if d is the HCF of 56 and 72,find x,y satisfying d=56x+72y. Also show that x and y are not unique
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Euclid division lemma:-
a = bq + r
0 ≤ r < b
a > b
72 > 56
72 = 56 × 1 + 16
56 = 16 × 3 + 8
16 = 8 × 2 + 0
As the remainder is 0,HCF is 8
HCF of 56 and 72 is 8
d = 8
d = 56x + 72y
8 = 56 - 16×3
= 56 - [72 - 56(1)]×3
= 56 - 72×3 + 56×3
= 56×4 - 72×3
= 56×4 + 72(-3)
= 56x + 72y
Therefore, x = 4 and y = -3
Hope it helps you:)
a = bq + r
0 ≤ r < b
a > b
72 > 56
72 = 56 × 1 + 16
56 = 16 × 3 + 8
16 = 8 × 2 + 0
As the remainder is 0,HCF is 8
HCF of 56 and 72 is 8
d = 8
d = 56x + 72y
8 = 56 - 16×3
= 56 - [72 - 56(1)]×3
= 56 - 72×3 + 56×3
= 56×4 - 72×3
= 56×4 + 72(-3)
= 56x + 72y
Therefore, x = 4 and y = -3
Hope it helps you:)
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