Question

When the HCF (420, 130) is expressed as a linear combination of 420 and 130 i.e. HCF (420, 130) = 420x + 130y, the values of x and y satisfying the above relation are:
(a)x= 3, y = 1
(c) x = 4, y = -13
(b) x=-4, y = 13
(d) x = 2, y = 3​

Answer 1

Answer:

We'll follow the Euclid Algorithm to solve this problem,

420 = 3*130 + 30...(1)

Now,

130 = 4*30 + 10 ...(2)

30 = 3*10+0...(3)

Hence the HCF of both these numbers will be 10.

From equation 2 :

HCF (420,130) = 10 = (130-4*30)

and, 30 = 420-(3*130)

So,

10 = (130-4*(420-3*130)) = 13*130 + (-4)*420...(4)

And hence we've shown that the GCD can be shown as a linear combination

To prove that it's not unique

Let's add and subtract the number

(420)*(130)*m

to equation 4

We get

10 = 13*130 + (-4)*420 + (420m)*130 - (130m)*420

=(13+420m)*130 + (-4-130m)*420

So, we can clearly see that on putting in different values of m as an integers we can get different ways of expressing the HCF as a linear combination of both the number

Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

When the HCF (420, 130) is expressed as a linear combination of 420 and 130 i.e. HCF (420, 130) = 420x + 130y, the values of x and y satisfying the above relation are:

(a) x = 3, y = 1

(b) x = - 4, y = 13

(c) x = 4, y = - 13

(d) x = 2, y = 3

EVALUATION

Here the given numbers are 420 , 130

Now

$\sf 420 = (3 \times 130) + 30$

$\sf 130= (4 \times 30) + 10$

Thus we have

$\sf 10$

$\sf = 130 - (4 \times 30)$

$\sf = 130 -4 \times [420 - 3 \times 130]$

$\sf = 130 -4 \times 420 + 12 \times 130$

$\sf =13 \times 130 -4 \times 420$

$\sf =-4 \times 420 + 13 \times 130$

So HCF = 10

Now HCF (420, 130) = 10 is expressed as a linear combination of 420 and 130 as below

$\sf 420= -4 \times 420 + 13 \times 130$

$\sf \therefore \: HCF (420, 130) =-4 \times 420 + 13 \times 130$

Here it is given that

HCF (420, 130) = 420x + 130y

So we have x = - 4 , y = 13

FINAL ANSWER

Hence the correct option is (b) x = - 4, y = 13

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