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System with homogeneous equation in denominator help

  1. Jun 5, 2013 #1
    1. The problem statement, all variables and given/known data
    The system is declared as follows:

    8/(2*x - y) - 7/(x + 2*y) = 1
    4/((2*x - y)^2) - 7/((x + 2*y)^2) = 3/28

    2. Relevant equations



    3. The attempt at a solution

    I define 'x' to equal k*y and I replace it inside the equation:

    8/(2*k*y^2) - 7(k*y + 2*y) = 1
    4(4*k^2*y^2 - 4*k*y^2 + y*2) - 7(k^2*y^2 + 4*k*y^2 + 4*y^2) = 3/28

    I know I have to divide the top one by the bottom one and take y^2 out of the brackets, however the way it is now, it would become impossibly complex if I divide them.How can I simplify them before I divide?Not sure how to proceed.Using least common multiple also results in a monstrosity.
     
  2. jcsd
  3. Jun 5, 2013 #2

    mfb

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    Staff: Mentor

    How do you get 2ky^2 if you replace x with ky in 2x - y?
    What happens to the fractions in equation 2?

    I doubt that this will work.

    I would substitute 2x-y and x+2y with new parameters, and get rid of the denominators quickly.
     
  4. Jun 5, 2013 #3
    By doing that I get:
    8*v - 7*u = u*v
    12*v^2 - 196*u^2 = 3*u^2*v^2

    I can't use any of them to express u = # or v = # :(
     
  5. Jun 5, 2013 #4

    mfb

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    You can solve the first equation for v or u and plug it into the second equation.
     
  6. Jun 6, 2013 #5
    First, shouldn't the ##12v^2## in your second equation be ##112v^2##? (Probably just a typo)

    Second, I suggest manipulating the first equation to get ##u = \frac{8v}{v+7}##(assuming that ##v \neq -7##, you should check that this cannot be true by substituting ##v = -7## into the system). Then, as mentioned above, substitute this expression into the second equation and solve for v.
     
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