Solve $\dfrac{x}{3} > \dfrac{x}{2} + 1$ for real $x$.
A$x > 6$
B$x < 6$
C$x < -6$
D$x > -6$
Answer & Solution
Correct answer: C. $x < -6$
Multiply both sides by 6 (positive — preserved): $2x > 3x + 6$.
Subtract $3x$: $-x > 6$.
Multiply by $-1$ (flip): $x < -6$.
Verify at $x = -12$: LHS $= -4$, RHS $= -6 + 1 = -5$. $-4 > -5$ ✓.
At $x = 0$: $0 > 1$? No ✓ (confirms $x < -6$).
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