Chapter 6: Linear Inequalities Exercise 6.3

Given 2x + y ≥ 4,

Putting values of x = 0 and y = 0 in the equation one by one, we get values of

y = 4 and x =2

The required points are (0, 4) and (2, 0)

Checking for origin (0, 0)

0 ≥ 4, this is not true

Hence, the origin doesn’t lie in the solution area of the line graph. The solution area would be given by the right side of the line graph.

x + y ≤ 3,

Putting values of x = 0 and y = 0 in the equation one by one, we get values of

y = 3 and x = 3

The required points are (0, 3) and (3, 0)

Checking for the origin (0, 0)

0 ≤ 3, this is true

Hence, the solution area would include the origin and hence would be on the left side of the line graph.

2x – 3y ≤ 6

Putting values of x = 0 and y = 0 in the equation one by one, we get values of

y = -2 and x = 3

The required points are (0, -2), (3, 0)

Checking for the origin (0, 0)

0 ≤ 6 this is true

So the origin lies in the solution area, and the area would be on the left of the line graph.

Hence, the shaded area in the graph is the required solution area for the given inequalities.

Given, x – 2y ≤ 3

Putting values of x = 0 and y = 0 in the equation one by one, we get values of

y = -3/2 = -1.5 and x = 3

The required points are (0, -1.5) and (3, 0)

Checking for the origin (0, 0)

0 ≤ 3, this is true.

Hence, the solution area would be on the left of the line graph

3x + 4y ≥ 12,

Putting values of x = 0 and y = 0 in the equation one by one, we get values of

y = 3 and x = 4

The required points are (0, 3) and (4, 0)

Checking for the origin (0, 0)

0 ≥ 12, this is not true

So the solution area would include the origin, and the required solution area would be on the right side of the line graph.

We have x ≥ 0,

For all the values of y, the value of x would be the same in the given inequality, which would be the region above the x-axis on the graph.

Consider, y ≥ 1

For all the values of x, the value of y would be the same in the given inequality.

The solution area of the line would not include the origin, as 0 ≥ 1 is not true.

The solution area would be on the left side of the line graph.

The shaded area in the graph is the required solution area which satisfies all the given inequalities at the same time.

Given, 4x + 3y ≤ 60,

Putting values of x = 0 and y = 0 in the equation one by one, we get values of

y = 20 and x = 15

The required points are (0, 20) and (15, 0)

Checking for the origin (0, 0)

0 ≤ 60, this is true.

Hence, the origin would lie in the solution area. The required area would be included on the left of the line graph.

We have y ≥ 2x,

Putting values of x = 0 and y = 0 in the equation one by one, we get values of

y = 0 and x = 0

Hence, the line would pass through the origin.

To check which side would be included in the line graph solution area, we would check for the point (15, 0)

⇒ 0 ≥ 15, this is not true, so the required solution area would be to the left of the line graph.

Consider, x ≥ 3,

For any value of y, the value of x would be the same.

Also, the origin (0, 0) doesn’t satisfy the inequality as 0 ≥ 3

So the origin doesn’t lie in the solution area; hence the required solution area would be the right of the line graph.

We have x, y ≥ 0

Since given both x and y are greater than 0

∴ the solution area would be in the first I^st quadrant only.

The shaded area in the graph shows the solution area for the given inequalities

Given, 3x + 2y ≤ 150

Putting values of x = 0 and y = 0 in the equation one by one, we get values of

y = 75 and x = 50

The required points are (0, 75) and (50, 0)

Checking for the origin (0, 0)

0 ≤ 150, this is true

Hence, the solution area for the line would be on the left side of the line graph, which would include the origin too.

We have x + 4y ≤ 80,

Putting values of x = 0 and y = 0 in the equation one by one, we get values of

y = 20 and x = 80

The required points are (0, 20) and (80, 0)

Checking for the origin (0, 0)

0 ≤ 80, this is also true, so the origin lies in the solution area.

The required solution area would be toward the left of the line graph.

Given x ≤ 15,

For all the values of y, x would be the same

Checking for the origin (0, 0)

0 ≤ 15, this is true, so the origin would be included in the solution area. The required solution area would be towards the left of the line graph.

Consider y ≥ 0, x ≥ 0

Since x and y are greater than 0, the solution would lie in the 1^st quadrant.

The shaded area in the graph satisfies all the given inequalities and hence is the solution area for given inequalities.

Given, x + 2y ≤ 10,

Putting values of x = 0 and y = 0 in the equation one by one, we get values of

y = 5 and x = 10

The required points are (0, 5) and (10, 0)

Checking for the origin (0, 0)

0 ≤ 10, this is true.

Hence, the solution area would be toward the origin, including the same. The solution area would be toward the left of the line graph.

We have x + y ≥ 1,

Putting values of x = 0 and y = 0 in the equation one by one, we get values of

y = 1 and x = 1

The required points are (0, 1) and (1, 0)

Checking for the origin (0, 0)

0 ≥ 1, this is not true.

Hence, the origin would not be included in the solution area. The required solution area would be toward the right of the line graph.

Consider x – y ≤ 0,

Putting values of x = 0 and y = 0 in the equation one by one, we get values of

y = 0 and x = 0

Hence, the origin would lie on the line.

To check which side of the line graph would be included in the solution area, we would check for the (10, 0)

10 ≤ 0, which is not true hence the solution area would be on the left side of the line graph.

Again we have x ≥ 0, y ≥ 0

Since both x and y are greater than 0, the solution area would be in the 1^st quadrant.

Hence, the solution area for the given inequalities would be the shaded area of the graph satisfying all the given inequalities.

Exercise 6.1 Solutions 26 Questions

Exercise 6.2 Solutions 10 Questions

Miscellaneous Exercise on Chapter 6 Solutions 14 Questions

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