Solve $3(x - 1) \leq 2(x - 3)$ for real $x$.
A$x \leq 3$
B$x \geq -3$
C$x \leq -3$
D$x \geq 3$
Answer & Solution
Correct answer: C. $x \leq -3$
Expand: $3x - 3 \leq 2x - 6$.
Subtract $2x$: $x - 3 \leq -6$.
Add 3: $x \leq -3$.
Verify at $x = -3$: $3(-4) = -12$ and $2(-6) = -12$. Equal ✓ (boundary).
At $x = -4$: $3(-5) = -15$, $2(-7) = -14$. $-15 \leq -14$ ✓.
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