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:
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.
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.
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.
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.
