Absolute value

The best way to try to understand the absolute-value is to look at the number line.
If we take any number on the number line, its distance from zero is its absolute value.

The absolute value of $x$ can be seen as follows:
For $x \geq 0$ :

$\left |x \right |=x$

For $x<0$ :

$\left|x\right |=-x$

Properties of absolute values:

$\left |\frac{x}{y} \right|=\frac{\left|x\right|}{\left |y\right |}$ as long as $y\neq 0$ .

Finally
$\left|-x\right|=\left|x\right|$

As an example, if we pick $-3$ and $3$ from the number line, they are equally away from the $0$ . The distance is 3 units.

Another fact is the distance between 2 points in the number line. This is simply the absolute value of their difference.

For points $4$ and $-2$ , we count a total of $6$ units.

$Dist=\left|4-(-2)\right|=\left|6\right|$ .

Practice:

When solving equations containing absolute values we should remember that:

For a number $n>0$ , the equation $\left|x\right|=n$ will have a solution of $-n$ and $n$ .

To understand this, we look at our number line example. The distance 3 units can be counted to left or to the right.

$\left|x\right|=n$ with $n<0$ . It’s a trap.

Absolute values are always positive.

Don’t waste time when a complete absolute value is assigned a number on the left of $0$ , meaning negative. Just answer: $NO\; SOLUTION$

Example: Solve for $x$ :

$\left|x-9\right|=-3$

No solution here. We just answer: NO SOLUTION

Now solve the following for $x$ :

$\left|x-9\right|=3$

Remember that the value inside the absolute value can be either positive or negative.

That is the beauty of these equations. We always check the two cases:

1. $x-9=3$ solving yields $x=12$
2. $-(x-9)=3$ or $x-9=-3$ The solution is the same $x=6$
Each solution satisfies our equation 3 units away from $0$ .
If you understand this, you’ll be set for bigger equations.

Now let’s solve for $x$ :

$\left|3x-4\right|=17$

Case 1:

$3x-4=17$
$3x-4+4=17+4$
$3x=21$
$\frac{3x}{3}=\frac{21}{3}$
$x=7$

Case 2:

$3x-4=-17$
$3x-4+4=-17+4$
$3x=-13$
$\frac{3x}{3}=\frac{-13}{3}$
$x= -\frac{13}{3}$

Our final example is more challenging:

Solve for $x$

$\left|\frac{6-8x}{5}\right|=\left|\frac{7+3x}{2}\right|$

We normally consider each case where we have the quantity inside the absolute value is positive or negative.

This is 4 cases right? Wrong. Each two cases make themselves one case. We are left with the two following cases:

1. Both sides inside of the absolute values are positive. Please note that the case when both are negative will give the same result.
2. One side inside of absolute value positive, the other side negative. Either way we get the same results. Now let’s solve for $x$

CASE 1:

$\frac{6-8x}{5}=\frac{7+3x}{2}$