multiplication and addition fractions Try to visualize process of multiplication fraction additio

Let us say you have 2 fractions you need to multiply, $\frac{2}{7}$ and $\frac{4}{5}$ for example.
Then the multiplication rule you explained above means that:
$\frac{2}{7}\times <!--\; \times \; -->\frac{4}{5}=\frac{2\times <!--\; \times \; -->4}{7\times <!--\; \times \; -->5}=\frac{8}{35}$
In general:
$\frac{a}{b}\times <!--\; \times \; -->\frac{c}{d}=\frac{ac}{bd}$
Now, to explain why this works, you need to consider what a fraction is or means. $\frac{2}{7}$ actually means $2÷<!--\; ÷\; -->7$.
So above, when I did $\frac{2}{7}\times <!--\; \times \; -->\frac{4}{5}$, I was actually doing the same thing as $(2÷<!--\; ÷\; -->7)\times <!--\; \times \; -->(4÷<!--\; ÷\; -->5)$
Now: visualise some cakes (or pizzas or pies). Let's say I have 3 cakes. If I multiply the number of cakes I have by 5, then I take 5 sets of 3 cakes to make 15 cakes. If I divide the number of cakes I have by 6, I am trying to split up the cakes into 6 groups of equal size. So I will end up with 6 halves.
Now notice what happens if I first multiply my 3 cakes by 5 and then divide by 6. I get the same answer as first dividing by 6 and then multiplying by 5.
So order does not matter when it comes to multiplication and division (Note you still have to pay attention to brackets though).
Therefore, we could rearrange what we had above:
$\frac{2}{7}\times <!--\; \times \; -->\frac{4}{5}=(2÷<!--\; ÷\; -->7)\times <!--\; \times \; -->(4÷<!--\; ÷\; -->5)=2÷<!--\; ÷\; -->7\times <!--\; \times \; -->4÷<!--\; ÷\; -->5=2\times <!--\; \times \; -->4÷<!--\; ÷\; -->7÷<!--\; ÷\; -->5=(2\times <!--\; \times \; -->4)÷<!--\; ÷\; -->(7\times <!--\; \times \; -->5)=\frac{2\times <!--\; \times \; -->4}{7\times <!--\; \times \; -->5}$
Another helpful way of understanding what I said above is to think about the analogy between addition/subtraction and multiplication/division.
What I said above is still true if I replace all of the + and − with × and ÷ respectively (ignoring all of the fractions, as there isn't a "fraction" sign for subtraction). The way in which these 2 pairs are related are very similar.
If I asked your original question, but with addition/subtraction instead, then it would have been something like:
Show me why this rule works:
$(5-<!--\; -\; -->3)+(4-<!--\; -\; -->9)=(5+4)-<!--\; -\; -->(3+9)$
$(a-<!--\; -\; -->b)+(c-<!--\; -\; -->d)=(a+c)-<!--\; -\; -->(b+d)$
It is because you can change around the order of things without changing the meaning. In mathematics, this means that an operation is "commutative".
Yet another way of understanding this is to consider the idea of "inverses". An inverse operation undoes an operation. So for example, if I add 5 to a number, the inverse operation to this would be subtracting 5 from the new number, as this gets me back to my original number.
Subtraction is the inverse operation of addition. Division is the inverse operation of multiplication.
With inverses, you can change around the order of which you do things. This is actually related to basic Group Theory. Group theory replaces the operations $+$ $-<!--\; -\; -->$ $\times <!--\; \times \; -->$ $÷<!--\; ÷\; -->$ with other symbols like $\ast <!--\; \ast \; -->$ to prove generic properties about these operations without it being necessary for you to know what that particular operation is.
I hope this helps!

I'm unsure if this helps, but:
$\frac{a}{b}=a\cdot <!--\; \cdot \; -->\frac{1}{b}$
$\frac{a}{b}\cdot <!--\; \cdot \; -->\frac{c}{d}=a\cdot <!--\; \cdot \; -->\frac{1}{b}\cdot <!--\; \cdot \; -->c\cdot <!--\; \cdot \; -->\frac{1}{d}$
$\begin{array}{}\text{(1)}& =ac\cdot <!--\; \cdot \; -->\frac{1}{b}\cdot <!--\; \cdot \; -->\frac{1}{d}\end{array}$
$\begin{array}{}\text{(2)}& =ac\cdot <!--\; \cdot \; -->\frac{1}{bd}\end{array}$
$=\frac{ac}{bd}$
Now the problem is really getting from (1) to (2), visually.
You can think of it as if you have a slice of pizza, $\frac{1}{b}$ out of the entire pizza. You cut that slice into $d$ pieces and now you have $\frac{1}{bd}$ of the whole slice.