# Finding the solution set for an absolute value equation algebraically

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.

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.

• "But why?": because $\lvert a\rvert=\begin{cases}a&\text{if }a\ge0\\ -a&\text{if }a<0\end{cases}$.
– user239203
Commented Mar 20, 2020 at 20:55
• How would you explain to a child that $|7| =7$ and $|-7| = 7$. And if $a = -7$ then $|a| = ....$ what. Does it equal $a = -7$? No it equals $7$ where is .. what.... compared to $a$. (Hint: $-(-7) = 7$.) Commented Mar 21, 2020 at 0:08

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.

Different approaches possible. One approach is more of a basic geometric approach: To define the absolute value of a number as the distance from that number to $$0$$ on the number line. So $$|8|=8$$ because $$8$$ is eight units away from $$0$$. And for exactly that reason we also have $$|-8|=8$$, because minus eight is also $$8$$ units away from zero. With this interpretation, we can solve your absolute value equation, if we for a moment consider $$|z|=3$$. From the definition it immediately follows that we have $$z=3$$ and $$z=-3$$. Now with $$z=x-2$$ you find the values of $$x$$

• the op said s/he understood it geometrically. S/he wanted the algebraic method explained. Commented Mar 21, 2020 at 0:19

Let

$$f\left(x\right)=\left|x-2\right|$$,

which is to say:

$$f\left(x\right)=\sqrt{\left(x-2\right)^{2}}$$.

Because $$u^{2}\ge0$$ for any real $$u$$, and $$\sqrt{u}\ge0$$ everywhere it's defined:

$$\left|u\right|\ge0$$

whether $$u$$ is positive or negative. Therefore, if

$$x-2\ge0$$,

the squaring and square root operations simply cancel to give:

$$f(x)=x-2$$.

However, if

$$x-2<0$$,

it must be made positive, necessitating:

$$f(x)=-(x-2)=2-x$$.

The reason they change it to a piecewise is because the resulting functions are much easier to work with and visualize in many cases.

Set : number $$A = x-2$$.

Saying that $$|A| = 3$$ means that the distance of number $$A$$ from $$0$$ is $$3$$.

This implies that : either number $$A = +3$$ or number $$A = -3$$. ( Only $$+3$$ and $$-3$$ are at a $$3$$-unit distance from $$0$$).

Hence two cases ( substituting $$x-2$$ for A) : (1) $$x-2 = 3$$ OR (2) $$x-2 = -3$$.

( More algebraically : $$(|A| = 3) \equiv (A=3 \lor -A = 3) \equiv (A=3 \lor A=-3)$$,

with symbol $$\equiv$$ meaning " is equivalent to" and symbol $$\lor$$ meaning " or")

In case (1) $$x-2 +(2) = x = 3+(2) = 5$$.

In case (2) $$x-2 +(2) =x = -3 +(2) = -1$$.

So, the solution set is $$S =$$ {$$-1, 5$$}