I am trying to complete the square on:
$xy - xz + yz = 1$.
I tried plugging in: $u = x + y$
If you plug in $u$ you get:
$-y^2 + uy - uz + 2yz = 1$
but I'm not sure where to go from here.
Asked
Active
Viewed 28 times
0
-
Does the following help?$$(xy-xz+yz)-1=(y-1)(1+x)(1+z)+(x-y+z)$$ I don't fully understand what you are trying to do here, but that looks like a nice factorisation $\uparrow$. – John Doe Apr 13 '18 at 03:04
-
I am trying to complete the square. For example, getting the equation of the form $x^2 + y^2 + z^2 = 1$ – mathematicallymathematic Apr 13 '18 at 03:12
-
It is still not clear exactly what you want, but here is another perhaps useful identity: $xy-xz+yz=(x-y)(y-z)+y^2$. – mzp Apr 13 '18 at 03:29
-
Basically, I want to get to the equations to an equivalent form but with squared terms. For example, say$ (x-y+z)^2 + (y-z)^2 + (x)^2 = 1$ – mathematicallymathematic Apr 13 '18 at 03:50
-
@mathematicallymathematic You can always use that $2ab = (a+b)^2-a^2-b^2$ if that's what it's about. – dxiv Apr 13 '18 at 04:25