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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 )

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  • $\begingroup$ Use that $|a|=b \ge 0 \iff a = \pm b\,$. $\endgroup$ – dxiv Mar 10 '18 at 21:18
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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.

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  • $\begingroup$ Perfect and clear, many thanks $\endgroup$ – Mattiu Mar 10 '18 at 21:27
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there are two cases: $$x-5=1$$ or $$-(x-5)=1$$

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  • $\begingroup$ Thanks Sonnhard $\endgroup$ – Mattiu Mar 10 '18 at 21:22
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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 $x=12, -2$.

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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, $$|x-a| < b\iff -b < x-a < b\iff a-b < x < a+b.$$

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