Question

If d is the HCF of 30 and 72. find the value of x and y satisfying d= 30x+ 72y​

Answer 1

Answer -

x = 5 and y = - 2

Step-by-step explanation -

Using Euclid's division lemma :-

a = bq + r

(72 > 32)

So,

→ 72 = 30(2) + 12

→ 30 = 12(2) + 12

→ 12 = 6(2) + 0

As, remainder is 0. So, divisor is 6.

Means, the HCF of 30 and 72 is 6.

→ 30 = 12(2) + 6

→ 30 - 12(2) = 6 ...(1)

Similarly,

→ 72 = 30(2) + 12

→ 72 - 30(2) = 12 ...(2)

Substitute value of equation (2) in (1)

→ 30 - [ {72 - 30(2)} (2)] = 6

→ 30 - [72(2) - 30(4)] = 6

→ 30 - 72(2) + 30(4) = 6

→ 30 + 72(-2) + 30(4) = 6

→ 30(5) + 72(-2) = 6 ...(3)

Now, compare equation (3) with d = 30x + 72y

On comparing we get,

→ x = 5 and y = -2

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Answer 2

Question :---- we have to Find value of x and y satisfying d = 30x + 72 y ..

Given :---

• d is the HCF of 30 & 72..

Formula to be used :---

• Given positive integers a and b , there exist unique integers q and r satisfying

a = bq + r

0 ≤ r < b .........

Solution :-------

Applying Euclid's division lemma to 30 and 72

Since 72 > 30

72 = 30 × 2 + 12 ------------ Equation ( 1 )

30 = 12 × 2 + 6 -------------Equation( 2 )

12 = 6 × 2 + 0 -------------Equation( 3 )

The remainder has now become zero....

Since the divisor at this stage is 6

The HCF of 30 and 72 = 6

now from Equation (2)

30 = 12 × 2 + 6

Rearrange this :-- 6 = 30 - 12 × 2

⇒ 6 = 30 - [ (72 - 30 × 2 ) × 2 ] { from ( 1 ) }

⇒ 6 = 30 - 72 × 2 + 4 × 30 [ using distributive property ]

⇒ 6 = 30 × ( 5 ) + 72 × ( -2 ) -------Equation ( 4 )

according to the problem ,

d = 30x + 72y --------------------------Equation ( 5 )

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compare Equation ( 4 ) and Equation ( 5 ) Now we get,

d = 30 × ( 5 ) + 72 × ( -2 ) = 30x + 72y

we get,

x = 5

y = (-2)

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(Hope it helps you)