Question:
$\frac{3 x-2}{5} \leq \frac{4 x-3}{2}$
Solution:
$\frac{3 x-2}{5} \leq \frac{4 x-3}{2}$
$\Rightarrow 2(3 x-2) \leq 5(4 x-3)$
$\Rightarrow 6 x-4 \leq 20 x-15$
$\Rightarrow-4+15 \leq 20 x-6 x \quad$ (Transposing $6 x$ to the RHS and $-15$ to the LHS)
$\Rightarrow 11 \leq 14 x$
$\Rightarrow 14 x \geq 11$
$\Rightarrow x \geq \frac{11}{14} \quad$ (Dividing both the sides by 14 )
Hence, the solution set of the given inequation is $\left[\frac{11}{14}, \infty\right)$.