I have this statement:
I) The equation: $|x - 5 | = 1$
I want to solve the equation, but I have a question.
When I have the absolute value of an incognita and a number, $ | x - 5 | $.
How do I proceed ? Should I eliminate the absolute value as a parenthesis, without changing signs? like: $x - 5 = 1$
Or I should leave it as: $x + 5 = 1$ ( changing all signs to positive )
Remember: $$|x| = \begin{cases} x, & \text{if } x \ge 0\\[0.2cm] -x, & \text{if } x < 0 \end{cases}$$
This is true more generally also, not just for $|x|$ but for things like $|x-5|$: $$ |x-5| = \begin{cases} x - 5, & \text{if } x-5 \ge 0 \\[0.2cm] -(x-5), & \text{if } x - 5 < 0 \end{cases} $$
This is why the equation $|x-5| = 1$ gets broken up into $x-5 = 1$ and $-(x-5) = 1$. Then both of these equations are solved individually.
Here is a slightly more complex way, if you are having trouble with the positive/negative, and want to completely avoid any cases.
In my opinion, it is also beneficial in equations like these:
$$|x+2|=|x-3|$$
In your question, square both sides to get rid of the absolute value.
$$|x-5|^2=1 \implies x^2-10x+25=1 \implies x^2-10x+24=0 \implies(x-12)(x+2)=0$$
From which you get the solutions $x=12, -2$.
You just have to remember that, for any $b\ge 0$, $$|x-a|=b\iff x-a=\pm b\iff x=a\pm b,$$ and similarly, for inequalities, $$|x-a| < b\iff -b < x-a < b\iff a-b < x < a+b.$$
