Fill in the rest of this joint distribution such that $X$ and $Y$ are independent. (There are multiple answers) $$\begin{array}{c|cc} &X=0&X=1 \\
\hline Y=0&?&3/16\\
Y=1&3/16&?\end{array}$$
I know that they are independent if $P(X=x,Y=y)=P(X=x)\cdot P(Y-y)$ but when I try to solve this I just get a mess of equations. Is there a specific way to solve these types of problems outside of playing around with algebra?
I got $x_0y_1=x_1y_0=3/16$ and $x_0+x_1=y_0+y_1=1$, but I can't make anything of them. 

$$
\begin{array}{c|cc} &X=0&X=1 \\
\hline Y=0& a &3/16\\
Y=1&3/16& b\end{array}
$$
$\Pr(X=0) = a + \dfrac 3 {16} = \Pr(Y=0)$, so
$$
a = \Pr(X=0\ \&\ Y=0) = \Pr(X=0) \cdot \Pr(Y=0) = \left( a + \frac 3 {16} \right)^2.
$$
$$
a = \left( a + \frac 3 {16} \right)^2
$$
$$
256 a^2 - 160a + 9 = 0
$$
etc.
