Question

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 ?

please explain each thing​

Answer 1

➵ 72 = 56 × 1 + 16 __[1]

➵ 56 = 16 × 3 + 8 __[2]

➵ 16 = 8 × 2 + 0

:. H.C.F = 8

______________________________

From eq[2],

➵ 56 - (16) × 3 = 8 __(i)

From eq[1],

➵ 72 - 56 = 16 __(ii)

By substituting (ii) in (i),

➵ 56 - (72 - 56) × 3 = 8

➵ 8 = 56 + 72(- 3) + 56(3)

➵ 8 = 56(4) + 72(- 3) __[3]

➵ d = 56x + 72y

Where, x = 4 and y = - 3

______________________________

By adding (72)(56) - (72)(56) in eq[3],

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

➵ 8 = 56(4 + 72) + 72(- 3 - 56)

➵ 8 = 56(76) + 72(- 59)

➵ d = 56x + 72y

Hence, x = 56 and y = - 59.

______________________________

Therefore, x and y are not unique.

Q.E.D

Answer 2

Answer:-

$72 = 56 \times 1 = 16 \\ 56 = 16 \times 3 + 8 \\ 16 = 8 \times 2 + 10$

Now L. C. M =8

$56 - (16) \times 3 = 18 \\ 72 - 56 = 16$

$56 - (72 - 56) \times 3 = 8 \\ 8 = 56 + 72( - 3) + 56(3) \\ 8 = 56(4) + 72( - 3) \\ d = 56 x + 72y$

$x = 4 \: and \: y = - 3 \\ (72)(56) - (72)(56)$

$8 = 56(4) + 72( - 3) + (72)(56) - (72)(56) \\ 8 = 56(4 + 72) + 72( - 3 - 56) \\ 8 = 56(76) + 72( - 59) \\ d = 56 x + 72y$

$therefore.. \\ x = 56 \: and \: y = \: 59 \\ x \: and \: ar \: not \: unique$