Question
Solve the following systems of linear inequations graphically: $\text{x}-\text{y}\leq1,\text{x}+2\text{y}\leq8,2\text{x}+\text{y}\geq2,\text{x}\geq0$ and $\text{y}\geq0$
✓
Answer
We have,
$\text{x}-\text{y}\leq1,\text{x}+2\text{y}\leq8,2\text{x}+\text{y}\geq2,\text{x}\geq0$ and $\text{y}\geq0$
Converting the inequations into equations, we obtain
x - y = 1, x + 2y = 8, 2x + y > 2,
x = 0 and y = 0.
Region represented by x - y = 1:
Putting x = 0 in x - y = 1,
we get y = -1
Putting y = 0 in X - Y = 1,
we get x = 1
$\therefore$ The line x - y = 1 meets the coordinate axes at (0, -1) and (1, 0). Draw a thick line joining these points.
Now, putting x = 0 and y = 0 in $\text{x}-\text{y}\leq1$
In $\text{x}-\text{y}\leq1$ we get $0\leq1$
Clearly, we find that (0, 0) satisfies inequation $\text{x}-\text{y}\leq1$
Region represented by $\text{x}+2\text{y}\leq8$
Putting x = 0 in x + 2y = 8,
we get, $\text{y}=\frac{8}{2}=4$
Putting y = 0 in x + 2y = 8,
we get x = 8,
$\therefore$ The line x + 2y = 8 meets the coordinate axes at (8, 0) and (0, 4). Draw a thick line joining these points.
Now, putting x = 0, y = 0 in x +2y < 8, we get 0 < 8
Clearly, we find that (0, 0) satisfies inequation $\text{x}+2\text{y}\leq8$
Region represented by 2x + y > 2
Putting x = 0 in 2x + y = 2, we get y = 2
Putting y = 0 in 2x + y = 2, we get $\text{x}=\frac{2}{2}=1$
The line 2x + y = 2 meets the coordinate axes at (0, 2) and (1, 0). Draw a thick line joining these points.
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