Question

can anyone solve this problem?
Question 29.

Answer 1

Answer:

Step-by-step explanation:

Given integers,

40 and 65

Applying Euclid's division lemma we get,

65=40×1+25......................................1

Applying Euclid's division lemma to 40 and 25 we get,

40=25×1+15......................................2

Applying Euclid's division lemma to 25 and 15 we get,

25=15×1+10........................................3

Applying Euclid's division lemma to 15 and 10 we get,

15=10×1+5..........................................4

Applying Euclid's division lemma to 10 and 5 we get,

10=5×2+0...........................................5

eq1,eq2,eq3,eq4 can also be written as,

65-40×1=25..................................…..6

40-25×1=15........................................7

25-15×1=10.........................................8

15-10×1=5............................................9

Eq9 we have,

5=15-10×1

By putting eq8 in above eq we get,

5=15-(25-15×1)1

5=15-25×1+15×1

5=15×2-25×1

By putting eq7 in above eq we get,

5=(40-25×1)2-25×1

5=40×2-25×2-25×1

5=40×2-25×3

By putting eq6 in above eq we get,

5=40×2-(65-40×1)3

5=40×2-65×3+40×3

5=40×5-65×3

5=40(5)+65(-3)

Given equation,

d=40x+65y

Here d is HCF.

Therefore , 5=40x+65y

On comparing we get,

x=5 and y=-3

Hence value of x=5 and value of y=-3