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: Turn each inequality into a bound on y.
Rewrite each condition so $y$ is isolated:
From $x - 3y \geq 0$: $y \leq \frac{x}{3}$.
From $x + y \geq -2$: $y \geq -2 - x$.
From $3x - y \leq -2$: $y \geq 3x + 2$.
So for a given $x$, $y$ must satisfy $\max(-2-x, \, 3x+2) \leq y \leq \frac{x}{3}$.

Step 2: Find which of the two lower bounds is larger, depending on x.
Compare $-2-x$ and $3x+2$: $-2-x \geq 3x+2 \iff -4 \geq 4x \iff x \leq -1$.
So for $x \leq -1$, the lower bound is $-2-x$; for $x > -1$, the lower bound is $3x+2$.

Step 3: Find the range of x for which the lower bound does not exceed the upper bound.
For $x \leq -1$: need $-2-x \leq \frac{x}{3}$, which gives $-6-3x \leq x$, so $x \geq -\frac{3}{2}$. Combined with $x \leq -1$, this gives $-\frac{3}{2} \leq x \leq -1$.
For $x > -1$: need $3x+2 \leq \frac{x}{3}$, which gives $9x+6 \leq x$, so $x \leq -\frac{3}{4}$. Combined with $x>-1$, this gives $-1 < x \leq -\frac{3}{4}$.
Putting both pieces together, $x$ ranges over the interval $\left[-\frac{3}{2}, -\frac{3}{4}\right]$, which is a proper interval containing infinitely many real values, not a single point.

Step 4: Check that each such x gives a real range of y, not just a single value.
Take $x = -1.2$, which lies inside the interval: lower bound $= -2-(-1.2) = -0.8$, upper bound $= \frac{-1.2}{3} = -0.4$. Since $-0.8 < -0.4$, $y$ can be any of the infinitely many real numbers between $-0.8$ and $-0.4$, and each choice gives a valid $(x,y)$ pair.

Step 5: Conclude.
Since there is a whole interval of valid $x$ values, and for each of these an entire interval of valid $y$ values, the total set of pairs $(x,y)$ satisfying all three inequalities is a two-dimensional region, not empty and not a finite list of isolated points.
\[ \boxed{\text{The solution set is infinite}} \]