Leren wrote:(5=8) r2c1 - r2c9 = (8-2) r3c8 = (2-5) r9c8 = (5) r9c7 => - 5 r2c7
Got it.
If r2c1 is 5 then r2c7 is not 5.
If r2c1 is 8 then r2c9 is not 8 and r3c8 becomes the 8 in that box.
So r9c8 becomes the 2 in that column.
And r9c7 is 5 and r2c7 is not 5 again.
=> - 5 r2c7
And 46 pair eliminates the 4 from r3c7.
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| 3 56 49 | 1 8 2 | 7 459 569 |
| 58 7 2 | 34 4569 569 | 46 13 13689 |
| 1 568 49 | 346 4569 7 | 25 2489 369 |
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| 28 28 7 | 46 3 1 | 9 45 56 |
| 49 13 6 | 7 459 459 | 8 13 2 |
| 49 13 5 | 8 2 469 | 46 7 13 |
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| 27 29 1 | 5 67 68 | 3 89 4 |
| 57 59 3 | 2 47 48 | 1 6 89 |
| 6 4 8 | 9 1 3 | 25 25 7 |
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