Revision history [back]
 | 1 | initial version |
You can divide by $2$ and solve for $w$. So the set of all solutions is parametrized by $(x,y,z)\in\mathbb{Z}^3$ arbitrary.
| | 1 | initial version |
You can divide by $2$ and solve for $w$. So the set of all solutions is parametrized by $(x,y,z)\in\mathbb{Z}^3$ arbitrary.
Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.