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
If $x$ is a real number and $|x|<3$, then