4g mixture of NaCl & NaBr contain 30%Na in the total mixture.Calculate the % composition of mixture

answer is 45.15%NaCl and 54.85%NaBr please tell me method how to solve this...

Amount of Na in the mixture = 4x30/100 = 1.2 g
Now let 'a' g be the amount of NaCl in the mixture, then the amount of NaBr in the mixture = (4 - a) g
Further amount of Na in 'a' g NaCl = 23xa/58.5 = 0.393 a g (Here 23 is the atomic mass of Na and 58.5 is the molar mass of NaCl)
And amount of Na in (4 - a) g NaBr = 23x(4 -a)/103 = 0.223(4 - a) g (Here 23 is the atomic mass of Na and 103 is the molar mass of NaBr)
Thus total mass of Na in the mixture = 0.393 a + 0.223(4 - a) which is equal to 1.2
i.e. 0.393 a+ 0.892 - 0.223 a = 1.2
OR 0.17 a = 1.200 - 0.892 = 0.308
OR a = 0.308/0.17 = 1.812 g
So the amount of NaCl in the mixture = 1.812 g
Therefore % of NaCl = 1.812x100/4 = 45.3
And % of NaBr = 100 - 45.3 = 54.7