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Two algebraic problems I can't solve

  1. Jul 24, 2012 #1
    1. The problem statement, all variables and given/known data
    1) [(1/y2)-x2]/[(1/x2)-y2]=?

    2) what is the sum of the solutions, x+y, in the following system of equations?
    x/y=1/2
    (x+2)/(y-9)=-4/5



    3. The attempt at a solution

    1) I started by separating the numerator of the initial big fractionk trying to simplify but failing. However, i did get to the point where I had to separated fractions with the same denominator, but it was a fail because I was just doing a round trip, lol.

    2) I defined x in terms of y and plugged it in into the (x+2)/(y-9)=-4/5 equation, being as careful as I could with having all addingg/subtracting terms with the same denominator, careful about signs etc. But I'm just doing something wrong.



    BTW, I'm already a college freshman, I already passed algebra back in 9th and 10th, but Im just loosing the rhythm for this stuff and I don't want that, so any help is appreciated, thanks (just tell me wich is the best path to start with if you can, thanks).
     
  2. jcsd
  3. Jul 24, 2012 #2

    micromass

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    It would be nice of you to actually type out what you did!
     
  4. Jul 24, 2012 #3

    SammyS

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    For #1, multiply the numerator & denominator by x2y2.
     
  5. Jul 24, 2012 #4

    Mentallic

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    If you're in college, have you learnt Linear Algebra? You know, using matrices to solve linear equations? They're not necessary here, but it would be good practice to use them to solve #2 if you should already know them.

    To start off, simplify both given equations into a linear combination of x and y, for example, [tex]\frac{x}{y}=\frac{1}{2}[/tex]
    [tex]2x=y[/tex]
    [tex]2x-y=0[/tex]

    Do the same for the second equation.

    Now, we have a 3rd equation to solve, x+y=? but rather than a question mark, we'll introduce another variable, namely z for consistency.
    You'll now have 3 linear equations with 3 unknowns. So either by using matrices or the conventional plugging and solving, try find the value of z.
     
  6. Jul 25, 2012 #5
    hey this is pretty interesting! Im starting college today actually! I will see linear algebra next semester, but thanks a lot
     
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