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I'm a bit stuck on one part of Bézout's identity when used with Euclid's algorithm.

The specific part of the equation I can't see is;

               3=27-4*(60-2*27)

Which reduces to

3=927-460

(How and why should we know to reduce the equation in this format?)

Then we backward substitute again to say

               3=9*(207-3*60)-4*60

Which reduces to

3=9x207-31*60

(Again, how and why do we reduce the equation in this manner?)

This is to satisfy the equation av+bw=d.

Many thanks!

Bill Dubuque
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CoedFoel
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    Huh? You mean $9\times 27 - 4\times 60$? You’re doing back substitution carefully to get $3=ma+nb$, where $a$ and $b$ are the integers you started with. This is explained clearly in every elementary number theory and algebra text. – Ted Shifrin Aug 23 '23 at 05:25
  • Please read the Note in an answer in the dupe for the key idea behind back-substitution. But you should stop doing it backwards and instead propagate forward the linear combinations, since that direction is much simpler and far less error-prone - as explained there. $\ \ $ – Bill Dubuque Aug 23 '23 at 07:27

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