http://www1.cs.columbia.edu/~tal/499...notes/ITC6.pdf
You are using the Extended Euclidean Algorithm. Refer to Example 1. It is a much simpler version of this problem. You use the equations computed in the Euclidean Algorithm to replace the value of the remainders until you are left with only the orginal values a & b.
Restating each of the equations above you get:
190 = 5(34) + 20 --> 20 = 190 - 5(34)
34 = 1(20) + 14 --> 14 = 34 - 1(20)
20 = 1(14) + 6 --> 6 = 20 - 1(14)
14 = 2(6) + 2 --> 2 = 14 - 2(6)
We use these to compute s & t as follows:
Start with the last one -> 2 = 14 - 2(6)
Use the next equation above it to replace 6 to get -> 2 = 14 - 2[20 - 1(14)]
Simplify -> 2 = 14 -2(20) + 2(14) --> 2 = 3(14) - 2(20)
Use the next equation above to replace 14 to get -> 2 = 3[34 - 1(20)] - 2(20)
Simplify -> 2 = 3(34) - 3(20) - 2(20) --> 2 = 3(34) - 5(20)
Now use the top equation to replace 20 to get -> 2 = 3(34) - 5[190 - 5(34)]
Simplify -> 2 = 3(34) - 5(190) + 25(34) --> 2 = 28(34) - 5(190)
since a = 190, s = -5
since b = 34, t = 28
Hope this helps