Sure, let's solve the equation [tex]\(| |x-1| - 2 | = |x - 3|\)[/tex] step-by-step.
### Step 1: Understanding Absolute Values
The equation involves absolute values, which means we need to consider the different cases where the expressions inside the absolute values change from positive to negative and vice versa.
### Step 2: Analyze Inner Absolute Value
First, let’s break down the inner absolute portions individually:
1. [tex]\(|x - 1| - 2\)[/tex]
2. [tex]\(|x - 3|\)[/tex]
We need to consider different scenarios for [tex]\(x\)[/tex] that affect these expressions:
1. [tex]\(x - 1 \geq 0\)[/tex] : [tex]\(x \geq 1\)[/tex]
2. [tex]\(x - 1 < 0\)[/tex] : [tex]\(x < 1\)[/tex]
3. [tex]\(x - 3 \geq 0\)[/tex] : [tex]\(x \geq 3\)[/tex]
4. [tex]\(x - 3 < 0\)[/tex] : [tex]\(x < 3\)[/tex]
### Step 3: Break Down the Cases
We'll evaluate the absolute value equation in piecewise sections based on these intervals. Here we will include necessary intervals:
#### Case 1: [tex]\(x \geq 3\)[/tex]
For [tex]\(x \geq 3\)[/tex]:
- [tex]\(|x - 1| = x - 1\)[/tex] since [tex]\(x - 1 \geq 0\)[/tex]
- So, [tex]\(|x - 1| - 2 = (x - 1) - 2 = x - 3\)[/tex]
- Also, [tex]\(|x - 3| = x - 3\)[/tex]
The equation becomes:
[tex]\[|x - 3| = |x - 3|\][/tex]
This is true for all [tex]\(x \geq 3\)[/tex]. Thus, all points in this interval satisfy the equation. Hence, one solution set is:
[tex]\[x \geq 3\][/tex]
#### Case 2: [tex]\(1 \leq x < 3\)[/tex]
For [tex]\(1 \leq x < 3\)[/tex]:
- [tex]\(|x - 1| = x - 1\)[/tex] since [tex]\(x - 1 \geq 0\)[/tex]
- So, [tex]\(|x - 1| - 2 = (x - 1) - 2 = x - 3\)[/tex]
- Also, [tex]\(|x - 3| = 3 - x\)[/tex] since [tex]\(x - 3 < 0\)[/tex]
The equation becomes:
[tex]\[|x - 3| = |3 - x|\][/tex]
Specifically:
[tex]\[|x - 3| = 3 - x = x - 3\][/tex]
If we solve this:
[tex]\[3 - x = x - 3\][/tex]
[tex]\[3 + 3 = x + x\][/tex]
[tex]\[6 = 2x\][/tex]
[tex]\[x = 3\][/tex]
This [tex]\(x = 3\)[/tex] falls under both intervals 1 ≤ x < 3, therefore this solution also holds. Therefore, another solution is:
[tex]\[x = 3\][/tex]
(Note that since [tex]\(3 \geq 3\)[/tex], this overlaps with the previous solution).
#### Case 3: [tex]\(x < 1\)[/tex]
For [tex]\(x < 1\)[/tex]:
- [tex]\(|x - 1| = 1 - x\)[/tex] since [tex]\(x - 1 < 0\)[/tex]
- So, [tex]\(|x - 1| - 2 = (1 - x) - 2 = -x - 1\)[/tex]
- [tex]\(|x - 3| = 3 - x\)[/tex] since [tex]\(x - 3 < 0\)[/tex]
The equation becomes:
[tex]\[|- x - 1| = |3 - x|\][/tex]
Specifically:
[tex]\[-(x + 1) = 3 - x\][/tex]
Solving this:
[tex]\[- (x + 1) = 3 - x\][/tex]
[tex]\[-x - 1 = 3 - x\][/tex]
This doesn't cancel out except for extraneous [tex]\(x\)[/tex]. Therefore, no valid [tex]\(x\)[/tex] is extracted.
### Final Answer
The solutions for the equation are:
[tex]\[x \geq 3.\][/tex]
Upon consolidating the acceptable ranges, we can conclude that [tex]\(x \geq 3\)[/tex]. Therefore, the full set of solutions for the equation [tex]\(||x-1|-2|=|x-3|\)[/tex] is [tex]\(x \geq 3\)[/tex], uncommon for complex algebraic absolute scenarios.
