Solubility of Ag3AsO4 in 0.02 M K3AsO4 Calculate the solubility of $\ce{Ag3AsO4}$ in $\pu{0.02M}~\ce{K3AsO4}$ neglecting the activity coefficients. Find the relative error. $K_\mathrm{sp}(\ce{Ag3AsO4}) = \pu{6.0e-23}$

I know how to calculate the relative error but I get a very complicated equation finding the concentration solubility product constant ($K'_\mathrm{sp}$). There should be a quicker way to solve this since it is a midterm question.
I tried this:
$$\ce{Ag3AsO4 -&gt; 3Ag+ + AsO4^3-}$$
$$K_\mathrm{sp} = 27x^4$$
$$x = \pu{2.78e-6}$$
$$[\ce{Ag+}] = 3x = \pu{8.34e-6M}$$
$$[\ce{AsO4^3-}] = x = \pu{2.78e-6M}$$
$0.02~\mathrm{M}\ \ce{AsO4^3-}$ comes from $\ce{K2AsO4}$. So there should be an equation like:
$$K_\mathrm{sp} = (\pu{8.34e-6} - 3x)^3 \times (\pu{2.78e-6} + 0.02 - x) = \pu{6.0e-6}$$
And things get complicated. After finding $x$, I will also have found the final concentrations of silver and $\ce{AsO4^3-}$ ions. Then I will read the activity coefficients of them from the appendix table.
I have two questions:

Is my method true or false?

What is an easier way of solving this problem?
You start of by calculating solubility with one approximation: that all $\ce{AsO4^3-}$ ions are obtained by dissolving the potassium salt:
Let $S$ be the molar solubility:
$$K_\mathrm{sp} = 0.02 \times 27 S^3$$
$$S = \pu{0.0223 M}$$
Then, you calculate the solubility by taking in account the concentration of arsenate ion obtained by dissolution of silver arsenate:
$$K_\mathrm{sp} = 27 S^3 \times (S+0.02) = 27 S^4 + 0.54S^3$$
Solving this equation by applying the iteration method gives the value of $S = \pu{0.018M}$.
So the relative error is $23.9~\%$.