Solve the following inequailities :
`abs(x-1)+2abs(x+1)+abs(x-2)le8`

Answer

To solve the inequality \( |x - 1| + 2|x + 1| + |x - 2| \leq 8 \), we will break it down into different cases based on the critical points where the expressions inside the absolute values change sign. The critical points here are \( x = -1, 1, \) and \( 2 \).

### Step 1: Identify the intervals
The critical points divide the number line into the following intervals:
1. \( (-\infty, -1] \)
2. \( (-1, 1] \)
3. \( (1, 2] \)
4. \( (2, \infty) \)

### Step 2: Solve the inequality in each interval

#### Case 1: \( x \leq -1 \)
In this interval:
- \( |x - 1| = -(x - 1) = -x + 1 \)
- \( |x + 1| = -(x + 1) = -x - 1 \)
- \( |x - 2| = -(x - 2) = -x + 2 \)

Substituting these into the inequality:
\[ -x + 1 + 2(-x - 1) + (-x + 2) \leq 8 \]
Simplifying:
\[ -x + 1 - 2x - 2 - x + 2 \leq 8 \]
\[ -4x + 1 \leq 8 \]
\[ -4x \leq 7 \quad \Rightarrow \quad x \geq -\frac{7}{4} \]
Since we are in the interval \( (-\infty, -1] \), we take the intersection:
\[ -\frac{7}{4} \leq x \leq -1 \]

#### Case 2: \( -1 < x \leq 1 \)
In this interval:
- \( |x - 1| = -(x - 1) = -x + 1 \)
- \( |x + 1| = x + 1 \)
- \( |x - 2| = -(x - 2) = -x + 2 \)

Substituting these into the inequality:
\[ -x + 1 + 2(x + 1) + (-x + 2) \leq 8 \]
Simplifying:
\[ -x + 1 + 2x + 2 - x + 2 \leq 8 \]
\[ 0x + 5 \leq 8 \]
This is always true, so:
\[ -1 < x \leq 1 \]

#### Case 3: \( 1 < x \leq 2 \)
In this interval:
- \( |x - 1| = x - 1 \)
- \( |x + 1| = x + 1 \)
- \( |x - 2| = -(x - 2) = -x + 2 \)

Substituting these into the inequality:
\[ (x - 1) + 2(x + 1) + (-x + 2) \leq 8 \]
Simplifying:
\[ x - 1 + 2x + 2 - x + 2 \leq 8 \]
\[ 2x + 3 \leq 8 \]
\[ 2x \leq 5 \quad \Rightarrow \quad x \leq \frac{5}{2} \]
Since we are in the interval \( (1, 2] \), we take:
\[ 1 < x \leq 2 \]

#### Case 4: \( x > 2 \)
In this interval:
- \( |x - 1| = x - 1 \)
- \( |x + 1| = x + 1 \)
- \( |x - 2| = x - 2 \)

Substituting these into the inequality:
\[ (x - 1) + 2(x + 1) + (x - 2) \leq 8 \]
Simplifying:
\[ x - 1 + 2x + 2 + x - 2 \leq 8 \]
\[ 4x - 1 \leq 8 \]
\[ 4x \leq 9 \quad \Rightarrow \quad x \leq \frac{9}{4} \]
Since we are in the interval \( (2, \infty) \), we take:
\[ x \leq \frac{9}{4} \]

### Step 3: Combine the solutions
Now we combine the results from all intervals:
1. From Case 1: \( -\frac{7}{4} \leq x \leq -1 \)
2. From Case 2: \( -1 < x \leq 1 \)
3. From Case 3: \( 1 < x \leq 2 \)
4. From Case 4: \( 2 < x \leq \frac{9}{4} \)

The final solution is:
\[ x \in \left[-\frac{7}{4}, 2\right] \cup \left(2, \frac{9}{4}\right] \]