Question:
Solve the given inequality for real x: $\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)$
Solution:
$\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)$
$\Rightarrow 3\left(\frac{3 x}{5}+4\right) \geq 2(x-6)$
$\Rightarrow \frac{9 x}{5}+12 \geq 2 x-12$
$\Rightarrow 12+12 \geq 2 x-\frac{9 x}{5}$
$\Rightarrow 24 \geq \frac{10 x-9 x}{5}$
$\Rightarrow 24 \geq \frac{x}{5}$
$\Rightarrow 120 \geq x$
Thus, all real numbers $x$, which are less than or equal to 120 , are the solutions of the given inequality.
Hence, the solution set of the given inequality is $(-\infty, 120]$.