calculate simultaneous solubility of AgCNS and AgBr in a solution of water.Ksp AgCNS=1.2*10^-12,Ksp AgBr=5*10^-13

$$\begin{gathered} AgCNS \to A{g}^{+} + CN{S}^{-} \\ x x \\ AgBr \to A{g}^{+} + B{r}^{-} \\ y y \end{gathered}$$

Now,
Total [Ag⁺] = x + y
Total [CNS^-] = x
Total [Br^-] = y

$$\begin{gathered} K_{sp}\left(AgCNS\right) = \left[A{g}^{+}\right]\left[CN{S}^{-}\right] = (x+y)x \\ (x+y)x= 1.2 x {10}^{-12 }-----\left(1\right) \\ K_{sp}\left(AgBr\right)=\left[A{g}^{+}\right]\left[B{r}^{-}\right] = (x+y)y \\ (x+y)y = 5 x {10}^{-13}-----\left(2\right) \\ Dividing \left(2\right) by \left(1\right) \\ \frac{y}{x} = 0.42 \\ y = 0.42x \\ Substituting in \left(1\right): \\ (x + 0.42x)x = 1.2 x {10}^{-12} \\ 1.42 x^2 = 1.2 x {10}^{-12} \\ x = \sqrt{\frac{1.2 x {10}^{-12}}{1.42}} \\ =9.2 x {10}^{-7} mol/L \\ y = 0.42 x 9.2 x {10}^{-7} \\ =3.86 x {10}^{-7} mol/L \end{gathered}$$

Hence, solubility of AgCNS = 9.2 x 10^-7 mol/L
  solubility of AgBr = 3.86 x 10^-7  mol/L