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Linear Inequations — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsLinear Inequations5 Marks
Question
Solve the following systems of inequations graphically:
$5\text{x}+\text{y}\leq10,2\text{x}+2\text{y}\geq12,\text{x}+4\text{y}\geq12,\text{x}\geq0,\text{y}\geq0$
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Answer



We have,

$5\text{x}+\text{y}\leq10,2\text{x}+2\text{y}\geq12,\text{x}+4\text{y}\geq12,\text{x}\geq0,\text{y}\geq0$

Converting the inequations into equations, we obtain,

5x + y = 10, 2x + 2y = 1, x + 4y = 12, x = 0 and y = 0

Region represented by $5\text{x}+\text{y}\geq1$

Putting x = 0 in 5x + y = 10, we get y = 10

Putting y = 0 in 5x + y = 10, we get $\text{x}=\frac{10}{5}=2$

$\therefore$ The line 5x + y = 10, meets the coordinate axes at (0, 10) and (2, 0). Join these points by a thick line.

Now, putting x = 0 and y = 0 in $5\text{x} + \text{y} \geq 10,$ we get $0\geq10,$ This is not possible.

$\therefore$ (0, 0) does not satisfies the inequality $5\text{x} + \text{y} \geq 10,$. so, the portion not containing the origin is represented by the inequation $5\text{x} + \text{y} \geq 10,$

Region represented by 2x + 2y > 12:

Putting x = 0 in 2x +2y = 12, we get $\text{y}=\frac{12}{2}=6$

Putting y = 0 in 2x + 2y = 12, we get $\text{x}=\frac{12}{2}=6$

The line 2x + 2y = 12 meets the coordinate axes at (0, 6) and (6, 0). Join these point by a thick line.

Now, putting x = 0 and y = 0 in 2x + 2y = 12, we get $0\geq12,$ which is not possible. Therefore, (0, 0) does not satisfies the inequality 2x + 2y = 12. so, the portion not containing the origin is represented by the inequation 2x + 2y = 12.

Region represented by $\text{x}+4\text{y}\geq12$

Putting x = 0 in x + 4y = 12, we get $\text{y}=\frac{12}{4}=3$

Putting y = 0 in x + 4y = 12, we get x = 12.

$\therefore$ The line x + 4y = 12 meets the coordinate axes at (0, 3) and (12, 0). Join these points by a thick line.

Now, putting x = 0 and y = 0 in x + 4y = 12, we get $0\geq12,$ which is not possible.

Therefore, (0, 0) does not satisfies the inequality $\text{x}+4\text{y}\geq12$ so, the portion not containing the origion is represented by the inequation $\text{x}+4\text{y}\geq12$

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