How did he simplify this problem like this? $$\left(\frac{x}{12}+\frac{x}{18}\right)t=x$$
$$\left(\frac{1}{12}+\frac{1}{18}\right)t=1$$
I want a solid rule, how did he simplify the problem in the image like this! 

If
$$\left(\frac{x}{12}+\frac{x}{18}\right)t=x\tag{1}$$
then in general it is not true that
$$\left(\frac{1}{12}+\frac{1}{18}\right)t=1\tag{2}$$
Counterexample: if $t=36$ and $x=0$, then you see that $(1)$ is equivalent to $0=0$ which is true, but $(2)$ is equivalent to $5=1$ which is not.
However, if you already know that $x\neq0$, then the implication is true since in that case
$$
\left(\frac{x}{12}+\frac{x}{18}\right)t=x
$$
is equivalent to (this is just a factorization)
$$
\left(\left(\frac{1}{12}+\frac{1}{18}\right)t\right)x=x
$$
and 
you can divide both sides (which are equal, after all!) by $x$ (since $x\neq0$... otherwise division by $0$ doesn't make sense!) to maintain equality and infer
$$
\frac{\left(\left(\frac{1}{12}+\frac{1}{18}\right)t\right)x}{x}=\frac{x}{x}
$$
which is the same as
$$
\left(\left(\frac{1}{12}+\frac{1}{18}\right)t\right)\frac{x}{x}=\frac{x}{x}
$$
or, since $\frac{x}{x}=1$ always,
$$
\left(\left(\frac{1}{12}+\frac{1}{18}\right)t\right)\cdot1=1
$$
that is
$$
\left(\frac{1}{12}+\frac{1}{18}\right)t=1
$$
Of course, at a certain level of mathematical maturity, all of these intermediate steps are very rarely all written out since they should be obvious.

Note that if you have
$$
\left(\frac{1}{12}+\frac{1}{18}\right)t=1\tag{3}
$$
then it must always be true that
$$
\left(\frac{x}{12}+\frac{x}{18}\right)t=x
$$
(just multiply both sides of $(3)$ by $x$... this will always be a valid operation, unlike division by an a priori arbitrary real number $x$ which is not valid if $x=0$).

In fact, a well-known false proof that $2=1$ makes use of an invalid division by $0$.
Suppose that $x=y\neq0$. Then
\begin{align}
x^2&=xy\\
x^2-y^2&=xy-y^2\\
(x+y)(x-y)&=y(x-y)\\
x+y&=y\\
2y&=y\\
2&=1
\end{align}
right? No! From
$$
(x+y)(x-y)=y(x-y)\tag{4}
$$
you can't infer that
$$
x+y=y
$$
because dividing both sides of $(4)$ by $(x-y)$ and simplifying doesn't make sense since $x-y=0$.
