### Lesson Objectives:

- Solve absolute value equation.- Solve absolute value inequality.

The absolute value of a number is its distance from 0 on a number line.

So for any positive value a, the `abs(x) = a` means that `x = a` or `x = -a`.

Solve `abs(x-1) + 4 = 10`.

So start by subtracting 4 from both sides. This gives us `abs(x-1) = 6`.

So `x-1 = 6` or `x-1 = -6`. So if we add 1 to both sides, we have `x = 7` or `x = -5`.

So let's check `x = 7`. `abs(7-1)+4 = 10`. That means `6+4 = 10` or `10 = 10`. So, this is a true statement.

Now let's check `x = -5`. We get `abs(-5-1)+4 = 10`, which is `6+4 = 10` or `10 = 10`. Another true statement.

So the solutions are 7 and -5.

Absolute Value Inequalities.

If `a > 0` and `abs(x) < a`, then `-a < x < a`.

If `a > 0` and `abs(x) > a`, then `x < -a` or `x > a`.

Similarly, if `a > 0` and `abs(x) <= a`, then `-a <= x <= a`.

And if `a > 0` and `abs(x) >= a`, then `x <= -a` or `x >= a`.

Solve and give your solution set in interval notation and then graph the solution set.

`abs(x+12) >= 15`. This means that `x+12 >= 15` or `x+12 <= -15`. So if we subtract 12 from both sides, we get `x >= 3` or if we subtract 12 over here, we get `x <= -27`.

We can graph the solution set by having a closed circle at -27 and a closed circle at 3. And `x <= -27`, so we shade to the left, and `>= +3`, so we shade to the right.

In interval notation, this is `(-oo, -27] cup [3, oo)`.