Answer:
c
Step-by-step explanation:
Answer:
Step-by-step explanation:
x = y - 2.........(1)
4x + y = 2......(2)
Substituting x in (1) in (2)
4(y - 2) + y = 2
4y - 8 + y = 2
5y = 2 + 8
5y = 10
y = 10/5
y = 2
Putting x in (1)
x = y - 2
x = 2 - 2
x = 0
Let a = 693, b = 567 and c = 441
Now first we will find HCF of 693 and 567 by using Euclid’s division algorithm as under
693 = 567 x 1 + 126
567 = 126 x 4 + 63
126 = 63 x 2 + 0
Hence, HCF of 693 and 567 is 63
Now we will find HCF of third number i.e., 441 with 63 So by Euclid’s division alogorithm for 441 and 63
441 = 63 x 7+0
=> HCF of 441 and 63 is 63.
Hence, HCF of 441, 567 and 693 is 63.
