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Question

If d is HCF of 56 and 72, then find x,y satisfying the equation d=56x+72y. Also show that x and y are not unique.


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Solution

Step 1: Calculate d by division.

Since d is the HCF of 56 and 72, it divides both the numbers evenly.

25672228362141879

Thus, HCF of 56 and 72 is 2×2×2=8.

Step 2: Express 8 in terms of 56 and 72.

Write 8 in terms of 56.

Using Euclid's division algorithm,

56=16×3+88=56-(16×3)

Similarly, write 16 in terms of 72.

8=56-(72-56)×316=72-568=56-72×3+56×3a(b+c)=ab+ac8=56(1+3)+72(-3)8=56(4)+72(-3)

Step 3: Compare with the given equation.

Comparing d=56x+72y and 8=56(4)+72(-3),

x=4 and y=-3.

So, the value of x is 4 and the value of y is -3.


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