Question
Show that the solution set of the following linear in equations is an unbounded set: $\text{x}+\text{y}\geq9,3\text{x}+\text{y}\geq12,\text{x}\geq0,\text{y}\geq0.$
✓
Answer
We have,
$\text{x}+\text{y}\geq9,3\text{x}+\text{y}\geq12,\text{x}\geq0,\text{y}\geq0.$
Converting the inequations into equations, we get
x + y = 9, 3x + y = 12, x = 0 and y = 0.
Region represented by $\text{x}+\text{y}\geq9.$
Putting x = 0 in x + y = 9, we get y = 9.
Putting y = 0 in x + y = 9, we get x = 9.
$\therefore$ The line x + y = 9 meets the coordinat axes at (0, 9) and (9, 0). Join these points by a thick line.
Now, putting x = 0 and y = 0 in $\text{x}+\text{y}\geq9.$ we get 0 9 This is not possible.
$\therefore$ We find that (0, 0) is not satisfies the inequation $\text{x}+\text{y}\geq9.$
So, the portion not containing the origin is represented by the given inequation.
Region represented by 3x + y > 12:
Putting x = 0 in 3x + y = 12, we get y = 12
Putting y = 0 in 3x + y = 12, we get $\text{x}=\frac{12}{3}=4$
$\therefore$ The line 3x + y = 12 meets the coordinate axes at (0, 12) and (4, 0). Joining these points by a thick line.
Now, putting x = 0 and y = 0 in $3\text{x}+\text{y}\geq12$ we get, $0\geq12$
This is not possible.
$\therefore$ We find that (0, 0) is not satisfies the inequation $3\text{x}+\text{y}\geq12.$ so the portion not containing the origin is represented by the given inequation.
Region represented by $\text{x}\geq0$ and $\text{y}\geq0$: clearly, $\text{x}\geq0$ and $\text{y}\geq0$ represent the first quadrant.
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