The number of real values of x satisfying the equation `3 |x-2| +|1-5x|+4|3x+1|=13` is:

Options

A. 1
B. 4
C. 2
D. 3

Answer

To solve the equation \(3 |x-2| + |1-5x| + 4|3x+1| = 13\), we will analyze different intervals based on the critical points where the expressions inside the absolute values change sign. The critical points are \(x = 2\), \(x = \frac{1}{5}\), and \(x = -\frac{1}{3}\).

### Step 1: Identify the intervals
The critical points divide the real line into the following intervals:
1. \(x < -\frac{1}{3}\)
2. \(-\frac{1}{3} \leq x < \frac{1}{5}\)
3. \(\frac{1}{5} \leq x < 2\)
4. \(x \geq 2\)

### Step 2: Solve for each interval

#### Interval 1: \(x < -\frac{1}{3}\)
In this interval:
- \( |x-2| = 2-x \)
- \( |1-5x| = 1-5x \)
- \( |3x+1| = -3x-1 \)

Substituting these into the equation:
\[ 3(2-x) + (1-5x) + 4(-3x-1) = 13 \]
This simplifies to:
\[ 6 - 3x + 1 - 5x - 12x - 4 = 13 \]
Combining like terms:
\[ -20x + 3 = 13 \]
Solving for \(x\):
\[ -20x = 10 \implies x = -\frac{1}{2} \]
Since \(-\frac{1}{2} < -\frac{1}{3}\), this solution is valid.

#### Interval 2: \(-\frac{1}{3} \leq x < \frac{1}{5}\)
In this interval:
- \( |x-2| = 2-x \)
- \( |1-5x| = 1-5x \)
- \( |3x+1| = 3x+1 \)

Substituting these into the equation:
\[ 3(2-x) + (1-5x) + 4(3x+1) = 13 \]
This simplifies to:
\[ 6 - 3x + 1 - 5x + 12x + 4 = 13 \]
Combining like terms:
\[ 4x + 11 = 13 \]
Solving for \(x\):
\[ 4x = 2 \implies x = \frac{1}{2} \]
Since \(\frac{1}{2} > \frac{1}{5}\), this solution is invalid.

#### Interval 3: \(\frac{1}{5} \leq x < 2\)
In this interval:
- \( |x-2| = 2-x \)
- \( |1-5x| = 5x-1 \)
- \( |3x+1| = 3x+1 \)

Substituting these into the equation:
\[ 3(2-x) + (5x-1) + 4(3x+1) = 13 \]
This simplifies to:
\[ 6 - 3x + 5x - 1 + 12x + 4 = 13 \]
Combining like terms:
\[ 14x + 9 = 13 \]
Solving for \(x\):
\[ 14x = 4 \implies x = \frac{2}{7} \]
Since \(\frac{2}{7} < \frac{1}{2}\), this solution is valid.

#### Interval 4: \(x \geq 2\)
In this interval:
- \( |x-2| = x-2 \)
- \( |1-5x| = 5x-1 \)
- \( |3x+1| = 3x+1 \)

Substituting these into the equation:
\[ 3(x-2) + (5x-1) + 4(3x+1) = 13 \]
This simplifies to:
\[ 3x - 6 + 5x - 1 + 12x + 4 = 13 \]
Combining like terms:
\[ 20x - 3 = 13 \]
Solving for \(x\):
\[ 20x = 16 \implies x = \frac{4}{5} \]
Since \(\frac{4}{5} < 2\), this solution is invalid.

### Conclusion
The valid solutions we found are:
1. \(x = -\frac{1}{2}\)
2. \(x = \frac{2}{7}\)

Thus, the number of real values of \(x\) satisfying the equation is **2**.