| Equation | x1 | x2 | x3 | x4 | x5 |
| (1) | 1 | 1 | 1 | ||
| (2) | 1 | 1 | |||
| (3) | 1 | 1 | 1 | ||
| (4) | 1 | 1 | |||
| (5) | 1 |
Rearranging and reordering the incidence table leaves us with:
| Equation | x5 | x1 | x3 | x4 | x2 |
| (5) | 1 | ||||
| (4) | 1 | 1 | |||
| (3) | 1 | 1 | |||
| (1) | 1 | 1 | 1 | ||
| (2) | 1 | 1 |
2. The incidence table for the equation set is:
| Equation | x1 | x2 | x3 | x4 |
| (1) | 1 | 1 | 1 | 1 |
| (2) | 1 | 1 | 1 | 1 |
| (3) | 1 | 1 | ||
| (4) | 1 | 1 |
Rearranging and reordering the incidence table leaves us with:
| Equation | x4 | x3 | x1 | x2 |
| (4) | 1 | 1 | ||
| (3) | 1 | 1 | ||
| (1) | 1 | 1 | 1 | 1 |
| (2) | 1 | 1 | 1 | 1 |
In this case there is no head or tail. Instead all of the equations form the partitions. Either of x2, x3 or x4 would be suitable for the tear variable here. x1 would not be suitable as estimating a value for x1 does not leave an equation with only one further unknown present.
3. The incidence table for the equation set is:
| Equation | x1 | x2 | x3 | x4 |
| (1) | 1 | 1 | ||
| (2) | 1 | 1 | 1 | |
| (3) | 1 | 1 | 1 | 1 |
| (4) | 1 | 1 |
Rearranging and reordering the incidence table leaves us with:
| Equation | x1 | x2 | x3 | x4 |
| (4) | 1 | 1 | ||
| (1) | 1 | 1 | ||
| (2) | 1 | 1 | 1 | |
| (3) | 1 | 1 | 1 | 1 |
Equations (4), (1) and (2) form the partition here. This leaves equation (3) to be the tail. There is no head. Either of x1, x2 or x3 would be suitable choices for the tear variable.
Return to Section 3.5.3: Non-linear Equations Questions