Show that the solution set of the following linear inequations is empty set:
(i) x − 2y ≥ 0, 2x − y ≤ −2, x ≥ 0, y ≥ 0
(ii) x + 2y ≤ 3, 3x + 4y ≥ 12, y ≥ 1, x ≥ 0, y ≥ 0
(i) Converting the inequations to equations, we obtain:
$x-2 y=0,2 x-y=-2, x=0, y=0$
$x-2 y=0$ : This line meets the $x$-axis at $(0,0)$ and $y$-axis at $(0,0)$. If $x=1$, then $y=1 / 2$,
so we have another point $(1,1 / 2)$. Draw a thick line through $(0,0)$ and $(1,1 / 2)$.
We see that the origin $(1,0)$ satisfies the inequation $x+2 y \leq 3$ So, the portion containing the $(1,0)$ represents the solution set of the inequation $x-2 y \leq 0$
$2 x-y=-2$ : This line meets the $x$-axis at $(-1,0)$ and $y$-axis at $(0,2)$. Draw a thick line joining these points.
We see that the origin $(0,0)$ does not satisfy the inequation $2 x-y \leq-2$ So, the portion not containing the origin represents the solution set of the inequation $2 x-y \leq-2$
Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
We see in the figure that there is no common region in all the lines. Hence, the solution set to the given set of inequations is empty.
(ii) Converting the inequations to equations, we obtain:
x + 2y = 3, 3x + 4y =12, y = 1
x + 2y = 3: This line meets the x-axis at (3, 0) and y-axis at (0, 3/2). Draw a thick line joining these points.
We see that the origin (0, 0) satisfies the inequation x + 2y ≤ 3. So, the portion containing the origin represents the solution set of the inequation x + 2y ≤ 3
3x + 4y =12: This line meets the x-axis at (4, 0) and y-axis at (0, 3). Draw a thick line joining these points.
We see that the origin $(0,0)$ does not satisfy the inequation $3 x+4 y \geq 12$ So, the portion not containing the origin represents the solution set of the inequation $3 x+4 y \geq 12$
y = 1: This line is parallel to x-axis at a distance of 1 unit from it.
We see that the origin $(0,0)$ does not satisfy the inequation $y \geq 1$ So, the portion not containing the origin represents the solution set of the inequation $y \geq 1$
Clearly, x ≥ 0, y ≥ 0 represents the first quadrant.
We see in the figure that there is no common region in all the lines. Hence, the solution set to the given set of inequations is empty..png)