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Need help to solve this type of equation

  1. Jul 12, 2005 #1
    I need to solve the equations of type
    ax+ay-xy=c (or) a(x+y)-xy=c

    In this equation a & c are known.Whether is it possible to find x & y values using a deterministic method other than trial & error method


    I need to find x & y from this equation
  2. jcsd
  3. Jul 12, 2005 #2


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    :confused: Do you mean to find solutions for x and y in integers or something?

    Because otherwise you need a value of x and y and just rearrange to get:

    [tex]x = \frac{c - ay}{a - y} \quad \text{or} \quad y = \frac{c - ax}{a - x}[/tex]
  4. Jul 12, 2005 #3
    In this equation both x , y & c are integers only.

    How we solve the equation by rearranging x or y by your method

    Its results a ambiguous equation equation.We can't able to solve it.
  5. Jul 14, 2005 #4
    Finding a solution in integers is easy if you first choose y=a-1 or a+1 then solve for x
    Last edited: Jul 14, 2005
  6. Jul 15, 2005 #5
    In this type of equation

    1.x and y are the possitve even no integers between (a-1) and
    2.a is always a odd no.
    3.c is always a even no.
    So Ramsey We can't use your method.Thanks a lot for your reply
  7. Jul 15, 2005 #6
    Hey, my solution solves the posted problem. Now you are adding further conditions!!
    I suggest that you take my method a step further and chose a even y dependent upon "a" such that a-y evenly divides c-ay
    Last edited: Jul 15, 2005
  8. Jul 15, 2005 #7
    Did you notice what he did?
    He simply made those denominators (in the equations given by Zurtex) as 1.
    This automatically makes x and y integers if a and c are integers.

    -- AI
  9. Jul 21, 2005 #8
    This can be simplified to

    [tex]x = a + \frac{c - a^2}{a - y} [/tex]

    Thus if either |a-x| or |a-y| can't = 1, |a-y| must be a factor of |a^2-c| between 1 and |a^2-c|. Since in the posted example, the factors of [tex]127^2[/tex] -12732 are 43 and 79, the only other solution set is |x| and |y| equal 127-43 or 84 and 127-79 or 48 respectively.
    Last edited: Jul 22, 2005
  10. Jul 27, 2005 #9

    how do u derive the equation

    [tex]x = a + \frac{c - a^2}{a - y} [/tex]
  11. Jul 27, 2005 #10


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    [tex]x = \frac{c - ay}{a - y} = \frac{a(a - y) - a^2 + c}{a - y} = a + \frac{c - a^2}{a - y}[/tex]
    Since you want to have an integer x, and you have a as an integer, this rearrangement shows that x is an integer if (c - a^2) is divisible by a - y. You have c, a, you can easily calculate c - a^2, from there you can solve for y, and then x.
    Viet Dao,
  12. Jul 28, 2005 #11
    Thank U very much

    Thanks a lot for your reply
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