4. \( \frac{3 x-1}{2}-\frac{3-x}{3}<\frac{x}{12}+2 \) and \( x \in \mathbb{R} \)

To solve the inequality \( \frac{3x-1}{2} - \frac{3-x}{3} < \frac{x}{12} + 2 \), we start by simplifying the left-hand side. First, we find a common denominator for the fractions on the left side. The common denominator for \(2\) and \(3\) is \(6\). Rewriting the fractions, we have: \[ \frac{3x-1}{2} = \frac{3(3x - 1)}{6} = \frac{9x - 3}{6} \] and \[ \frac{3-x}{3} = \frac{2(3 - x)}{6} = \frac{6 - 2x}{6} \] Now we can combine them: \[ \frac{9x - 3 - (6 - 2x)}{6} = \frac{9x - 3 - 6 + 2x}{6} = \frac{11x - 9}{6} \] Thus, our inequality becomes: \[ \frac{11x - 9}{6} < \frac{x}{12} + 2 \] Next, we rewrite the right side with a common denominator of \(12\): \[ 2 = \frac{24}{12} \quad \text{so} \quad \frac{x}{12} + 2 = \frac{x}{12} + \frac{24}{12} = \frac{x + 24}{12} \] Now the inequality is: \[ \frac{11x - 9}{6} < \frac{x + 24}{12} \] To eliminate the denominators, we can multiply through by \(12\) (keeping in mind that \(12 > 0\) preserves the inequality): \[ 12 \cdot \frac{11x - 9}{6} < x + 24 \] This simplifies to: \[ 2(11x - 9) < x + 24 \] Expanding the left side gives us: \[ 22x - 18 < x + 24 \] Now, we will move \(x\) to the left side and the constant terms to the right side: \[ 22x - x < 24 + 18 \] This simplifies to: \[ 21x < 42 \] Dividing both sides by \(21\): \[ x < 2 \] Thus, the solution to the inequality is: \[ x < 2 \quad (x \in \mathbb{R}) \]