Solution of `|4x +3| + |3x - 4| = 12 ` is

Options

A. `x = - 7/3, 3/7`
B. `x = - 5/2 , 2/5`
C. `x = - 11/7 , 13/7`
D. `x = - 3/7 , 7/5`

Answer

To solve the equation \( |4x + 3| + |3x - 4| = 12 \), we need to consider the different cases based on the expressions inside the absolute values.

### Step 1: Identify the critical points
The expressions inside the absolute values change sign at certain points. We need to find these points:
1. \( 4x + 3 = 0 \) gives \( x = -\frac{3}{4} \)
2. \( 3x - 4 = 0 \) gives \( x = \frac{4}{3} \)

These points divide the number line into three intervals:
1. \( (-\infty, -\frac{3}{4}) \)
2. \( [-\frac{3}{4}, \frac{4}{3}) \)
3. \( [\frac{4}{3}, \infty) \)

### Step 2: Solve for each interval

#### Interval 1: \( (-\infty, -\frac{3}{4}) \)
In this interval, both expressions are negative:
- \( |4x + 3| = -(4x + 3) = -4x - 3 \)
- \( |3x - 4| = -(3x - 4) = -3x + 4 \)

Substituting these into the equation:
\[ -4x - 3 - 3x + 4 = 12 \]
This simplifies to:
\[ -7x + 1 = 12 \]
\[ -7x = 11 \implies x = -\frac{11}{7} \]

#### Interval 2: \( [-\frac{3}{4}, \frac{4}{3}) \)
In this interval, \( 4x + 3 \) is non-negative and \( 3x - 4 \) is negative:
- \( |4x + 3| = 4x + 3 \)
- \( |3x - 4| = -(3x - 4) = -3x + 4 \)

Substituting these into the equation:
\[ 4x + 3 - 3x + 4 = 12 \]
This simplifies to:
\[ x + 7 = 12 \]
\[ x = 5 \]
However, \( x = 5 \) is not in the interval \( [-\frac{3}{4}, \frac{4}{3}) \), so we discard this solution.

#### Interval 3: \( [\frac{4}{3}, \infty) \)
In this interval, both expressions are non-negative:
- \( |4x + 3| = 4x + 3 \)
- \( |3x - 4| = 3x - 4 \)

Substituting these into the equation:
\[ 4x + 3 + 3x - 4 = 12 \]
This simplifies to:
\[ 7x - 1 = 12 \]
\[ 7x = 13 \implies x = \frac{13}{7} \]

### Step 3: Collect solutions
The solutions we found are:
1. From Interval 1: \( x = -\frac{11}{7} \)
2. From Interval 3: \( x = \frac{13}{7} \)

### Final Solutions
The complete solution set is:
\[ x = -\frac{11}{7}, \quad x = \frac{13}{7} \]