Solve for \( x \) \[ \frac{x-3}{4}+\frac{x}{8} \geq \frac{2 x-3}{6} \]

To solve the inequality \[ \frac{x-3}{4}+\frac{x}{8} \geq \frac{2 x-3}{6}, \] let’s start by eliminating the fractions. We can find the least common multiple (LCM) of the denominators \(4\), \(8\), and \(6\). The LCM of these numbers is \(24\). Now, we can multiply each term in the inequality by \(24\) to eliminate the denominators: \[ 24 \cdot \left(\frac{x-3}{4}\right) + 24 \cdot \left(\frac{x}{8}\right) \geq 24 \cdot \left(\frac{2x-3}{6}\right). \] This simplifies to: \[ 6(x-3) + 3x \geq 4(2x-3). \] Now distribute each term: \[ 6x - 18 + 3x \geq 8x - 12. \] Combining like terms on the left side gives: \[ 9x - 18 \geq 8x - 12. \] Next, subtract \(8x\) from both sides: \[ 9x - 8x - 18 \geq -12, \] which simplifies to: \[ x - 18 \geq -12. \] Now add \(18\) to both sides: \[ x \geq 6. \] Thus, the solution to the inequality is \[ \boxed{x \geq 6}. \]