Question

if d is HCF of 56 and 72 find x and y satisfying d=56x+72y

Answer 1

given no 72 and 56
72=56*1+16........[1]
56=16*3+8...[2]
16=8*2+0.....[3]
the HCF of 56 and 72 is d=8

now,56=16*3+8 ...[from1]
        8=56-16*3
        8=56-[72-56]*3
        8=56-72*3+56*3
        8=56+56*3-72*3
        8=56*[1+3]-73*3
        8=56*4-72*3
56x+72y=56*4+72*3
 hence, x=4 and y=3

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Answer 2

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)
 x = 4 and y = -3