Can You Solve ax+ay-xy=c for x and y Deterministically?

[aravindsubramanian]

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


ex
127x+127y-xy=12732

I need to find x & y from this equation

 

[Zurtex]

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:

$$x = \frac{c - ay}{a - y} \quad \text{or} \quad y = \frac{c - ax}{a - x}$$

 

[aravindsubramanian]

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.

 

[ramsey2879]

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:

$$x = \frac{c - ay}{a - y} \quad \text{or} \quad y = \frac{c - ax}{a - x}$$

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

 

[aravindsubramanian]

In this type of equation

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

 

[ramsey2879]

In this type of equation

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

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

 

[TenaliRaman]

So Ramsey We can't use your method.Thanks a lot for your reply

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

 

[ramsey2879]

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:

$$x = \frac{c - ay}{a - y} \quad \text{or} \quad y = \frac{c - ax}{a - x}$$

This can be simplified to

$$x = a + \frac{c - a^2}{a - y}$$

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 $$127^2$$ -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.

 

[aravindsubramanian]

This can be simplified to

$$x = a + \frac{c - a^2}{a - y}$$

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 $$127^2$$ -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


$$x = a + \frac{c - a^2}{a - y}$$

 

[VietDao29]

$$x = \frac{c - ay}{a - y} = \frac{a(a - y) - a^2 + c}{a - y} = a + \frac{c - a^2}{a - y}$$
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,

 

[aravindsubramanian]

Thank U very much

Thanks a lot for your reply