This is an exercise in using Euclid's algorithm. (If you haven't come across this, Google it to see why the following calculations are being done.)

Now work through those equations in the reverse order to get

So one solution to the equation 42x+55y=1 is x=–17, y= 13. You get the other solutions by adding a multiple of 55 to x, and subtracting the same multiple of 42 from y. Thus

(you can see that the multiples of k cancel, to leave you with the result from the previous displayed equation).

Now all you have to do is to choose k so that –17+55k lies between 100 and 200.

**For this one, start by finding x and y such that 85x+194y=1 (using Euclid's algorithm, as for Q1). Multiply the result by 12 on both sides, to give you a solution to 85X – 12 = (multiple of 19). Then add suitable multiples of 19 to find the two smallest positive values for X.**