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2 variables!

  1. Jul 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Find x and y in terms of a and b from the following equations

    x * √x + y * √y = a
    x * √y + y * √x = b

    2. Relevant equations
    the general algebra equations


    3. The attempt at a solution

    adding the equations

    (x + y) * (√x + √y) = a + b
    x + y =
    a + b
    _______
    √x + √y​

    x =
    a + b + y√x + y√y)
    ______________________
    √x + √y​

    Now substituting in eqn 1

    a + b + y√x + y√y
    _________________ + y√y = a
    √x + √y

    a + b + y√x + y√y + y√x√y + y2 = a√x + a√y


    ... Now I dunno how to proceed..
     
  2. jcsd
  3. Jul 14, 2010 #2

    ehild

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    Gold Member

    To begin, add 3 times the second equation to the first one.

    ehild
     
  4. Jul 14, 2010 #3
    I tried it out..

    x * √x + y * √y = a
    3*x * √y + 3*y * √x = b

    adding and using (p + q)3 formula

    (√x + √y)3 = 3b + a

    therefore

    √x = (3√3b + a) - √y
    so
    x√x = {3(√3b + a) - √y}3

    Further solving

    I get y =
    -(3√3b + a)2 +- √(3√3b + a)4 + 4b * (3√3b + a)
    _________________________________________________________________
    2 * (3√3b + a)​


    Is it right?(Because I havent got its solutions.. and I have never got such complex answers!!)
     
  5. Jul 14, 2010 #4

    ehild

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    Correct up to here, but all the others are wrong. What is x+y? The whole left side should be under the cubic root.

    More: You can factor out √x √y from the second equation. As you know
    (√x + √y), you get an equation for √x √y.

    There will be a system of two equations, one for (√x + √y), the other for √x √y, which is easy to solve.

    ehild
     
  6. Jul 14, 2010 #5
    Whew!
    I tried it out your way and got the answer!(which I dont think I want to key down)

    Though u say that the whole left side should be under cube root... actually it is(that is cube root of (√x + √y)3 so it will be just √x + √y)... so anything wrong? ( maybe I went wrong in the calculations)
     
  7. Jul 14, 2010 #6

    ehild

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    I mixed my hands... :) I meant the right side.

    √x = (3√3b + a) - √y--this is wrong. You have the cubic root of 3 alone.

    [tex]\sqrt{x}+\sqrt{y}=\sqrt[3]{3b+a}[/tex]



    ehild
     
  8. Jul 14, 2010 #7

    Mentallic

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    I think that's what he implied by the use of brackets (even though it wasn't correctly expressed) :smile:
     
  9. Jul 15, 2010 #8
    Ok thanks ehild!
    And also Mentallic got me right! I juggled up my use of brackets!!
     
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