Q.Solve $\dfrac{3(x-2)}{5} \ge \dfrac{5(2-x)}{3}$ for real $x$.

[!TLDR] Clear both denominators by multiplying by their LCM, 15 (a positive number), then collect terms. [!ANSWER] $x \ge 2$, i.e. $x \in [2, \infty)$.

Start with $\dfrac{3(x-2)}{5} \ge \dfrac{5(2-x)}{3}$. Multiply both sides by $15$, the LCM of the denominators $5$ and $3$ (a positive number, so no flip): the left side becomes $15 \cdot \dfrac{3(x-2)}{5} = 9(x-2) = 9x - 18$, and the right side becomes $15 \cdot \dfrac{5(2-x)}{3} = 25(2-x) = 50 - 25x$. So $9x - 18 \ge 50 - 25x$. Add $25x$ to both sides: $34x - 18 \ge 50$. Add $18$: $34x \ge 68$. Divide by the positive number $34$: $x \ge 2$. [!ANSWER] $x \ge 2$, i.e. $x \in [2, \infty)$.

Clear the fractions first by multiplying through by the LCM of the denominators, then expand the brackets and collect like terms.

A common mistake is multiplying only one side by the LCM, or forgetting to distribute the multiplication into the bracket $(x-2)$ or $(2-x)$ correctly.