To solve the given equation \( | \frac{7}{5}X + \frac{3}{2} | = | \frac{4}{3}X - \frac{1}{4} | \), we need to consider two cases based on the properties of absolute values.

### Case 1:
Both sides are positive or both are negative.
\[
\frac{7}{5}X + \frac{3}{2} = \frac{4}{3}X - \frac{1}{4}
\]

### Case 2:
One side is positive and the other one is negative.
\[
\frac{7}{5}X + \frac{3}{2} = -\left(\frac{4}{3}X - \frac{1}{4}\right)
\]

### Solving Case 1:

1. Start with:
\[
\frac{7}{5}X + \frac{3}{2} = \frac{4}{3}X - \frac{1}{4}
\]

2. Get all terms involving \(X\) on one side and constant terms on the other side:
\[
\frac{7}{5}X - \frac{4}{3}X = -\frac{1}{4} - \frac{3}{2}
\]

3. Find a common denominator for the \(X\) coefficients. The LCM of 5 and 3 is 15. Thus,
\[
\frac{7}{5}X = \frac{21}{15}X \quad \text{and} \quad \frac{4}{3}X = \frac{20}{15}X
\]
Hence,
\[
\frac{21}{15}X - \frac{20}{15}X = \frac{1}{15}X
\]

4. Calculate the right side:
\[
-\frac{1}{4} - \frac{3}{2} = -\frac{1}{4} - \frac{6}{4} = -\frac{7}{4}
\]

Therefore,
\[
\frac{1}{15}X = -\frac{7}{4}
\]

5. Multiplying both sides by 15 gives:
\[
X = -\frac{7}{4} \times 15 = -\frac{105}{4}
\]

### Solving Case 2:

1. Start with:
\[
\frac{7}{5}X + \frac{3}{2} = -\left(\frac{4}{3}X - \frac{1}{4}\right)
\]

2. Expanding the right side:
\[
\frac{7}{5}X + \frac{3}{2} = -\frac{4}{3}X + \frac{1}{4}
\]

3. Get all \(X\) terms on one side and constant terms on the other side:
\[
\frac{7}{5}X + \frac{4}{3}X = \frac{1}{4} - \frac{3}{2}
\]

4. Again, using a common denominator for the \(X\) coefficients:
\[
\frac{7}{5}X = \frac{21}{15}X \quad \text{and} \quad \frac{4}{3}X = \frac{20}{15}X
\]
Therefore,
\[
\frac{21}{15}X + \frac{20}{15}X = \frac{41}{15}X
\]

5. Calculate the right side:
\[
\frac{1}{4} - \frac{3}{2} = \frac{1}{4} - \frac{6}{4} = -\frac{5}{4}
\]
Thus,
\[
\frac{41}{15}X = -\frac{5}{4}
\]

6. Multiplying both sides by \(\frac{15}{41}\) gives:
\[
X = -\frac{5}{4} \cdot \frac{15}{41} = -\frac{75}{164}
\]

### Final Solutions:

The solutions to the equation \( | \frac{7}{5}X + \frac{3}{2} | = | \frac{4}{3}X - \frac{1}{4} | \) are:
\[
X = -\frac{105}{4} \quad \text{and} \quad X = -\frac{75}{164}
\]