Converting polar to cartesian?

So far I got

\begin{align} r & = 7 / (4 - 2 \cos\theta) \\ r (4 & - 2\cos\theta) = 7 \\ r (4 & - 2( x / r ) ) = 7 \end{align}

You are on the right track. So we get $$4r-2x=7.$$ This implies that $$4r=7+2x.$$ Taking square on both sides, we get $$16r^2=(7+2x)^2.$$ Now note that $x^2+y^2=r^2$. Put this into the last equation, we have $$16(x^2+y^2)=(7+2x)^2.$$