When do we specify restrictions in rational expressions or equations In the following equation $x \neq 0$ apply's according to wolframalpha (expanded form).
$$\frac{x^3 + x^2}{2x} = \frac{x^2 + x}{2}$$
Yet in the following equation it does not...
$$\frac{ab^2}{b} = 5 \iff ab = 5$$
Why don't we specify the restriction $b \neq 0$ here? And when do we specify restrictions in general?
 A: I believe it must be implicitly understood that $b\ne0$ since $ab=5$ has no solutions when $b=0$.  Consider instead the equation $\frac{ab}b=5$.  In this case, the solution is indeed $a=5,b\ne0$.
A: or $\frac{ab^2}{b}-5b=0$ is equivalent to $b(ab-5)=0$ the solution $b=0$ is senceless and we obtain $ab=5$.
And $\frac{x^3+x^2}{2x}-\frac{x^2+x}{2}=0$ if $x\ne 0$ sometimes a CAS ignore this case.
A: Note that the expression $e(x)=\frac {x^2}x$ is strictly undefined for $x=0$. But defining $e(0)=0$ we recover $e(x)=x$ for all $x$ rather than just for $x\neq 0$ and this value is required to maintain continuity.
In the second case, you could say that $b\neq 0$ is implicit in the equation, because $ab=5$ implies $b\neq 0$.
The first case is an identity except when $x=0$, so this needs to be stated, even though the gap can be repaired.
A: I think that
$$\dfrac{x^3 + x^2}{2x} = \dfrac{x^2 + x}{2} \Leftrightarrow 
\begin{cases}
2x\cdot \dfrac{x^3 + x^2}{2x}=2x\cdot \dfrac{x^2 + x}{2}\\
2x\neq 0
\end{cases}$$
Also $$\dfrac{ab^2}{b} = 5 \iff 
\begin{cases}
b\cdot\dfrac{ab^2}{b}=b\cdot 5\\
b\neq 0 \end{cases}\iff
\begin{cases}
ab^2=5b\\
b\neq 0 \end{cases}\iff\begin{cases}
 b\cdot (ab-5)=0\\
b\neq 0\end{cases}\iff ab=5$$
I a general way: $$\dfrac{A}{B}=C\iff \begin{cases}
B\cdot \dfrac{A}{B}=B\cdot C\\
B\neq 0 \end{cases}$$
