How to answer this

Question

Antimony has two naturally occuring isotopes, Sb121 and Sb123 . Sb121 has an atomic mass of 120.9038 u , and Sb123 has an atomic mass of 122.9042 u . Antimony has an average atomic mass of 121.7601 u . What is the percent natural abundance of each isotope?

Answer 1

let x = fraction that is Sb121

Then 1 - x = fraction that is Sb123

(x * 120.9038) + (1-x * 122.9042) = 121.7701

120.9038x + 122.9042 - 122.9042x = 121.7701

2.0004x = 1.1341

x = 0.5659

Therefore,

%Sb121 = 56.59%

%Sb123 = 43.41%