Keep a denominator zero separate from allowed endpoints

$10/x-19+6x\le0$ gives $(10-19x+6x^2)/x\le0$, hence $(3x-2)(2x-5)/x\le0$. The domain excludes x=0.

Mark critical values 0,2/3,5/2. The interval signs are −,+,−,+. Choose negative intervals and valid zeros: x<0 or 2/3≤x≤5/2. The numerator zeros are included because the inequality is non-strict; the denominator zero is always excluded.

The speech bubble asks whether x can simply be removed by cross-multiplying. The source remark explains that x may be positive or negative: multiplication by a positive x preserves the sign, while multiplication by a negative x reverses it. A sign diagram handles both cases together without silently dropping the negative-x solutions.