Given integers,40 and 65Applying Euclid's division lemma we get,65=40×1+25......................................1Applying Euclid's division lemma to 40 and 25 we get,40=25×1+15......................................2Applying Euclid's division lemma to 25 and 15 we get,25=15×1+10........................................3Applying Euclid's division lemma to 15 and 10 we get,15=10×1+5..........................................4Applying Euclid's division lemma to 10 and 5 we get,10=5×2+0...........................................5eq1,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............................................9Eq9 we have,5=15-10×1By putting eq8 in above eq we get,5=15-(25-15×1)1

5=15-25×1+15×1

5=15×2-25×1By putting eq7 in above eq we get,5=(40-25×1)2-25×1

5=40×2-25×2-25×1

5=40×2-25×3By 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+65yHere d is HCF.Therefore , 5=40x+65yOn comparing we get,x=5 and y=-3Hence value of x=5 and value of y=-3