`if (5x)/(4) + (3x)/(8) gt (39)/(8) " and " (2x-1)/(12)-(x-1)/(3)lt(3x+1)/(4)`, then x belongs to the internal

Options

A. `(3, oo)`
B. `(0, oo)`
C. `(-oo, 3)`
D. `(-oo, 0)`

Explanation

To solve the given inequalities step by step, we will break down each inequality and find the solution for \( x \).

### Step 1: Solve the first inequality
The first inequality is given as:

\[ \frac{5x}{4} + \frac{3x}{8} > \frac{39}{8} \]

**Finding a common denominator:**
The least common multiple (LCM) of 4 and 8 is 8. We can rewrite the left side:

\[ \frac{5x \cdot 2}{8} + \frac{3x}{8} > \frac{39}{8} \]

This simplifies to:

\[ \frac{10x + 3x}{8} > \frac{39}{8} \]

Combining the terms gives:

\[ \frac{13x}{8} > \frac{39}{8} \]

**Eliminating the denominator:**
Multiply both sides by 8 (since 8 is positive, the inequality remains the same):

\[ 13x > 39 \]

**Dividing by 13:**
\[ x > 3 \]

### Step 2: Solve the second inequality
The second inequality is given as:

\[ \frac{2x - 1}{12} - \frac{x - 1}{3} < \frac{3x + 1}{4} \]

**Finding a common denominator:**
The LCM of 12, 3, and 4 is 12. We can rewrite each term:

\[ \frac{2x - 1}{12} - \frac{4(x - 1)}{12} < \frac{3(3x + 1)}{12} \]

This simplifies to:

\[ \frac{2x - 1 - 4x + 4}{12} < \frac{9x + 3}{12} \]

Combining the terms gives:

\[ \frac{-2x + 3}{12} < \frac{9x + 3}{12} \]

**Eliminating the denominator:**
Multiply both sides by 12:

\[ -2x + 3 < 9x + 3 \]

**Rearranging the terms:**
Add \( 2x \) to both sides:

\[ 3 < 11x + 3 \]

Subtract 3 from both sides:

\[ 0 < 11x \]

**Dividing by 11:**
\[ 0 < x \]

### Step 3: Combine the results
From the first inequality, we found:

\[ x > 3 \]

From the second inequality, we found:

\[ x > 0 \]

The more restrictive condition is \( x > 3 \).

### Conclusion
Thus, the solution for \( x \) is:

\[ x \in (3, \infty) \]