Question

if d is the hcf of 56 and 72 find x,y satisfyind 56x+72y and also show that xand y are not unique ​

Answer 1

Step-by-step explanation:

According to Euclid division Lemma,

72=56*1+1656

=16*3+816

=8*2+0…….

So, 8 is the HCF.

Now 8=56-(16*3)…..

8 is expressed in terms of 56 and 16, but it should be in the form of 56 and 72.

Hence by splliting 16 we get,

8=56-(72-56*1)(3)

8 =56-72*3+56*3…….

Since we have one more 56 is grouped as

8 =56*4+72(-3).

So,X and Y are 4,3 respectively ..

Therefore X and Y are not equal.

Answer 2

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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

HENCE , x and y are not unique.

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