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Solve the simultaneous equation:

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data
    x+2y=3
    x^2-4y^2=-33


    2. Relevant equations



    3. The attempt at a solution
    x-2y=root-33?
     
  2. jcsd
  3. Oct 23, 2008 #2

    HallsofIvy

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    First, that's not a solution, you haven't found a value for x and y! Second, sqrt(a- B) is NOT sqrt(A)- sqrt(B).

    Probably it is simplest to solve the first equation for x, replace the "x" in the second equation by that expression.
     
  4. Nov 20, 2008 #3
    Try this:
    x+2y=3
    => x=3-2y

    << detailed solution hints deleted by berkeman >>
     
    Last edited by a moderator: Nov 20, 2008
  5. Nov 20, 2008 #4
    LOL dude, use a^2 - b^2 = (a+b)(a-b)
    and u have the a+b eqn. with you..
    Get the a-b eqn from the above equation, and solve a+b and a-b for a and b.
     
  6. Nov 21, 2008 #5

    HallsofIvy

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    Oh, that's cute.
     
  7. Nov 21, 2008 #6

    Mentallic

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    Spidey I wasn't able to follow your reasoning :confused:

    It is always easiest (imo) to make x or y the subject in the linear equation: i.e. the first equation becomes [tex]x=3-2y[/tex] or [tex]y=\frac{3-x}{2}[/tex] and then substitute either of these values into the quadratic equation so that becomes: [tex](3-2y)^2-4y^2=-33[/tex] or if you made y the subject: [tex]x^2-4(\frac{3-x}{2})^2=-33[/tex]

    Oh and remember you can't take the square root of a negative value.
     
  8. Nov 21, 2008 #7
    x+2y=3
    x^2-4y^2=-33
    -33 = (x+2y)(x-2y)
    got it now?
     
  9. Nov 21, 2008 #8

    Mentallic

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    Ahh the 4y2 slipped my mind. Didn't think difference of 2 squares would've had a factor equal to the linear equation.
     
  10. Nov 21, 2008 #9
    I know, I wouldn't have found it until I was halfway through the problem. It was a nice catch.
     
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