Answer
The solution set $\left[ \frac{21}{8},\infty \right)$ consist of all real numbers.
Work Step by Step
Considered the inequality,
$\frac{x}{6}+\frac{1}{8}\le \frac{x}{2}-\frac{3}{4}$ ,
Add via LCM $\left( 6,8,2,4:\text{ }24 \right)$:
$\begin{align}
& \frac{x}{6}+\frac{1}{8}\le \frac{x}{2}-\frac{3}{4} \\
& \frac{4x+1}{24}\le \frac{12x-18}{24} \\
\end{align}$
Multiply $24$ on both sides,
$4x+3\le 12x-18$
Subtract $3$ and $12x$ from both sides,
$\begin{align}
& 4x+3-3-12x\le 12x-18-3-12x \\
& -8x\le -2
\end{align}$
Multiply both sides by $-1$ (reverse the inequality right)
$\left( -8x \right)\left( -1 \right)\ge \left( -21 \right)\left( -1 \right)$
Divide both sides by $8$
$\begin{align}
& \frac{8x}{8}\ge \frac{21}{8} \\
& x\ge \frac{21}{8}
\end{align}$
The solution set consists of all real numbers greater than or equal to $\frac{21}{8}$.
The graph on the number line is shown below.
