Show that the following system of linear inequalities has no solution $x+2 y \leq 3,3 x+4 y \geq 12, x \geq 0, y \geq 1$.
Consider the inequation $x+2 y \leq 3$ as an equation, we have
$$\begin{array}{r} & x+2 y =3 \\ \Rightarrow & x =3-2 y \\ \Rightarrow & 2 y =3-x \end{array}$$
Now, $(0,0)$ satisfy the inequation $x+2 y \leq 3$.
So, half plane contains $(0,0)$ as the solution and the line $x+2 y=3$ intersect the coordinate axis at $(3,0)$ and $(0,3 / 2)$.
Consider the inequation $3 x+4 y \geq 12$ as an equation, we have $3 x+4 y=12$ $\Rightarrow \quad 4 y=12-3 x$
Thus, coordinate axis intersected by the line $3 x+4 y=12$ at points $(4,0)$ and $(0,3)$.
Now, $(0,0)$ does not satisfy the inequation $3 x+4 y=12$.
Therefore, half plane of the solution does not contained $(0,0)$.
Consider the inequation $y \geq 1$ as an equation, we have
$$y=1 .$$
It represents a straight line parallel to $X$-axis passing through point $(0,1)$.
Now, $(0,0)$ does not satisfy the inequation $y \geq 1$.
Therefore, half plane of the solution does not contains $(0,0)$.
Clearly $x \geq 0$ represents the region lying on the right side of $Y$-axis.
The solution set of the given linear constraints will be the intersection of the above region.
It is clear from the graph the shaded portions do not have common region. So, solution set is null set.
Solve the following system of linear inequalities $3 x+2 y \geq 24,3 x+y \leq 15, x \geq 4$.
Consider the inequation $3 x+2 y \geq 24$ as an equation, we have $3 x+2 y=24$.
$$\Rightarrow \quad 2 y=24-3 x$$
| $x$ | 0 | 8 | 4 |
|---|---|---|---|
| $y$ | 12 | 0 | 6 |
Hence, line $3 x+y=24$ intersect coordinate axes at points $(8,0)$ and $(0,12)$.
Now, $(0,0)$ does not satisfy the inequation $3 x+2 y \geq 24$.
Therefore, half plane of the solution set does not contains $(0,0)$.
Consider the inequation $3 x+y \leq 15$ as an equation, we have
$$\begin{aligned} & 3 x+y=15 \\ \Rightarrow \quad & y=15-3 x \end{aligned}$$
| $x$ | 0 | 5 | 3 |
|---|---|---|---|
| $y$ | 15 | 0 | 6 |
Line $3 x+y=15$ intersects coordinate axes at points $(5,0)$ and $(0,15)$.
Now, point $(0,0)$ satisfy the inequation $3 x+y \leq 15$.
Therefore, the half plane of the solution contain origin.
Consider the inequality $x \geq 4$ as an equation, we have
$$x=4$$
It represents a straight line parallel to $Y$-axis passing through $(4,0)$. Now, point $(0,0)$ does not satisfy the inequation $x \geq 4$.
Therefore, half plane does not contains $(0,0)$,
The graph of the above inequations is given below.
It is clear from the graph that there is no common region corresponding to these inequality. Hence, the given system of inequalities have no solution.
Show that the solution set of the following system of linear inequalities is an unbounded region $2 x+y \geq 8, x+2 y \geq 10, x \geq 0, y \geq 0$.
Consider the inequation $2 x+y \geq 8$ as an equation, we have
$$\begin{aligned} 2 x+y & =8 \\ \Rightarrow \quad y & =8-2 x \end{aligned}$$
| $x$ | 0 | 4 | 3 |
|---|---|---|---|
| $y$ | 8 | 0 | 2 |
The line $2 x+y=8$ intersects coordinate axes at $(4,0)$ and $(0,8)$. Now, point $(0,0)$ does not satisfy the inequation $2 x+y \geq 8$. Therefore, half plane does not contain origin.
Consider the inequation $x+2 y \geq 10$, as an equation, we have
$$\begin{aligned} & x+2 y=10 \\ \Rightarrow \quad & 2 y=10-x \end{aligned}$$
| $x$ | 10 | 0 | 8 |
|---|---|---|---|
| $y$ | 0 | 5 | 1 |
The line $2 x+y=8$ intersects the coordinate axes at $(10,0)$ and $(0,5)$.
Now, point $(0,0)$ does not satisfy the inequation $x+2 y \geq 10$.
Therefore, half plane does not contain $(0,0)$.
Consider the inequation $x \geq 0$ and $y \geq 0$ clearly, it represents the region in first quadrant.
The graph of the above inequations is given below
It is clear from the graph that common shaded portion is unbounded.
If $x<5$, then
If $x, y$ and $b$ are real numbers and $x< y, b<0$, then