Solve each of the following in equations and represent the solution set on the number line.
$\frac{1}{2}\left(\frac{2}{3} x+1\right) \geq \frac{1}{3}(x-2)$ where $x \in \mathbf{R}$.
Given:
$\frac{1}{2}\left(\frac{2}{3} x+1\right) \geq \frac{1}{3}(x-2)$, where $x \in R$
$\frac{1}{2}\left(\frac{2 x}{3}\right)+\frac{1}{2}(1) \geq \frac{1}{3}(x)-\frac{1}{3}(2)$
$\frac{x}{3}+\frac{1}{2} \geq \frac{x}{3}-\frac{2}{3}$
Now, subtracting $\frac{1}{2}$ from both the sides in the above equation
$\frac{x}{3}+\frac{1}{2}-\frac{1}{2} \geq \frac{x}{3}-\frac{2}{3}-\frac{1}{2}$
$\frac{x}{3} \geq \frac{2 x-4-3}{6}$
$\frac{x}{3} \geq \frac{2 x-7}{6}$
$\frac{x}{3} \geq \frac{x}{3}-\frac{7}{6}$
Now, subtracting $\frac{\mathrm{x}}{3}$ from both the sides in the above equation,
$\frac{x}{3}-\frac{x}{3} \geq \frac{x}{3}-\frac{7}{6}-\frac{x}{3}$
$0 \geq-\frac{7}{6}$
Therefore, the solution is: true for all values of x.