Re: A very simple question (I think!)
- From: Jean-Claude Arbaut <jean-claude.arbaut@xxxxxxxxxxx>
- Date: Mon, 27 Jun 2005 15:35:01 +0200
On 27/06/2005 14:53, jimal1971@xxxxxxxxxxx wrote:
> Hi,
>
> I am dealing with the following equation:
>
> a(x/y) + b(y/x) = 1
>
> All the variables are non-zero real numbers.
> The only other stipulation is that a + b = 1.
a*x^2+(1-a)*y^2-x*y=0 with x<>0 and y<>0
You can write:
(1-a)*y^2 - x*y + a*x^2 = 0
If a=1/2, equation can be written (x-y)^2=0,
solution is y=x and x<>0.
Otherwise:
Assume a<>1 and a<>0
Solve for y: discriminant is x^2 - 4*x^2*a*(1-a) = x^2*alpha with alpha > 0,
y = (x +- sqrt(alpha) * |x|) / (2*(1-a))
The equations of the two lines.
You can simplify, alpha = 1 - 4*a + 4*a^2 = (2*a-1)^2, etc.
So general solution is y=a/b*x or y=x, x<>0, y<>0.
If a=1, equation is x^2-x*y=0 <=> y=x (since x<>0)
If a=0, solution is y=x (x<>0)
> Can I therefore state that the "general" solution
> to this equation is that x/y = b/a ??
Replace in your equation:
a*(b/a) + b*(a/b) = b+a = 1
So it's *one* solution. But you missed another line, y=x.
And what would you do if a=0 ?
> I have been told that this is NOT a "general"
> solution and I cannot make this statement.
>
> Or is this some special solution or class of
> solutions?
>
> By the way, if this kind of equation has a name
> or belongs to a familly that has a name then any
> background information would be really appreciated!
>
> Thank you very much in advance!
>
> J.
>
.
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