5 6 = 9 [/tex]

    cancel

    [tex] 2 * 2 x - 5 = 9 [/tex]

    [tex] 4x - 5 = 9 [/tex]

    [tex] 4x = 14 [/tex]
    Now finally divide both sides by 4 to isolate x

    [tex] \frac {4x} 4 = \frac {14} 4 [/tex]

    [tex] x = \frac 7 2 [/tex]
     
  12. May 9, 2007 #11
    double post.
     
    Last edited: May 9, 2007
  13. May 9, 2007 #12
    thank you! thats what i wanted to know, now i got it. maybe it seems stupid but all i needed to know was why 4 was the denominator, it all makes sense now. will spend some time to learn latex.
     
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