Recurring decimals to fraction 
*

*$.\overline{36}=\frac{36}{99}$  

*$2.1\overline{36}=2\frac3{22}$


The part I do not understand however, is "you could used 1) to speed up the working of 2)" which is written in the book. 
How would I use 1) to help me work out 2)? 
Thanks.
 A: $$2.1\overline{39}=2+0.1+\frac1{10}\cdot0.\overline{39}$$
So, $$2.1\overline{39}=2+\frac1{10}+\frac1{10}\cdot\frac{36}{99}=2+\frac1{10}\cdot\left(1+\frac4{11}\right)=2+\frac1{10}\cdot\frac{15}{11}=2+\frac3{22}$$ 
A: *

*$.\overline{36}=\frac{36}{99}$

*$2.1\overline{36}=2\frac3{22}$
part 1.
suppose
$$\begin {equation}\dfrac pq=0.\overline{36}\end {equation}$$
this is eqn (1).
mulipltly this eqn with 100
$$\begin {equation}100\dfrac pq=36.\overline{36}\end {equation}$$
this is eqn (2).Now subtract (1) from (2).
$$\begin {equation}100\dfrac pq-\frac pq=36.\overline{36}-0.\overline{36}\end {equation}$$
$$99\dfrac pq=36.0$$
$$\dfrac pq=\frac {36}{99}$$
part 2
$2.1\overline{36}=2+0.1\overline{36}$
$$\dfrac pq=0.1\overline{36}\implies 10\dfrac pq=1.\overline{36}$$this is eqn (1).
multilpy with 100 in eqn(1)
$$1000\dfrac pq=136.\overline{36}$$ then subtract (1)  from (2).
$$990\dfrac pq=135.0$$
$$\dfrac pq=\dfrac {135}{990}\implies\dfrac {3}{22}$$
so  $2.1\overline{36}\implies 2+0.1\overline{36}\implies 2+\dfrac{3}{22}\implies 2\dfrac3{22}$
