When you divide, you are implicitly assuming that the number you are dividing by is not equal to zero. By dividing, you are excluding the possibility that the number in question is zero, and as such you may be eliminating correct answers.
For a very simple example, consider the case of the equation $x^2-x=0$.
There are two answers: $x=0$, and $x=1$. However, if you "divide by the variable", you can end up doing this:
$$\begin{align*}
x^2 - x & = 0\\
x^2 &= x &&\text{(adding }x\text{ to both sides)}\\
\frac{x^2}{x} &= \frac{x}{x} &&\text{(divide by }x\text{, which assumes }x\neq 0)\\
x &= 1.
\end{align*}$$
So you "lost" the solution $x=0$, because when you divided by $x$, you implicitly were saying "and $x\neq 0$". In order to "recover" this solution, you would have to consider "What happens if what I divided by is equal to $0$?"
For a more extreme example, consider something like
$$(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)=0.$$
Since a product is equal to $0$ if and only if one of the factors is equal to $0$, there are six solutions to this equation: $x=1$, $x=2$, $x=3$, $x=4$, $x=5$, and $x=6$. Divide both sides by $x-1$, and you lose the solution $x=1$; divide both sides by $x-2$, you lose $x=2$. Continue this way until you are left with $x-6=0$, and you lost five of the six solutions. And if then you go ahead and divide by $x-6$, you get $1=0$, which has no solutions at all!
Whenever you divide by something, you are asserting that something is not zero; but if setting it equal to $0$ gives a solution to the original equation, you will be excluding that solution from consideration, and so "eliminate" that answer from your final tally.
Best Answer
If you have 2x - 1 = -x + 1, you can simply add +x and +1 to both sides. This gives you 2x - 1 + x + 1 = -x + 1 + x + 1, which leads to 3x = 2 and therefore x = 2/3.