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
If $-3 x+17<-13$, then