Question

if d is the hcf of 56 and 72 find the values of x and y which satisfies d=56x+72y

Answer 1

1/28=x and y=1/36 and hence this is the answer

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)

= 56x + 72y

Therefore, x = 4 and y = -3

Hope it helps you:)