Solve : `|x|+2|x-1|-3|x-2|=4 , x in R `

Answer

To solve the equation \( |x| + 2|x-1| - 3|x-2| = 4 \), we will analyze it by considering different cases based on the critical points where the expressions inside the absolute values change their signs. The critical points are \( x = 0 \), \( x = 1 \), and \( x = 2 \).

### Step 1: Identify Cases
We will break the problem into four cases based on the intervals defined by the critical points.

1. **Case 1**: \( x \geq 2 \)
2. **Case 2**: \( 1 \leq x < 2 \)
3. **Case 3**: \( 0 \leq x < 1 \)
4. **Case 4**: \( x < 0 \)

### Step 2: Solve Each Case

#### Case 1: \( x \geq 2 \)
In this case, all absolute values are positive:
\[ |x| = x, \quad |x-1| = x-1, \quad |x-2| = x-2 \]
Substituting these into the equation:
\[ x + 2(x-1) - 3(x-2) = 4 \]
Simplifying:
\[ x + 2x - 2 - 3x + 6 = 4 \]
\[ 0 + 4 = 4 \]
This is true for all \( x \geq 2 \). Thus, the solution for this case is:
\[ x \in [2, \infty) \]

#### Case 2: \( 1 \leq x < 2 \)
In this case, \( |x| = x \), \( |x-1| = x-1 \), and \( |x-2| = 2-x \):
\[ x + 2(x-1) - 3(2-x) = 4 \]
Simplifying:
\[ x + 2x - 2 - 6 + 3x = 4 \]
\[ 6x - 8 = 4 \]
\[ 6x = 12 \implies x = 2 \]
Since \( x = 2 \) is at the boundary of this case, it is included in the solution.

#### Case 3: \( 0 \leq x < 1 \)
In this case, \( |x| = x \), \( |x-1| = 1-x \), and \( |x-2| = 2-x \):
\[ x + 2(1-x) - 3(2-x) = 4 \]
Simplifying:
\[ x + 2 - 2x - 6 + 3x = 4 \]
\[ 2x - 4 = 4 \]
\[ 2x = 8 \implies x = 4 \]
Since \( x = 4 \) is not in the interval \( [0, 1) \), there is no solution from this case.

#### Case 4: \( x < 0 \)
In this case, all absolute values are negative:
\[ |x| = -x, \quad |x-1| = -x + 1, \quad |x-2| = -x + 2 \]
Substituting these into the equation:
\[ -x + 2(-x + 1) - 3(-x + 2) = 4 \]
Simplifying:
\[ -x - 2x + 2 + 3x - 6 = 4 \]
\[ 0 - 4 = 4 \]
This is not true, so there are no solutions from this case.

### Final Solution
Combining the solutions from Case 1 and Case 2, we find:
\[ \text{The solution is } x \in [2, \infty) \]