Two bivalue cells with the same candidates are connected via a conjugate pair; their common digit is eliminated from cells seeing both.
Two cells share the same two candidates {X,Z}. A conjugate pair of digit X connects them: if one cell takes X, the bridge forces the other cell to take Z, and vice versa. Either way, Z appears in one of the two endpoint cells — so any cell seeing both can have Z eliminated.
| 5 | 9 | 6 | 3 | 1 | 7 | 2 | 4 | 8 |
| 1 | 2 | 48 | 4 | 58 | 68 | 6 | 7 | 9 |
| 4 | 7 | 8 | 2 | 6 | 9 | 1 | 3 | 5 |
| 2 | 1 | 4 | 5 | 3 | 6 | 8 | 9 | 7 |
| 3 | 5 | 4 | 8 | 9 | 4 | 4 | 2 | 6 |
| 6 | 8 | 9 | 7 | 2 | 4 | 3 | 5 | 1 |
| 7 | 3 | 48 | 6 | 4 | 48 | 9 | 8 | 2 |
| 8 | 6 | 2 | 9 | 7 | 3 | 5 | 1 | 4 |
| 9 | 4 | 5 | 1 | 8 | 2 | 7 | 6 | 3 |
R2C3={4,8}, R7C6={4,8}. Digit 4 has a conjugate pair in Row 5 at R5C3 and R5C6. → Any cell seeing both R2C3 and R7C6 can have 8 eliminated.
This 9×9 puzzle is solver-verified to require this technique on its solution path.