Let us first find the HCF of 210 and 55.

Applying Euclid division lemna on 210 and 55, we get

210 = 55 × 3 + 45 ....(1)

Since the remainder 45 ≠ 0. So, again applying the Euclid division lemna on 55 and 45, we get

55 = 45 × 1 + 10 .... (2)

Again, considering the divisor 45 and remainder 10 and applying division lemna, we get

45 = 4 × 10 + 5 .... (3)

We now, consider the divisor 10 and remainder 5 and applying division lemna to get

10 = 5 × 2 + 0 .... (4)

We observe that the remainder at this stage is zero. So, the last divisor i.e., 5 is the HCF of 210 and 55.

∴ 5 = 210 × 5 + 55y

⇒ 55y = 5 - 1050 = -1045

⇒ y = -19

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