Given, integers 96 and 336 

Again, 336 > 96

Applying division lemma to 96 and 336, we get

336 = 96 * 3 + 48  ..............1

Since, remainder = 48 which is not equal to 0, So we apply division lemma to divisor 98

and remainder 48, we get

96 = 48 * 2 + 0 

Here, remainder is zero.

So, the last divisor or the non-zero remainder at the earlier stage is 48

So, HCF(98, 336) = 48

From, equation 1, we have

      48 = 336 - 96 * 3

=> 48 = 336 * 1 + 96 * (-3)

=> 48 = 336m + 96n

=> 96m + 336n = 48   where m = 1 and n = -3

This is the required linear combination of 96 and 336