Question

Set of real numbers 'x, y', satisfying the inequations \(x - 3y \geq 0\), \(x + y \geq -2\) and \(3x - y \leq -2\) is:

• A. Empty
• B. Finite
• C. Infinite
• D. Cannot be determined

Hint:

Convert each inequality into a bound on y in terms of x, then check whether the resulting range of valid x and y values is empty, a single point, or a full region.

Answer 1

The Correct Option is C

Step 1: Understand what is being asked.
We need to check whether the set of points \((x,y)\) satisfying all three inequalities \(x-3y \geq 0\), \(x+y \geq -2\), and \(3x-y \leq -2\) together is empty, has finitely many points, or has infinitely many points.

Step 2: Find where the boundary lines meet.
Each inequality has a boundary line: \(L_1: x-3y=0\), \(L_2: x+y=-2\), \(L_3: 3x-y=-2\). Their pairwise intersections give the corner points of the region.
\(L_1 \cap L_2\): from \(L_1\), \(x=3y\). Put into \(L_2\): \(3y+y=-2 \implies y=-\frac{1}{2}, x=-\frac{3}{2}\). Corner point \(\left(-\frac{3}{2},-\frac{1}{2}\right)\).
\(L_1 \cap L_3\): put \(x=3y\) into \(L_3\): \(9y-y=-2 \implies y=-\frac{1}{4}, x=-\frac{3}{4}\). Corner point \(\left(-\frac{3}{4},-\frac{1}{4}\right)\).
\(L_2 \cap L_3\): from \(L_2\), \(x=-2-y\). Put into \(L_3\): \(3(-2-y)-y=-2 \implies -6-4y=-2 \implies y=-1, x=-1\). Corner point \((-1,-1)\).

Step 3: Check that these three points form a genuine triangle.
The three corner points \(\left(-\frac{3}{2},-\frac{1}{2}\right)\), \(\left(-\frac{3}{4},-\frac{1}{4}\right)\), and \((-1,-1)\) are not collinear: the line joining the first two has slope \(\frac{1}{3}\), while the line joining the first and third has slope \(-1\). Three non-collinear points always form a triangle with nonzero area.

Step 4: Confirm a point inside this triangle satisfies all three original inequalities.
Take the centroid of the triangle, approximately \((-1.08,-0.58)\), and check it in each inequality:
\(x-3y \approx -1.08-3(-0.58) = 0.67 \geq 0\) holds.
\(x+y \approx -1.08-0.58 = -1.67 \geq -2\) holds.
\(3x-y \approx 3(-1.08)-(-0.58) = -2.67 \leq -2\) holds.
All three hold, so the centroid, and by continuity a whole neighbourhood around it, lies in the solution set.

Step 5: Conclude.
The solution set is not empty, since we just found points that satisfy all three conditions, and it is a two-dimensional triangular region, not a single point or a short list of points. Every point inside and on this triangle is a valid solution, so there are infinitely many such points.

Final Answer:
The set of real solutions \((x,y)\) is infinite. \[ \boxed{\text{Infinite}} \]