Cfrgtkky

For example, in the case of this algebraic expression

$${7x^2+14xover2x+4}$$

It is provided that $x ne -2$, since division by 0 is not defined. It is understandable that replacing $-2$ will put forth an undefined solution. However on further simplification i.e.

$${7x^2+14xover2x+4} = {7x(x+2)over2(x+2)} = {7xover2}$$

we can also include $-2$ as a solvable value for $x$. My confusion here is, if $x$ cannot be $-2$, how is it possible that it can be used in a reduced form, shouldn't it be not usable in all forms of the algebraic expression?

No, this is known as a removable discontinuity. By simplifying the original function, you reached a new function with that gap removed. Note that the original function and the simplified function are NOT the same because their domains aren't the same. As another example:
$$y = frac{x(x+2)}{x+2}; quad y = x$$

If you simplify the first function, you reach the second function. However, they aren't the same since the second function is defined for all $x in mathbb{R}$ whereas the first is defined for all $x neq -2$. If you graph them, they're the exact same, except the first has a gap at $x = -2$, while the second does not. (By simplifying, you removed that gap, or discontinuity at $x = -2$.) The same applies to your example.

The two expressions are equivalent for $xneq -2$ that is

$${7x^2+14xover2x+4} = {7xover2}$$

but for $x=-2$ the LHS is not defined.

In that case we can assign a value to the LHS also for $x=-2$, notably if we define

$${7x^2+14xover2x+4} = -7, quad x=-2$$

the two expression become completely equivalent (in that case we define it a removable discontinuity).

I would draw a very rough sketch...
Say,you have some functions like

$f_1(x)=x^2$

$f_2(x)=dfrac{3x}{13}$

$f_3(x)=sqrt{x-1}$

now multiply $(x+2)$ with both the numerator and the denominator of those function .
Now they appear to be -

$f_1(x)=dfrac{x^2(x+2)}{(x+2)}$

$f_2(x)=dfrac{3x.(x+2)}{13(x+2)}$

$f_3(x)=dfrac{sqrt{x-1}.(x+2)}{13(x+2)}$

Now what do you say,each of the function is not valid at $x=-2$ !!!!
does it make any sense?Clearly this $x=-2$ point has nothing to do with those functions. This is purely irrelevant .so we must have to remove them.like this,in every function at first we should look for a removable discontinuity.if it exists then we must have to remove it to get the actual domain of the function.