Question

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

Answer 1

Step-by-step explanation:

BY APPLYING EUCLID's DIVISION LEMMA TO 56 AND 72 : 

72 = 56 × 1 +16 ---- (1) 

56 = 16 ×3 + 8 ---- (2) 

16 = 8×2 + 0 ---- (3) 

THEREFORE HCF OF 56 AND 72 = 8 

FROM (2) -- 8 = 56 - 16 ×3 

8 = 56 - (72-56×1) ×3 (from 1)

8 = 56 - 3 ×72 + 56 ×3 

8 = 56×4 + (-3) * 72 

therefore , x = 4 and y = -3 

Now, 8 = 56 ×4 + (-3) × 72 

8 = 56 ×4 + (-3) ×72 - 56 ×72 + 56 ×72 

8 = 56 × 4 -56 ×72 + (-3) × 72 + 56 ×72 

8 = 56 × (4-72) + {(-3) + 56} × 72 

8 = 56 × (-68) + (53) ×72 

therefore x = -68 and y = 53 

Answer 2

Answer:

x=1/7 y=0

Step-by-step explanation:

HCF(56,72)=8

56x + 72y = 8 ----> (1)

on div by 4

14x + 9y = 2

or, 14x = 2 - 9y

or, x = 2 - 9y / 14

on sub. in (1)

4(2 - 9y) + 72y = 8

or, 8 -36y + 72y = 8

or, 36y = 0

or, y=0

x =2/14 = 1/7