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so x = 5 and y = -19 ============================ another way : to find an infinite number of solutions.

210 x + 55 y = 5 42 x + 11 y = 1 11 (4 x + y) = 1 + 2 x

y is an odd number as RHS is odd and so LHS has to be odd. let y = 2 z+1. one of x and y is negative and the other is positive.

11 (4x + 2z+1) = 1 + 2x 42 x + 22 z + 10 = 0 21 x + 11 z + 5 = 0 --- (1) 11 (x+z) = - 5 (2 x + 1)

x+z is a multiple of -5 and perhaps (2x+1) /11 2 x +1 is an odd multiple of 11. so let 2x + 1 = (2 v + 1) 11 2 x = 22 v + 10 x = 11 v + 5 then by use of (1), 11 z = - 5 - 21 * (11 v + 5) = - 5 - 231 v - 105 = - 231 v - 110 z = - 21 v - 10 y = - 42 v - 19

5 = HCF = 210 * (11 v + 5) - 55 * (42 v + 19) x = 11 v + 5 y = - (42 v + 19)

v x y 210 * x + 55 * y 0 5 -19 5 -1 -6 23 5 1 16 -61 5