The solution set of the inequality \( 3x+4y\lt 12 \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → An open half plane containing the origin
The line 3x + 4y = 12 divides the plane into two half-planes. Putting the origin into the inequality:
\( \displaystyle 3(0)+4(0)=0\lt 12\ \text{(true)} \)
So the solution set is the half-plane containing the origin.
The inequality is strict (<), so the points of the line itself are not included: the half-plane is open.
Why the other options are incorrect
Option (1) An open half-plane not containing the origin — Incorrect. The origin satisfies 0 < 12, so it is in the solution set.
Option (3) The whole \( xy \)-plane not containing the line \( 3x+4y=12 \) — Incorrect. Points on the far side of the line, such as (4, 4) with 3(4) + 4(4) = 28, are not solutions, so the set is not the whole plane minus the line.
Option (4) A closed half-plane containing the origin — Incorrect. A closed half-plane includes the line, which needs ≤; the given inequality is strict.