If d is the hcf of 56 and 72 find x and y satisfying d=56x+72y
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Now L. C. M =8
72−56=16
56−(72−56)×3=8
8=56+72(−3)+56(3)
8=56(4)+72(−3)
d=56x+72y
x=4andy=−3
(72)(56)−(72)(56)
8=56(4)+72(−3)+(72)(56)−(72)(56)
8=56(4+72)+72(−3−56)
8=56(76)+72(−59)
d=56x+72y
Therefore, x=56 andy = 59 x and are not unique
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