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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