Let us omit these red lines and go several steps at a time, otherwise it gets all too boring. Find a pair of 7's and two 8's and a 9:
This 9 hits a pair of 3's, killing one possibility, so that the 3 must be in the other place. That also determines the fate of the other pair of 3's, and then also that of the intersecting pair of 6's. (Noting down these pairs was not necessary, but is a good speedup.)
We find two more 8's and a 2. (That used the pair because we have it anyway, but solving without it would have been just as easy.)
An 8 bumped into a pair of 7's, and we find a 7, and then a 6, and then the other 6.
We find another 2 and complete the second row (find 7, then 1, then 4).
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