Simultaneous solubility of CaF2 and SrF2
Calculate the simultaneous solubility of $\ce{CaF_2}$ and $\ce{SrF_2}$ given that:

$\mathrm {K_{sp}(\ce CaF_2)= 3.9 \times 10^{-11}}$ and $\mathrm {K_{sp}(\ce SrF_2)= 2.9 \times 10^{-9}}$


Let $\ce{[Ca^{2+}]} = x$ and $\ce{[Sr^{2+}]} = y$
So the dissolving $\ce{CaF_2}$ contributes $2x$ of $\ce{F^-}$ and the dissolving $\ce{SrF_2}$ contributes $2y$ of $\ce{F^-}$. So the total $\ce{[F^-]} = 2x + 2y = 2(x+y)$
Substituting into the two $\text{K}_{sp}$ equations leads to &quot;messy&quot; cubic equations which could be solved numerically since there are two equations with two unknowns.
$3.9\times10^{-11} = x(2(x+y))^2 = 4x(x+y)^2$ 
$2.9 \times 10^{-9} = y(2(x+y))^2 = 4y(x+y)^2$
But going back to the original $\text{K}_{sp}$ equations we have:
$\ce{[Ca^{2+}][F^-]^2 = 3.9 \times 10^{-11}}$ 
$\ce{[Sr^{2+}][F^-]^2 = 2.9 \times 10^{-9}}$
and by dividing the two we get $\ce{[Ca^{2+}] = 0.0134 [Sr^{2+}]}$ or $x = 0.0134y$. Substituting this into the second equation we get:
$2.9 \times 10^{-9} = 4y(x+y)^2 = 4y(1.0134y)^2 = 4.108y^3$ or $y=8.9\times10^{-4}$
So:
$\ce{[Sr^{2+}]} = 8.9\times10^{-4}$ 
$\ce{[Ca^{2+}]} = (0.0134)(8.9\times10^{-4}) = 1.2\times10^{-5} $
$\ce{[F^-]} = 2(8.9\times10^{-4} + 1.2\times10^{-5}) = 1.8\times10^{-3}$
Note that this can also be simplified using significant figures a different way. Since the two $\mathrm {K_{sp}}$ values only have two significant figures and  $\ce{[Ca^{2+}] = 0.0134 [Sr^{2+}]}$, we can assume that for all practical purposes that all the $\ce{F^-}$ comes from the dissolution of the $\ce{SrF_2}$. So:
$\ce{4[Sr^{2+}]^3} = 2.9\times10^{-9}$
Thus
$\ce{[Sr^{2+}]} = 9.0 \times10^{-4}$
$\ce{[F^-]} = \sqrt{\frac{2.9\times10^{-9}}{9.0 \times10^{-4}}} = 1.8\times10^{-3}$
$\ce{[Ca^{2+}]} = \dfrac{3.9\times10^{-11}}{(1.8\times10^{-3})^2}=1.2\times10^{-5}$