Chemistry help!!! please!

A 0.9157-g mixture of CaBr2 and NaBr is dissolved in water, and AgNO3 is added to the solution to form AgBr precipitate. If the mass of the precipitate is 1.6840 g, what is the percent by mass of NaBr in the original mixture?

using molar masses, find grams of Br:
(1.6840 g AgBr) ( 79.904 g/mol Br) / (187.77 g/mol AgBr) = 0.71661 total grams of Br present


if we say that "X" = grams of CaBr2, then (0.9157 - X) is the grams of NaBr

use molar masses to convert grams of CaBr2 into grams of Br
(X) (159.81 g / mol Br2) / (199.89 g/mol CaBr2) = 0.7995 X grams of Br from CaBr2

us molar masses to convert grams of NaBr into grams of Br
(0.9157 - X) ( 79.904 g/mol Br) / (102.89 g/mol NaBr) = (0.9157 - X) (0.7766) grams of Br from NaBr

add them
0.7995 X grams & (0.9157 - X) (0.7766) grams = 0.7166 total grams of Br

0.7995 X & 0.7111 - 0.7766X = 0.7166

0.7995 X - 0.7766X = 0.7166 - 0.7111

0.02290X = 0.005471

X = 0.2389 grams of CaBr2

find %
(0.2389 grams of CaBr2) / (0.9157-g mixture) = 26.09 % CaBr2
so that would make it
73.91% NaBr