Q.Solve $2x+3\ge 5x-6$ for $x\in\mathbb{R}$.

Collect the variable on one side; keeping its coefficient positive avoids a sign flip.

$2x+3\ge5x-6\Rightarrow 9\ge3x\Rightarrow 3\ge x$, i.e. $x\le3$.

To avoid dividing by a negative, move the $x$-terms to the side that keeps the coefficient positive.

Subtract $2x$ and add $6$ to both sides:

Divide by the positive $3$ (sign unchanged):

The solution set is $(-\infty,3]$ — a closed bracket at $3$ because $\ge$ includes the boundary.

Check (independent verification, keeping $x$ on the left instead): $2x-5x\ge-6-3\Rightarrow -3x\ge-9$; dividing by the negative $-3$ reverses the sign: $x\le3$ — the same answer, confirming the sign-flip rule was handled correctly. Test $x=0$: $3\ge-6$ ✓; test $x=4$: $11\ge14$ is false ✓ excluded.

Alternative: keep $x$ on the left — $-3x\ge-9$; dividing by $-3$ flips $\ge$ to $\le$, giving $x\le3$ (same result).

Dividing $-3x\ge-9$ by $-3$ without flipping the sign (wrongly getting $x\ge3$). Whenever you divide by a negative, reverse the inequality.