# Thread: Finding solution (X, Y, Z) that satisfies EQW

1. ## Finding solution (X, Y, Z) that satisfies EQW

Hey i've been strugling with this question for quite some time now, almost long enough for me to give up. Is there ANYONE out there who can solve it?

Find all the solutions (X, y, Z) of Real Numbers which satisfy the following EQW

Xy = Z – X – Y
XZ = Y – X – Z
YZ = X – Y – Z

2. Originally Posted by Mathmen
Hey i've been strugling with this question for quite some time now, almost long enough for me to give up. Is there ANYONE out there who can solve it?

Find all the solutions (X, y, Z) of Real Numbers which satisfy the following EQW

Xy = Z – X – Y
XZ = Y – X – Z
YZ = X – Y – Z
Use the first equation to get Z in terms of X and Y, then plug that Z into the other two equations, those two will then become 2 equations in 2 unknowns and will be solveable for X and Y. Once you have X and Y, you can use them in the first equation to solve for Z.

3. Y2X + XY + Y2 + Y = X
X2Y + X2 +XY +X = y

I know but this is what i ended up with:

Y2X + XY + Y2 + Y (X2Y + X2 + XY + X) = Z -(Y2X + XY + Y2 + Y) -(X2Y + X2 +XY +X)
Does anyone know if i'm on the right track. I'm not really sure how to simplify this =/

4. Originally Posted by Mathmen
Find all the solutions (X, y, Z) of Real Numbers which satisfy the following EQW
Xy = Z – X – Y
XZ = Y – X – Z
YZ = X – Y – Z
Notice that $\displaystyle (0,0,0)$ is a solution.

If $\displaystyle Z=0~\&~XY\not=0\, \Rightarrow \,\left\{ \begin{array}{l} x - y = 0 \\ y - x = 0 \\ \end{array} \right.\, \Rightarrow \,x = y\, \Rightarrow \,x^2 = - 2x$.
We can rule that out.

Suppose that $\displaystyle XYZ \not=0$.
$\displaystyle \left\{ \begin{array}{l} XY = Z - X - Y \\ XZ = Y - X - Z \\ YZ = X - Y - Z \\ \end{array} \right.\, \Rightarrow \,XY + XZ = - 2X\, \Rightarrow \,Y + Z = - 2$
Now carry on.

5. You should have obtained (I prefer little variables)

$\displaystyle z = xy + x + y$

from which substituting into the remaining two equations

$\displaystyle x(xy + x + 2y + 2) = 0,$
$\displaystyle y(xy + 2x + y + 2) = 0,$

Now consider the four cases x,y = 0, x not 0, y not zero, both x,y not zero.

6. Originally Posted by danny arrigo
You should have obtained (I prefer little variables)

$\displaystyle z = xy + x + y$

from which substituting into the remaining two equations

$\displaystyle x(xy + x + 2y + 2) = 0,$
$\displaystyle y(xy + 2x + y + 2) = 0,$

Now consider the four cases x,y = 0, x not 0, y not zero, both x,y not zero.
The thing is, it is exactly from here i'm really not sure of what i'm doing or what algrebraic rules work. Anyone who could help me get it? hope it is not too much to ask

Regards Mathmen

7. Originally Posted by Mathmen
The thing is, it is exactly from here i'm really not sure of what i'm doing or what algrebraic rules work. Anyone who could help me get it? hope it is not too much to ask
Have you looked into my approach?

8. Originally Posted by Plato
Have you looked into my approach?
Yes i analysed your approach and it did make sense but do i plug in the statements in the equation and from there work out what x.y, and z is?

if it is possible could you show me where your approach leads to, because now i have just been getting different solutions can't really figure out which way is the right way

9. From the point I said "carry on", you can do the same steps and get:
$\displaystyle \left\{ \begin{array}{l} X + Y = - 2 \\ X + Z = - 2 \\ \end{array} \right.$.

What system of three is easy to solve.

10. Originally Posted by Plato
From the point I said "carry on", you can do the same steps and get:
$\displaystyle \left\{ \begin{array}{l} X + Y = - 2 \\ X + Z = - 2 \\ \end{array} \right.$.

What system of three is easy to solve.
Is this as simplified as the equation can get or can you find out what one unknown is equal to?

11. Originally Posted by Mathmen
Is this as simplified as the equation can get or can you find out what one unknown is equal to?
Can you solve a system of three equations in three unknowns?
It cannot be any easier than that!

12. Originally Posted by Plato
Can you solve a system of three equations in three unknowns?
It cannot be any easier than that!
If that is the answer i guess i finally understood it Thanks a lot for all your help Plato and the others, you sure did help me.

Best regards Mathmen