Question

Express the GCD of 48 and 18 as a linear combination.

Answer 1

A=bq+r, where   o ≤  r < b
48=18x2+12
18=12x1+6
12=6x2+0 ∴ HCF  (18,48) = 6 now   6= 18-12x1
           6= 18-(48-18x2)
  6= 18-48x1+18x2
  6= 18x3-48x1   6= 18x3+48x(-1)
i.e.       6= 18x +48y
     
   ∴   

6= 18×3 +48×(-1)
=18×3 +48×(-1) + 18×48-18×48
=18(3+48)+48(-1-18)
=18×51+48×(-19)
6=18x+48y
 
   ∴     


Hence, x and y are not unique.

Answer 2

a = bq + r

$\bf\huge 48 = 18\times 2 + 12$

$\bf\huge 18 = 12\times 1 + 6$

$\bf\huge 12 = 6\times 2 + 0$

HCF (18 , 48) = 6

$\bf\huge 6 = 18 - 12\times 1$

$\bf\huge 6 = 18 - (48 - 18\times 2$

$\bf\huge 6 = 18 - 48\times 1 + 18\times 2$

$\bf\huge 6 = 18\times 3 + 48\times 1$

$\bf\huge 6 = 18\times 3 + 48\times (-1)$

$\bf\huge 6 = 18x + 48y$

Here x = 3 and y = -1

$\bf\huge 6 = 18\times 3 + 48\times (-1)$

$\bf\huge 6 = 18\times 3 + 48\times (-1) + 18\times 48 - 18\times 48$

$\bf\huge 6 = 18 (3 + 48) + 48(-1 - 18)$

$\bf\huge 6 = 18\times 51 + 48\times (-19)$

$\bf\huge 6 = 18x + 48y$

Here we have x = 51 and y = -19

Therefore x and y are not unique