Solving an equation has restricted values for being undefined but is not applicable to all forms For example, in the case of this algebraic expression
$${7x^2+14x\over2x+4}$$
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.
$${7x^2+14x\over2x+4} = {7x(x+2)\over2(x+2)} = {7x\over2}$$
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?
 A: 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.
A: The two expressions are equivalent for $x\neq -2$ that is
$${7x^2+14x\over2x+4} = {7x\over2}$$
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
$${7x^2+14x\over2x+4} = -7, \quad x=-2$$
the two expression become completely equivalent (in that case we define it a removable discontinuity).
A: 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.
