Express $1.3\overline{2}+0.\overline{35}$ as a fraction in the simplest form.

Hint:
A fraction can be defined as a part of a whole. A fraction consists of two parts: numerator and denominator. The simplest form of a fraction is the state when numerator and denominator cannot be divided any further, while still being whole numbers.

Complete step by step solution:
Let $x=1.3\overline{2}=1.322222.......\text{ }.....\left( 1 \right)$
Now, multiplying equation (1) by $10$
$\Rightarrow 10x=13.22222.......\text{ }.....\text{(2)}$
Now again, multiplying equation (2) by
$\Rightarrow 10\times 10x=132.22222........\text{ }.....\left( 3 \right)$
Here, subtracting equation (2) from (3)
$\begin{align}
  & \Rightarrow 100x-10x=132.22222-13.22222 \\
 & \Rightarrow 90x=119 \\
 & \therefore x=\dfrac{119}{90} \\
\end{align}$
Now again, let $y=0.\overline{35}=0.353535........\text{ }.....\left( 4 \right)$
Here, multiplying equation (4) by $100$
$\Rightarrow 100y=35.353535........\text{ }.....\left( 5 \right)$
Subtracting equation (4) from equation (5)
$\begin{align}
  & \Rightarrow 100y-y=35.353535-0.353535 \\
 & \Rightarrow 99y=35 \\
 & \therefore y=\dfrac{35}{99} \\
\end{align}$
We know that,
$\Rightarrow 1.3\overline{2}+0.\overline{35}=x+y$
Hence,
$\begin{align}
  & \Rightarrow 1.3\overline{2}+0.\overline{35}=\dfrac{119}{90}+\dfrac{35}{99} \\
 & \Rightarrow 1.3\overline{2}+0.\overline{35}=\dfrac{119\times 11+35\times 10}{990} \\
 & \Rightarrow 1.3\overline{2}+0.\overline{35}=\dfrac{1309+350}{990} \\
 & \Rightarrow 1.3\overline{2}+0.\overline{35}=\dfrac{1659}{990} \\
 & \therefore1.3\overline{2}+0.\overline{35}=\dfrac{553}{330} \\
\end{align}$

Hence, the value is $\dfrac{553}{330}$.

Note:
Always keep in mind that the digits needed to be multiplied by $10$ till all the required digits are on the left side. You only want one “set” of repeating digits on the left side of the decimal. For example, in this question the first digit has $2$ as the repeating digit, thus you only want one $2$ on the left of the decimal. In the second number repeating digits are $35$, thus you’d only want one set of $35$ on the left side.