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Chapter 15: Linear Inequations – Exercise 15.6

Linear Inequations – Exercise – 15.6 – Q.1(i)

We have,

2x + 3y ≤ 6,    3x + 2y ≤ 6,    x ≥ 0,  y ≥ 0

Converting the given inequation into equations, the inequations reduce to 2x+ 3y = 6,

3y + 2y = 6, x = 0 and y = 0.

Region represented by 2x + 3y ≤ 6:

Putting x = 0 inequation  2x + 3y = 6

we get y = 6/3 = 2.

Putting y = 0 in the equation 2x + 3y = 6,

we get x = 6/3 = 3. 

∴ This line 2x + 3y = 6 meets the coordinate axes at (0, 2) and (3, 0). Draw a thick line joining these points. we find that (0, 0) satisfies inequation 2x + 3y ≤ 6.

Region represented by 3x+ 2y ≤ 6:

Putting x = 0 in the equation

3x + y = 6, we get y = 6/2 = 3.

Putting y = 0 in the equation

3x + y = 6, we get x = 6/2 = 2.

∴ This line 3x + 2y = 6 meets the coordinate axes at (0, 3) and (2, 0). Draw a thick line joining these points. we find that (0,0) satisfies inequation 3x +2y ≤ 6.

Region represented by x ≥ 0 and y ≥ 0:

This line 3x + 2y = 6 meets the coordinate axes at (0, 3) and (2, 0).

Clearly x ≥ 0 and y ≥ 0 represent the first quadrant.

 

Linear Inequations – Exercise – 15.6 – Q.1(ii)

We have,

2x+ 3y ≤ 6,    x + 4y ≤ 4, x ≥ 0, y ≥ 0

Converting the inequations into equations, the inequations reduce to 2x+ 3y = 6,

x + 4y = 4, x = 0 and y = 0.

Region represented by 2x+ 3y ≤ 6:

Putting x = 0 in 2x + 3y = 6,

we get y = 6/3 = 2

Putting y = 0 in 2x + 3y = 6,

we get x = 6/2 = 3.

∴ The line 2x + 3y = 6 meets the coordinate axes at (0, 2) and (3, 0). Draw a thick line joining these points. Now, putting x = 0 and y = 0 in 2x+ 3y 6 ⟹ 0 < 6

Clearly, we find that (0, 0) satisfies inequation 2x+ 3y ≤ 6

Region represented by x+ 4y ≤ 4

Putting x = 0 in x + 4y = 4

we get, y = 4/4 = 1

Putting y = 0 in x + 4y = 4,

we get x = 4

∴ The line x + 4y = 4 meets the coordinate axes at (0, 1) and (4, 0). Draw a thick line joining these points. Now, putting x = 0, y = 0

in x + 4y ≤ 4, we get 0 ≤ 4

Clearly, we find that (0, 0) satisfies inequation x + 4y ≤ 4.

Region represented by x ≥ 0 and y ≥ 0:

Clearly x ≥ 0 and y ≥ 0 represent the first quadrant.

The line x + 4y = 4 meets the coordinate axes at (0, 1) and (4, 0)

 

Linear Inequations – Exercise – 15.6 – Q.1(iii)

We have,

x - y ≤ 1,  x + 2y ≤ 8,   2x + y ≥ 2,  x ≥ 0 and y ≥ 0

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

∴ 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 x - y ≤ 1

in x - y ≤ 1, we get, 0 ≤ 1

Clearly, we find that (0,0) satisfies inequation x - y ≤ 1

Region represented by x + 2 y ≤ 8,

Putting x = 0 in x + 2y = 8,

we get, y = 8/2 = 4

Putting y = 0 in x + 2y = 8,

we get x = 8,

∴ 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 inequationx + 2y ≤ 8.

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 x = 2/2 = 1.

The line 2x+ y = 2meek the coordinate axes at (0, 2) and (1, 0). Draw a thick line joining these points.

The line 2x+ y = 2meek the coordinate axes at (0, 2) and (1, 0).

 

Linear Inequations – Exercise – 15.6 – Q.1(iv)

We have,

x + y ≥ 1,   7x + 9y ≤ 63, x ≤ 6,  y ≤ 5 x ≥ 0 and y ≥ 0

Converting the inequations into equations, we obtain

x + y = 1, 7x + 9y = 63   x = 6, y = 5, 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

∴ The line x + y = 1 meets the coordinate axes at (0,1) and (1,0). join these point by a thick line.

Now putting x = 0 and y = 0 in x + y ≥ 1, we get 0 ≥ 1

This is not possible

∴  (0, 0) is not satisfies the inequality x + y ≥ 1. So, the portion not containing the origin is represented by the inequation x + y ≥ 1.

Region represented by 7X + 9y ≤ 63

Putting x = 0 in 7x + 9y = 63, we get, y = 63/9 = 7.

Putting y = 0 in 7x + 9y = 63, we get x = 63/7 = 9.

∴ The line 7x + 9y = 63 meets the coordinete axes of (0, 7) and (9, 0). Join these points by a thick line. Now putting x = 0 and y = 0

in 7x + 9y ≤ 63, we get, 0 ≤ 63

∴ we find (0, 0) satisfies the inequality 7x + 9y ≤ 63. So, the portion containing the origin represents the solution set of the inequation 7x + 9y ≤ 63.

Region represented by x ≤ 6: Clearly, x = 6 is a line parallel to y-axis at a distance of 6 units from the origin. Since (0, 0) satisfies the inequation x ≤ 6. so, the portion lying on the left side of x = 6 is the region represented by x ≤ 6.

Region represented by y ≤ 5: Clearly, y = 5 is a line parallel to x-axis at a distance 5 from it. since (0, 0) satisfies by the given inequation.

Region represented by x ≥ 0 and y ≥ 0: dearly, x ≥ 0 and y ≥ 0 represent the first quadrant.

The common region of the above six regions represents the solution set of the given inequation as shown below.

The common region of the above six regions represents the solution set of the given inequation

 

Linear Inequations – Exercise – 15.6 – Q.1(v)

We have,

2x + 3y ≤ 35,  y ≥ 3,  x ≥ 2, x ≥ 0 and y ≥ 0

Converting the inequations into equations, we get

2x + 3y = 35, y = 3, x = 2, x = 0 and y = 0.

Region represented by 2x + 3y ≤ 35:

Putting x = 0 in 2x + 3y = 35, we get y = 35/3

Putting y = 0 in 2x + 3y = 35, we get x = 35/2

The line 2u +3y = 35 meets the coordinate axes at (0, 35/3)and (35/2, 0). joining these point by 

a thick line.

Now, putting x = 0 and y = 0 in 2x + 3y ≤ 35, we get 0 ≤ 35.

Clearly, (0, 0) satisfies the inequality 2x +3y ≤ 35. So, the portion containing the origin represents the solution 2x + 3y ≤ 35.

Region represented by y ≥ 3.

Clearly, y = 3 is a line parallel to x-axis at a distance 3 units from the origin. Since (0,0) does not satisfies the inequation y ≥ 3.

So, the portion not containing the origin is represented by the y ≥ 3.

Region represented by x ≥ 2

Clearly, x = 2 is a line parallel to y-axis at a distance of 2 units from the origin. Since (0,0) does not satisfies the inequation x ≥ 2. so, the portion not containing the origin is represented by the given inequation.

Region represented by x ≥ 0 and y ≥ 0: clearly, x ≥ 0 and y ≥ 0 represent the first quadrant.

The common region of the above five regions represents the solution set of the given inequations as shown below.

The common region of the above five regions represents the solution set of the given inequations


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