# Thread: [SOLVED] Planes of some sort?

1. ## [SOLVED] Planes of some sort?

So in my calculus and vectors course, I was assigned the following question:

Determine ALL real values of x, y and z, that satisfy the following
system of equations:
1)x^2 - yz + xy + zx = 82
2)y^2 - zx + xy + yz = -18
3)z^2 - xy + zx + yz = 18

We have been learning about planes, so I am pretty sure I have to apply what I have learned about planes to this problem. But there are so many variables, and whatever I do (I've tried elimination and rearranging things) seems to lead me back to one of these equations! please help

2. Originally Posted by jenna0012
So in my calculus and vectors course, I was assigned the following question:

Determine ALL real values of x, y and z, that satisfy the following
system of equations:
1)x^2 - yz + xy + zx = 82
2)y^2 - zx + xy + yz = -18
3)z^2 - xy + zx + yz = 18

We have been learning about planes, so I am pretty sure I have to apply what I have learned about planes to this problem. But there are so many variables, and whatever I do (I've tried elimination and rearranging things) seems to lead me back to one of these equations! please help
Try completing the square on equation 3...

$z^2 + (x + y)z - xy = 18$

$z^2 + (x + y)z + \left(\frac{x + y}{2}\right)^2 - \left(\frac{x + y}{2}\right)^2 - xy = 18$

$\left(z + \frac{x + y}{2}\right)^2 - \left(\frac{x + y}{2}\right)^2 - xy = 18$.

Can you now solve for z and substitute these into the other equations to solve for x and y?

3. Originally Posted by Prove It
Try completing the square on equation 3...

$z^2 + (x + y)z - xy = 18$

$z^2 + (x + y)z + \left(\frac{x + y}{2}\right)^2 - \left(\frac{x + y}{2}\right)^2 - xy = 18$

$\left(z + \frac{x + y}{2}\right)^2 - \left(\frac{x + y}{2}\right)^2 - xy = 18$.

Can you now solve for z and substitute these into the other equations to solve for x and y?
hmm I see where you are going.. but I'm a bit confused at how you arrived at your third step
$\left(z + \frac{x + y}{2}\right)^2 - \left(\frac{x + y}{2}\right)^2 - xy = 18$.

4. Originally Posted by jenna0012
hmm I see where you are going.. but I'm a bit confused at how you arrived at your third step
$\left(z + \frac{x + y}{2}\right)^2 - \left(\frac{x + y}{2}\right)^2 - xy = 18$.
It's a standard completing the square step...

For a quadratic of the form $x^2 + bx + c$

$= x^2 + bx + \left(\frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$

$= \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$