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Math Help - How to solve this two equations simultaneously?

  1. #1
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    How to solve this two equations simultaneously?

    I have two equations with two unknowns x and y.
    (r-yp)x^2 {y^2 (b-a)-2yq+(a+c)}x+(y^2+1)( yp -r)=0 (1)
    (q-xp)y^2 {x^2 (a-b)-2xr+(b+c)}y+(x^2+1)( xp -q)=0 (2)
    here a,b,c,p,q,r are some constants.
    How to solve this ? Pls help me.




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  2. #2
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    Why post it 3 times...expect 3 answers ?

    Seems to be limitless solutions.
    Here's one of them:
    a=1, b=1, c=2, p=1, q=2, r=2, x=1, y=1
    Any comments?
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  3. #3
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    I didn't post it 3 times, it may happen due to disconnection of network while submitting. Sorry for the inconvenience.
    However, I don't need typical solution, I need algebraic solution of x, y in terms of a,b,c,p,q,r. Can u suggest any such solution ?
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  4. #4
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    Quote Originally Posted by Robin View Post
    I didn't post it 3 times, it may happen due to disconnection of network while submitting. Sorry for the inconvenience.
    However, I don't need typical solution, I need algebraic solution of x, y in terms of a,b,c,p,q,r. Can u suggest any such solution ?
    All I can see is this:

    Your equations:
    (r-yp)x^2 [y^2 (b-a) - 2yq + (a+c)]x + (y^2+1)(yp-r) = 0 (1)
    (q-xp)y^2 [x^2 (a-b) - 2xr + (b+c)]y + (x^2+1)(xp-q) = 0 (2)

    I'm replacing p,q,r with d,e,f:
    (f-dy)x^2 [y^2 (b-a) - 2ey + a+c]x + (y^2+1)(dy-f) = 0 (1)
    (e-dx)y^2 [x^2 (a-b) - 2fx + b+c]y + (x^2+1)(dx-e) = 0 (2)

    Let u = f-dy ; then -u = dy-f
    Let v = e-dx ; then -v = dx-e
    Let w = b-a ; then -w = a-b

    ux^2 (wy^2 - 2ey + a + c)x - u(y^2 + 1) = 0 (1)
    vy^2 (-wx^2 - 2fx + b + c)y - v(x^2 + 1) = 0 (2)

    Rewriting (2):
    (wy - v)x^2 + (2fy)x + v(y^2 - 1) - y(b + c) = 0 (3)

    So we now have:
    ux^2 (wy^2 - 2ey + a + c)x - u(y^2 + 1) = 0 (1)
    (wy - v)x^2 + (2fy)x + v(y^2 - 1) - y(b + c) = 0 ... ...(3)

    Both can be solved for x using the quadratic formula.

    I'll stop here for now.
    Follow all that? Any comments?
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  5. #5
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    Will this form give a correct root of x while you have taken v = e-dx ? without this x how would I get the root of x ?
    Then how will I get the roots of y while you have taken u = f-dy ?
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  6. #6
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    My bad; missed fact that 1st term of (3) which is (wy - v)x^2
    will result in a cubic : dx^3 + (wy - e)x^2

    Well, hope someone else can help you with this mess...I can't
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