Finding the solution set for an absolute value equation algebraically

absolute valuealgebra-precalculus

e.g. the equation $|x-2|=3$, geometrically it's easy to solve drawing the number line. I get the solution set {-1,5} which I understand.

but I was looking at some notes I found online of how to solve this algebraically but I don't understand the logic of this method.

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OK, I see that they have put the condition $if$ $x-2 \geq 0$ but why ? and also they multiplied $x-2$ by $-1$ and then come out with the condition $if$ $ x-2<0.$ Why ?

After that if $ x \geq 2$ then $3 = x-2$ which gives us the result of $5$ and the same thing happens if $x<2$ the $3=x-2$ which in turns gives -1.

I'm just trying understand the logic of this method in relations to absolute value and distance, any help would be greatly appreciated.

Best Answer

If $a\ge 0$ then what is $|a|$? It is $a$.

And if $a < 0$ then what is $|a|$? It is "the positive absolute value that is the size of $a$" (more or less). But what is that in terms of $a$? As $a =(1)\cdot a $ is negative but $-a = (-1)\cdot a$ is ... positive. So $|a| = -a$ if $a < 0$.

Because.... if $a < 0$ then $0 < -a$ and $-a$ is the positive value so

Definition: The absolute value of $a$ which is written as $|a|$ is a non-negative real number so that $|a|=\begin{cases}a& \text{if }a \ge 0\\-a&\text{if }a < 0\end{cases}$.

(In either case $|a| \ge 0$.)

With that in mind the logic is obvious.

...

$|x-2| =3$.

There are two possibilities: either $x-2 \ge 0$ or .... it isn't .....

If $x-2 \ge 0$ then $|x-2| =x-2$ and $|x-2| =3$ means $x-2 = 3$ which means $x=5$.

And if $x-2$ is not greater of equal to $0$ then $x-2 < 0$ and if $x-2 < 0$, then $|x-2| = -(x-2) = 2-x$. And therefore $|x-2|=3$ means $|x-2| = 2-x =3$ and so $x=-1$.

It's that simple and obvious.

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