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Linear Inequations — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLinear Inequations5 Marks
Question
Solve the following systems of inequations graphically:
$\text{x}+2\text{y}\leq40,3\text{x}+\text{y}\geq30,4\text{x}+3\text{y}\geq60,\text{x}\geq0,\text{y}\geq0$
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Answer



We have,

$\text{x}+2\text{y}\leq40,3\text{x}+\text{y}\geq30,4\text{x}+3\text{y}\geq60,\text{x}\geq0,\text{y}\geq0$

Converting the inequations into equations, we obtain,

x + 2y = 40, 3x + y = 30, 4x + 3y = 60, x = 0 and y = 0

Region represented by $\text{x} + 2\text{y} \leq 40:$

Putting x = 0 in x +2y = 40, we get $\text{y}=\frac{40}{2}=20$

Putting y - 0 in x + 2y = 40, we get x = 40

$\therefore$ The line x + 2y = 40, meets the coordinate axes at (0, 20) and (40, 0). Join these points by a thick line.

Now, putting x = 0 and y = 0 in $\text{x} + 2\text{y} \leq 40$ we get $0\leq40$

Therefore, (0, 0) satisfies the inequality $\text{x} + 2\text{y} \leq 40$ so, the portion containing the origin represents the solution set of the inequation $\text{x} + 2\text{y} \leq 40$

Region represented by $3\text{x} + \text{y} \geq 30$

Putting x = 0 in $3\text{x} + \text{y} \leq 30$ we get y = 30

Putting y = 0 in 3x + y = 30, we get, $\text{x}=\frac{30}{3}=10$

$\therefore$ The line 3x + y = 30 meets the coordinate axes at (0, 30) and (10, 0). Joining these points by a thick line.

Now, putting x = 0 and y = 0 in $3\text{x}+\text{y}\geq30$ we get, $0\geq30$ This is not possible.

Therefore (0, 0) does not satisfies the inequality $3\text{x} +\text {y} \geq 30.$ so, the portion not containing the origin is represented by the inequation $3\text{x} +\text {y} \geq 30.$

Region represented by 4x + 3y > 60:

Putting x = 0 in 4x + 3y = 60, we get, $\text{y}=\frac{60}{3}=20$

Putting y = 0 in 4x + 3y = 60, we get, $\text{x}=\frac{60}{4}=15.$

$\therefore$ The line 4x + 3y = 60 meets the coordinate axes at (0, 20) and (15, 0). Join these points by a thick line.

Now, putting x = 0, y = 0 in $4\text{x} + 3\text{y} \geq260,$ we get $0\geq60.$

This is not possible. Therefore, (0, 0) does not satisfies the inequality $4\text{x}+3\text{y}\geq60$ so, the portion not containing the origin is represented by the inequation $4\text{x}+3\text{y}\geq60$

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