18x + 48y = 6, where x = 3, y = –1
Solution:
Given numbers are 48 and 18.
Write it using linear combination, a = bq + r, where 0 ≤ r < b
48 = 18 × 2 + 12
⇒ 18 = 12 × 1 + 6 (write 18 using the previous remainder 12)
⇒ 12 = 6 × 2 + 0 (write 12 using the previous remainder 6)
Remainder is zero. So, the process stops.
∴ HCF(18, 48) = 6
Now, Express 6 using 18 and 48.
6 = 18 – 12 × 1
6 = 18 – (48 – 18 × 2)
6 = 18 – 48 + 18 × 2
6 = 18 × 3 – 48 × 1
6 = 18 × 3 + 48 × (–1)
6 = 18x + 48y
18x + 48y = 6, where x = 3, y = –1
Hence 18x + 48y = 6 as the linear combination.
