If the HCF of 210 and 55 is expressible in the form 210×5+55y then fi

1.	If the HCF of 210 and 55 is expressible in the form 210×5+55y then find y..
Answer» 210=55×3+45...........(1)55=45×1+10..............(2)45=10×4+5................(3)10=5×2+0So here HCF (210,55)=5From equation (3), 5 can be written as5=45—10×4..........(4)Now from (2) 10=55—45×1,now this value substitute in equation (4) in place of 10Then we get,5=45—(55—45×1)×4After simplifying we will get5=45×5—55×4.......(5)Now,from (1) 45=210—55×3, this value substitute in equation (5)5=(210—55×3)×5—55×4, after simplifying we will get5=210×(5)+55×(-19)....Now this one compare with given equation that 5=210×5+55×xThen X=–19 is the answer

Discussion