Is the equation before applying additive and multiplicative properties the same equation as the equation after application? When solving equations, one can use the additive and multiplicative properties of equality to transform a true equation, but what is the relationship between the original equation and the new equation?
 A: This is an excellent question. To answer it clearly, we need to focus on the different senses of "implication".
Suppose it holds that $a=b$ (where $a$ and $b$ are the two sides of the equation, whatever they may be). Then clearly, for any function $f$, it holds that $f(a)=f(b)$. We gave $f$ the same input, so we must get the same output. That is, $$\text{If}\,a=b,\,\text{then}\,f(a)=f(b).$$
So any solution of the original equation will also be a solution of the new equation.
On the other hand, it is not necessarily true that any solution of the new equation will be a solution of the old one. For example, say we want to solve $$x+5=8.$$ We decide on the (unorthodox) course of action of multiplying both sides by $0$ to get $$0=0.$$ That equation is clearly true for all values of $x$, as $x$ does not even appear. But the original equation is emphatically not true for all $x$.

So what you are getting at is: how do we know when we have "preserved" the equation? You ask about addition and multiplication, but what about other operations (squaring, taking roots, dividing, etc.)? You need to determine when the implication goes both ways. So in the case of addition, $$a=b\;\;\text{implies}\;\;a+x=b+x$$ and $$a+x=b+x\;\;\text{implies}\;\;a=b$$ regardless of what $x$ is. Hence addition always preserves the solutions to the equation.
For multiplication, $$a=b\;\;\text{implies}\;\;a\cdot x=b\cdot x$$ but $$a\cdot x=b\cdot x\;\;\text{implies}\;\;a=b$$ only when $x$ is not equal to $0$ (as illustrated above).
For other cases, the considerations are similar. $$a^2=b^2\;\;\text{implies}\;\;a=b$$
only when we are told that $a$ and $b$ have the same sign. Otherwise the best we can deduce is that $a=\pm b$.
Hope that helps!
A: The relationship is that the set of solutions to one is the set of solutions to the other.
A: What you're looking for is a logic statement, namely the implication.
Let's say that you have an equation of the form $f_1(x)=g_1(x)$ which is true only for certain values of $x$. Define $P(x)$ to be the boolean function that is true when $f_1(x)=g_1(x)$ is true and is false when $f_1(x)=g_1(x)$ is false. Let's further say that by manipulating $f_1(x)=g_1(x)$, you can demonstrate that $f_2(x)=g_2(x)$. Define $Q(x)$ to be the boolean function that represents the truth value of this equation.
You know that evaluated at a certain $x$, if $P(x)$ is true, then $Q(x)$ has to be true, because you cannot manipulate $f_1(x)=g_1(x)$ in such a way that $f_2(x)=g_2(x)$ is false because adding to and multiplying both sides preserves equality.
If at a certain $x$, however, $P(x)$ is false, you know nothing about $Q(x)$, because your $f_2(x)=g_2(x)$ can just as easily be as true as it could be false - take, for example, what happens when you multiply both sides of $1=2$ by $0$.
Thus you have that $P(x)\rightarrow Q(x)$. This tells you absolutely nothing about the relationship $Q(x)\rightarrow P(x)$, which is why you have to check the solution set of $Q(x)$ against $P(x)$.
