Question

if d is the HCF of 32 and 72 find the value of x & y satisfying d = 30x + 72y​

Answer 1

Answer :

• Value of x = 5.
• Value of y = -2.

Step-by-step explanation :

Using Euclid’s division Lemma, we get,

$:\implies\rm{72=30\times{}2+12\longrightarrow{}(1)}$

$:\implies\rm{30=12\times{}2+6\longrightarrow{}(2)}$

$:\implies\rm{12=6\times{}2+0\longrightarrow{}(3)}$

Thus, HCF(30, 72) = 6.

$\rule{350}{3}$

Now,

$:\implies\rm{6=30-12\times{}2\longrightarrow{}From\:(2)}$

$:\implies\rm{6=30-(72-30\times{}2)\times{}2\longrightarrow{}From\:(1)}$

$:\implies\rm{6=30-72\times{}2+30\times{}4}$

$:\implies\rm{6=30(1+4)-72\times{}2}$

$:\implies\rm{6=30\times{}5+72\times{}(-2)}$

$:\implies\rm{6=30x+72y}$

Therefore, x = 5, y = -2.

[NOTE : Value of x & y are not unique]

Answer 2

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Given:

• We have been given that d is the HCF of 32 and 72.

To Find:

• We need to find the value of x & y satisfying d = 30x + 72y.

Solution:

We need to apply the Euclid's Division Lemma.

Clearly 72>30

72 = 30 × 2 + 12_____(1)

30 = 12 × 2 + 6______(2)

12 = 6 × 2 + 0_______(3)

We got 6 as the HCF of 72 and 30.

From equation 2, we have 30 = 12 × 2 + 6

or 6 = 30 - 12 × 2

=> 6 = 30 - [ (72 - 30 × 2 ) × 2 ] [From equation 1]

=> 6 = 30 × ( 5 ) + 72 × ( -2 )____(4)

According to the question we have,

d = 30x + 72y ______________(5)

Now, on comparing equation 4 and 5 we get,

x = 5 and y = -2

But the values of x and y are not unique.