Why does (1.3)(.7) != 1? This is the context: 
1.3(10) = 13
13 - 10 = 3
|3| / 10 = 0.3 

0.7(10) = 7
7 - 10 = -3
|-3| / 10 = 0.3


(1.3)(0.7) != 1

If the difference between both real numbers is 0.3, when multiplied together, why does the number < 1 hold greater influence over the product (i.e. 1.3*0.7 = 0.91) ? 
 A: Here is $0.7 × 1.3$:

And here it is again, with the green rectangle in a different place:

A: Note that $$1.3\cdot 0.7=1\cdot 0.7+0.3\cdot 0.7$$ is the sum of $0.7$ and $0.3$ of $0.7$.  That is, when we "add back in" the $0.3$ by multiplying by $1.3$, we're adding only $0.3$ times $0.7$, not $0.3$ times the original $1$ we started with.
Alternatively, if you think of it as $$1.3\cdot 0.7=1.3\cdot 1-1.3\cdot 0.3,$$ we're subtracting $0.3$ times $1.3$, not $0.3$ times our original $1$.  So we're subtracting more than we originally added to get $1.3$ from $1$, and so we end up smaller than $1$.
A: The two numbers you have stated simply have 1 as their mean. The same applies for this pair (1.1, 0.9) but the product of these yields 0.99 getting me closer to one. As the difference from 1 decreases, the closer I'll get to 1.
Multiplication is commutative and no number gets individual priority over another. I hope this clears your question? 
A mathematical analysis of this would be - 
In your example, the difference from 1 is 0.3. 
You're carrying out (1.3)*(0.7)
= (1+0.3)*(1-0.3) 
=(1)-(0.3)^2 
Therefore your final product will ALWAYS be less than 1 unless your difference was an imaginary number. 
A: If I sell you a shirt that is on sale for $30\%$ off but then I charge you a tax of $30\%$, the two percents don't cancel each other out. This is because they are charged sequentially (not at the same time). The tax is applied after the discount, and so the extra amount that is added back to the discounted price is reduced since there is less of a price to tax.
A: Perhaps you expect $1.3\times 0.7$ to be 1 because 1 times 1 is 1;  you removed 0.3 from one number and attached that to other, and expect the product to remain the same. 
Let me generalise the question
That is, are the products $A\times B$ and $(A-x)\times (B+x)$ the same?
The second product is $AB+(A-B)x-x^2$. For this to be equal to $AB$, we should have $(A-B)x=x^2$ which is the same as requiring $A-B= x$ Or $A-x=B$ (and hence $B+x=A$). So that in the new product the original two terms are interchanged,; only in that case the product will remain the same.
Imagine a long procession of soldiers marching on road with 5 soldiers side by side and 100 in each line. This makes it 500 soldiers. Next day if the 6 soldiers go in each line it does not mean the lines will all be 99 long.
