Least positive inegral value of `x` satisfying `|4x + 3| + |3x - 4| = |7x - 1|` is

Explanation

To solve the equation \( |4x + 3| + |3x - 4| = |7x - 1| \), we will analyze the absolute value expressions by considering different cases based on the critical points where each expression inside the absolute values changes sign.

### Step 1: Identify critical points
The expressions inside the absolute values change sign at:
- \( 4x + 3 = 0 \) → \( x = -\frac{3}{4} \)
- \( 3x - 4 = 0 \) → \( x = \frac{4}{3} \)
- \( 7x - 1 = 0 \) → \( x = \frac{1}{7} \)

The critical points are \( x = -\frac{3}{4}, \frac{1}{7}, \frac{4}{3} \). We will analyze the equation in the intervals defined by these points.

### Step 2: Analyze intervals
The intervals to consider are:
1. \( x < -\frac{3}{4} \)
2. \( -\frac{3}{4} \leq x < \frac{1}{7} \)
3. \( \frac{1}{7} \leq x < \frac{4}{3} \)
4. \( x \geq \frac{4}{3} \)

#### Case 1: \( x < -\frac{3}{4} \)
In this interval:
- \( |4x + 3| = -(4x + 3) = -4x - 3 \)
- \( |3x - 4| = -(3x - 4) = -3x + 4 \)
- \( |7x - 1| = -(7x - 1) = -7x + 1 \)

Substituting these into the equation:
\[ -4x - 3 - 3x + 4 = -7x + 1 \]
Simplifying:
\[ -7x + 1 = -7x + 1 \]
This is true for all \( x < -\frac{3}{4} \).

#### Case 2: \( -\frac{3}{4} \leq x < \frac{1}{7} \)
In this interval:
- \( |4x + 3| = 4x + 3 \)
- \( |3x - 4| = -(3x - 4) = -3x + 4 \)
- \( |7x - 1| = -(7x - 1) = -7x + 1 \)

Substituting these into the equation:
\[ 4x + 3 - 3x + 4 = -7x + 1 \]
Simplifying:
\[ x + 7 = -7x + 1 \]
\[ 8x = -6 \quad \Rightarrow \quad x = -\frac{3}{4} \]
This is valid in the interval.

#### Case 3: \( \frac{1}{7} \leq x < \frac{4}{3} \)
In this interval:
- \( |4x + 3| = 4x + 3 \)
- \( |3x - 4| = -(3x - 4) = -3x + 4 \)
- \( |7x - 1| = 7x - 1 \)

Substituting these into the equation:
\[ 4x + 3 - 3x + 4 = 7x - 1 \]
Simplifying:
\[ x + 7 = 7x - 1 \]
\[ 6 = 6x \quad \Rightarrow \quad x = 1 \]
This is valid in the interval.

#### Case 4: \( x \geq \frac{4}{3} \)
In this interval:
- \( |4x + 3| = 4x + 3 \)
- \( |3x - 4| = 3x - 4 \)
- \( |7x - 1| = 7x - 1 \)

Substituting these into the equation:
\[ 4x + 3 + 3x - 4 = 7x - 1 \]
Simplifying:
\[ 7x - 1 = 7x - 1 \]
This is true for all \( x \geq \frac{4}{3} \).

### Step 3: Find least positive integral value
The solutions we found are:
- From Case 1: All \( x < -\frac{3}{4} \)
- From Case 2: \( x = -\frac{3}{4} \)
- From Case 3: \( x = 1 \)
- From Case 4: All \( x \geq \frac{4}{3} \)

The least positive integral value satisfying the equation is \( x = 1 \).

### Final Answer
The least positive integral value of \( x \) satisfying the equation is \( \boxed{1} \).