2. \( \frac{3 x+1}{5}-\frac{x}{3} \geq \frac{x+12}{15} \) and \( x \in \mathbb{N} \)

To solve the inequality \( \frac{3x+1}{5} - \frac{x}{3} \geq \frac{x+12}{15} \), we first need a common denominator. The least common multiple of 5, 3, and 15 is 15. Rewriting each term: 1. \( \frac{3x+1}{5} = \frac{3(3x+1)}{15} = \frac{9x + 3}{15} \) 2. \( \frac{x}{3} = \frac{5x}{15} \) 3. \( \frac{x+12}{15} \) remains the same. Now plug these into the inequality: \[ \frac{9x + 3}{15} - \frac{5x}{15} \geq \frac{x + 12}{15} \] Combining the left side: \[ \frac{(9x + 3 - 5x)}{15} \geq \frac{x + 12}{15} \] This simplifies to: \[ \frac{4x + 3}{15} \geq \frac{x + 12}{15} \] Next, multiply both sides by 15 (since 15 is positive, the inequality sign stays): \[ 4x + 3 \geq x + 12 \] Subtract \(x\) from both sides: \[ 3x + 3 \geq 12 \] Subtract 3 from both sides: \[ 3x \geq 9 \] Divide by 3: \[ x \geq 3 \] Since \(x\) must be in the natural numbers (\(\mathbb{N}\)), the possible values for \(x\) are 3, 4, 5, etc. Thus, the solution in natural numbers is: \[ x \in \{ 3, 4, 5, \ldots \} \]