- #1

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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

ex

127x+127y-xy=12732

I need to find x & y from this equation

- Thread starter aravindsubramanian
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- #1

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- 0

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

ex

127x+127y-xy=12732

I need to find x & y from this equation

- #2

Zurtex

Science Advisor

Homework Helper

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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]

- #3

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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.

- #4

- 841

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Finding a solution in integers is easy if you first choose y=a-1 or a+1 then solve for xZurtex said:

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]

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- #5

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1.x and y are the possitve even no integers between (a-1) and

(a-1)/2.

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

- #6

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Hey, my solution solves the posted problem. Now you are adding further conditions!!aravindsubramanian said:

1.x and y are the possitve even no integers between (a-1) and

(a-1)/2.

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

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

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- #7

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Did you notice what he did?aravindsubramanian said:So Ramsey We can't use your method.Thanks a lot for your reply

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

- #8

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This can be simplified toZurtex said:

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]

[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.

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- #9

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ramsey2879 said: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.

how do u derive the equation

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

- #10

VietDao29

Homework Helper

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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,

- #11

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Thanks a lot for your reply

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