Solving Triples with xy=a(x+y) and b = xy+xz+yz

[Yann]

Is it possible to find triples that satifsy;

$$xy=a(x+y)$$

$$xz=a(x+z)$$

$$yz=a(y+z)$$

$$b = xy+xz+yz$$

 

[AKG]

If a = 0, then (x = 0 or y = 0) AND (x = 0 or z = 0) AND (y = 0 or z = 0) so it's easy to see that b = 0. Any triple with at least two of x, y, and z being 0 will satisfy the given equations.

If a is nonzero and anyone of x, y, or z is 0, then it turns out that all three must be 0, from the first three equations. E.g. if x = 0, then 0 = xy = a(x+y) = ay, and since a is non-zero, y must be zero; then we get 0 = yz = a(y+z) = az and since a is non-zero, z too must be zero. So we only get a solution if b is zero, and that solution is (0,0,0).

If a is nonzero and none of the x, y, z are 0, then we get:

xy = a(x+y)
1/a = 1/x + 1/y [dividing both sides by axy]

Likewise, 1/a = 1/x + 1/z and 1/a = 1/y + 1/z. It's easy to see that this gives 1/x = 1/y = 1/z, i.e. x = y = z. x² = xy = a(x+y) = 2ax, so x = 2a. So we'd get b = 12a².

Summary:
(a,b) = (0,0) has solutions (0,0,z), (0,y,0), and (x,0,0) for any x, y, and z
(a,b) = (a,0) with a non-zero has the solution (0,0,0)
(a,b) = (a,12a²) with a non-zero has the solution (2a, 2a, 2a)
(a,b) = anything else has no solution

 

[tacman]

Yes, it is possible to find triples that satisfy these equations. To solve for the triples, we can use substitution and elimination to find the values of x, y, and z that satisfy the equations.

First, we can rearrange the equations to get x, y, and z on one side and a and b on the other side:

xy - ax - ay = 0

xz - ax - az = 0

yz - ay - az = 0

b - xy - xz - yz = 0

Next, we can use substitution to eliminate one variable at a time. We can substitute the value of x from the first equation into the second and third equations, and then substitute the values of y and z from the second and third equations into the fourth equation. This will give us a single equation in terms of only one variable, which we can solve to find the value of that variable.

For example, substituting x = ay/a from the first equation into the second equation gives us:

ayz/a - aay/a - az = 0

Simplifying, we get:

y^2 - ay - az = 0

Similarly, substituting x = az/a from the first equation into the third equation gives us:

yz - ay - aaz/a = 0

Simplifying, we get:

yz - ay - z^2 = 0

Now, we can substitute the value of y from the second equation into the third equation to eliminate y:

yz - a(y^2 - az) - z^2 = 0

Simplifying, we get:

yz - ay^2 - az^2 = 0

Finally, we can substitute the values of y and z from the second and third equations into the fourth equation to get a single equation in terms of only a:

b - a(y^2 - ay - az) - a(yz - ay - z^2) - yz = 0

Simplifying, we get:

b - ay^2 - az^2 - yz = 0

This is a quadratic equation in terms of a, which we can solve to find the values of a that satisfy the equations. Once we have the value(s) of a, we can substitute them back into the equations to find the corresponding values of x, y, and z.

In summary