20. Determine the value of the integer $y$ given that $y=3 x^{2}$ and
$$
\frac{2 x}{5}=\frac{1}{1-\frac{2}{3+\frac{1}{4-\frac{5}{6-x}}}}
$$
Solution
147
Note first that
$$
4-\frac{5}{6-x}=\frac{24-4 x-5}{6-x}=\frac{19-4 x}{6-x}=F \text {, say. }
$$

Then,
$$
3+\frac{1}{F}=3+\frac{6-x}{19-4 x}=\frac{57-12 x+6-x}{19-4 x}=\frac{63-13 x}{19-4 x}=G, \text { say. }
$$

Then,
$$
1-\frac{2}{G}=1-\frac{2 \times(19-4 x)}{63-13 x}=\frac{63-13 x-(38-8 x)}{63-13 x}=\frac{25-5 x}{63-13 x}=H \text {, say. }
$$

So we get the equation $\frac{2 x}{5}=\frac{1}{H}=\frac{63-13 x}{25-5 x}$ which simplifies to $2 x^{2}-23 x+63=0$. This has solutions $x=4.5$ and $x=7$; the corresponding values of $y$ are 60.75 and 147 respectively. Therefore $y=147$, since we are told $y$ is an integer.