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)`.