NCERT Solutions for Class 11 Math Chapter 6 - Linear Inequalities

Answer:

Solution set is {1, 2, 3, 4, 5, 6}.

Answer:

Solution set is {....... –3, –2, –1, 0, 1, 2, 3, 4, 5, 6}.

Answer:

We have : 5x – 3 < 3x + 1
$\Rightarrow$ 5x – 3 + 3 < 3x + 1 + 3 $\Rightarrow$ 5x < 3x + 4
$\Rightarrow$ 5x – 3x < 3x + 4 – 3x $\Rightarrow$ 2x < 4
$\Rightarrow$ x < 2.
Solution set is {......... – 3, –2, –1, 0, 1}.

Answer:

We have : 5x – 3 < 3x + 1
$\Rightarrow$ 5x – 3 + 3 < 3x + 1 + 3 $\Rightarrow$ 5x < 3x + 4
$\Rightarrow$ 5x – 3x < 3x + 4 – 3x $\Rightarrow$ 2x < 4
$\Rightarrow$ x < 2.
Solution set is (–$\infty$, 2).

Answer:

We have : 4x + 3 < 6x + 7
$\Rightarrow$ 4x + 3 – 3 < 6x + 7 – 3
$\Rightarrow$ 4x < 6x + 4 $\Rightarrow$ 4x – 6x < 6x + 4 – 6x
$\Rightarrow$ – 2x < 4 $\Rightarrow$ x > – 2. Hence, the solution set is (– 2, $\infty$).

Answer:

$\Rightarrow$ 2(5 – 2x) $\le$ x – 30 [Multiplying both sides by 6]
$\Rightarrow$ 10 – 4x $\le$ x – 30
$\Rightarrow$ – 5x $\le$ – 40 $\Rightarrow$ x $\ge$ 8.
Hence, the solution set is [8, $\infty$).

Answer:

We have : 7x + 3 < 5x + 9
$\Rightarrow$ 2x < 6 $\Rightarrow$ x < 3.
Hence, the solution set is (– $\infty$, 3),
which is shown on the number line as below :

Answer:

Answer:

Let ‘x’ be the smaller of two consecutive odd natural
numbers. Then the larger number = x + 2.
By the question, x > 10 ...(1)
and x + (x + 2) < 40 ...(2)
From (2), 2x + 2 < 40
$\Rightarrow$ 2x < 38
$\Rightarrow$ x < 19 ...(3)
Combining (1) and (3), 10 < x < 19.
Hence, the possible pairs are (11, 13), (13, 15), (15, 17) and
(17, 19).

Answer:

Let ‘x’ be the marks in the annual examination.

$\Rightarrow$ 110 + x $\ge$ 180
$\Rightarrow$ x $\ge$ 70.
Hence, the number of minimum marks = 70.

Answer:

The given inequation is 3x + 2y > 6 ...(1)
The corresponding equation is 3x + 2y = 6.
This meets x-axis at (2, 0) and y-axis at (0, 3).
Joining these points, we get the straight line 3x + 2y = 6.
Clearly the plane is divided into two regions.

Take (0, 0) as an arbitrary point.
Putting in (1), 0 + 0 > 6 i.e. 0 > 6, which is not true.
$\therefore$ O(0, 0) does not lie on the graph.
Hence, each part in the shaded region (excluding all points
on the line) determines the solution.

Answer:

We have : 3x – 6 $\ge$ 0 ...(1)
Converting it into an equation, we get :
3x – 6 = 0 $\Rightarrow$ 3x = 6
$\Rightarrow$ x = 2.
Clearly it is a st. line parallel to y-axis at a distance of 2 units on the right of origin. This divides the xy-plane into two parts.
Select the arbitrary point ; say (0, 0).

This does not satisfy (1). [_ 0 $\ge$ 2 is not true] Thus the shaded region is the required solution set.
Hence, each point on the RHS of the line (including all points on the line) determines the solution of the given inequation.

Answer:

We have : y < 2 ...(1)
Converting it into an equation, we get : y = 2.
Clearly it is a st. line parallel to x-axis at a distance of 2 units above the origin.
This divides the XY-plane into two parts.
Select the arbitrary point ; say (0, 0).

This does not satisfy (1) [_ 0 < 2 is not true] Thus the shaded region is the required solution set.
Hence, each point below the line (excluding all points on the line) determines the solution of the given inequation.

Answer:

The given inequations are x + y $\ge$ 5 ...(1)
and x – y $\le$ 3 ...(2)
The graph of the equation x + y = 5 is the line joining the points A (5, 0) and B (0, 5).
The solution of the inequation (1) is represented by the shaded region away from the origin (including points as the line).
The graph of the equation x – y = 3 is the line joining the points C(3, 0) and D(0, –3)

The solution of the inequation (2) is represented by the shaded region towards the origin including the points on the line.
Hence, the double shaded region, common to (1) and (2) is the required solution.

Answer:

The given inequalities are :
5x + 4y $\le$ 40 ...(1)
x $\ge$ 2 ...(2)
and y $\ge$ 3 ...(3)
Let us draw the graphs of 5x + 4y = 40, x = 2 and y = 3.
The st. line 5x + 4y = 40 passes through (8, 0) and (0, 10).
Let AB be this line

Shaded portion is towards the origin.
x = 2 and y = 3 are also shown.
Hence, the feasible region (shown shaded) is the required solution set.

Answer:

The given inequalities are :
8x + 3y $\le$ 100 ...(1)
x $\ge$ 0 ...(2)
and y $\ge$ 0 ...(3)
Draw the graph of 8x + 3y = 100.

The shaded portion is towards the origin.
x = 0 and y = 0 are also shown.
Hence, the feasible region (shown shaded) is the required solution set.

Answer:

We have : x + 2y $\le$ 8 ...(1)
2x + y $\le$ 8 ...(2)
x $\ge$ 0, y $\ge$ 0 ...(3)
Draw the lines x + 2y = 8, 2x + y = 8 and x = 0, y = 0.
The line x + 2y = 8 meets x = axis at A (8, 0)
and y-axis at B(0, 4),
This is the line AB.
Shaded portion is towards the origin.
The line 2x + y = 8 meets x-axis at C(4, 0)
and y-axis at (0, 8).
This is the line CD.
Shaded portion is towards the origin.
x = 0, y = 0 are also shown.

Hence, the feasible region (shown shaded) is the required solution set.

Answer:

Here – 8 $\le$ 5x – 3 < 7
$\Rightarrow$ – 8 + 3 $\le$ 5x – 3 + 3 < 7 + 3 [Adding 3]
$\Rightarrow$ – 5 $\le$ 5x < 10
$\Rightarrow$ – 1 $\le$ x < 2. [Dividing by 5]
Hence, the solution set is [– 1, 2). < Here – 8 $\le$ 5x – 3 < 7
$\Rightarrow$ – 8 + 3 $\le$ 5x – 3 + 3 < 7 + 3 [Adding 3]
$\Rightarrow$ – 5 $\le$ 5x < 10
$\Rightarrow$ – 1 $\le$ x < 2. [Dividing by 5]
Hence, the solution set is [– 1, 2)./p>

Answer:

Answer:

Answer:

Answer:

Let ‘x’ be the number of litres of 30% acid solution.
Then total mixture = (x + 600) litres.
By the question,
30% of x + 12% of (600) > 15$°$ of (x + 600)
and 30% of x + 12% of (600) < 18$°$ of (x + 600)

$\Rightarrow$30x + 7200 > 15x + 9000
and 30x + 7200 < 18x + 10800
$\Rightarrow$ 15x > 1800 and 12x < 3600
$\Rightarrow$ x > 120 and x < 300
$\Rightarrow$ 120 < x < 300.
Hence, the number of litres of 30% solution of acid will have to be more than 120 but less than 300.

Answer:

We have : 24x < 100 ...(1)

Hence, (1) is true when :
Solution set is {1, 2, 3, 4}.

Answer:

We have : 24x < 100 ...(1)

Hence, (1) is true when :
Solution set is
{............ – 3, – 2, – 1, 0, 1, 2, 3, ...............}.

Answer:

We have : – 12x > 30 ...(1)

Hence, (1) is true when :
Solution set is $\mathrm{\varphi }$ i.e. there is no solution.

Answer:

We have : – 12x > 30 ...(1)

Hence, (1) is true when :
Solution set is {............, – 4, – 3}.

Answer:

We have : 5x – 3 < 7
$\Rightarrow$ 5x < 7 + 3
[Adding 3 on both sides]
$\Rightarrow$ 5x < 10 $\Rightarrow$ x < 2.
[Dividing by 2]
The solution set is {........ – 2, – 1, 0, 1}.

Answer:

We have : 5x – 3 < 7
$\Rightarrow$ 5x < 7 + 3
[Adding 3 on both sides]
$\Rightarrow$ 5x < 10 $\Rightarrow$ x < 2.
[Dividing by 2]
The solution set is (– $\infty$, 2).

Answer:

We have : 3x + 8 > 2
$\Rightarrow$ 3x > 2 – 8
$\Rightarrow$ 3x > – 6
$\Rightarrow$ x > – 2. [Dividing by 2]
The solution set is {– 1, 0, 1, 2, 3, ........}.

Answer:

We have : 3x + 8 > 2
$\Rightarrow$ 3x > 2 – 8
$\Rightarrow$ 3x > – 6
$\Rightarrow$ x > – 2. [Dividing by 2]
The solution set is (– 2, $\infty$).

Answer:

We have : 4x + 3 < 6x + 7
$\Rightarrow$ 4x – 6x < 7 – 3 $\Rightarrow$ – 2x < 4.
Dividing by – 2, x > – 2.
Hence, the solution set is (– 2, $\infty$).

Answer:

We have : 3x – 7 > 5x – 1
$\Rightarrow$ 3x – 5x > 7 – 1 $\Rightarrow$ – 2x > 6
Dividing by – 2, x < – 3.
Hence, the solution set is (– $\infty$, – 3).

Answer:

We have : 3 (x – 1) $\le$ 2 (x – 3)
$\Rightarrow$ 3x – 3 $\le$ 2x – 6 $\Rightarrow$ 3x – 2x $\le$ – 6 + 3
$\Rightarrow$ x $\le$ – 3.
Hence, the solution set is (– $\infty$, – 3]

Answer:

We have : 3 (2 – x) $\ge$ 2 (1 – x)
$\Rightarrow$ 6 – 3x $\ge$ 2 – 2x $\Rightarrow$ – 3x + 2x $\ge$ 2 – 6
$\Rightarrow$ – x $\ge$ – 4 $\Rightarrow$ x $\le$ 4
[Dividing by – 1]
Hence, the solution set is (– $\infty$, 4].

Answer:

Answer:

Answer:

Answer:

Answer:

We have : 2 (2x + 3) – 10 < 6 (x – 2)
$\Rightarrow$ 4x + 6 – 10 < 6x – 12 $\Rightarrow$ 4x – 4 < 6x – 12
$\Rightarrow$ 4x – 6x < 4 – 12 $\Rightarrow$ – 2x < – 8
$\Rightarrow$ x > 4. [Dividing by – 2]
Hence, the solution set is (4, $\infty$).

Answer:

We have : 37 – (3x + 5) $\ge$ 9x – 8 (x – 3)
$\Rightarrow$37 – 3x – 5 $\ge$ 9x – 8x + 24
$\Rightarrow$ 32 – 3x $\ge$ x + 24 $\Rightarrow$ – 3x – x $\ge$ 24 – 32
$\Rightarrow$ – 4x $\ge$ – 8 $\Rightarrow$ x $\le$ 2. [Dividing by – 4]
Hence, the solution set is (– $\infty$, 2].

Answer:

Answer:

Answer:

We have : 3x – 2 < 2x + 1
$\Rightarrow$ 3x – 2x < 1 + 2 $\Rightarrow$ x < 3.
Hence, the solution set is (– $\infty$, 3).

Answer:

We have : 5x – 3 $\ge$ 3x – 5
$\Rightarrow$ 5x – 3x $\ge$ 3 – 5
$\Rightarrow$ 2x $\ge$ – 2
$\Rightarrow$ x $\ge$ – 1. [Dividing by 2]
Hence, the solution set is [– 1, $\infty$).

Answer:

We have : 3 (1 – x) < 2 (x + 4)
$\Rightarrow$ 3 – 3x < 2x + 8 $\Rightarrow$ – 3x – 2x < 8 – 3
$\Rightarrow$ – 5x < 5 $\Rightarrow$ x > – 1.
[Dividing by – 1]
Hence, the solution set is (– 1, $\infty$).

Answer:

Answer:

Let ‘x’ be the marks obtained by Ravi in the third test.

$\Rightarrow$ 145 + x $\ge$ 180
$\Rightarrow$ x $\ge$ 35.
Hence, Ravi must obtain a minimum of 35 marks to get an average of at least 60 marks.

Answer:

Let Sunita obtain ‘x’ marks in the fifth examination.

$\Rightarrow$ 368 + x $\ge$ 450 $\Rightarrow$ x $\ge$ 82.
Hence, minimum marks that Sunita must obtain = 82.

Answer:

Let ‘x’ be the smaller of two consecutive positive integers. Then the other number = x + 2.
By the question, x < 10 ...(1)
and x + (x + 2) > 11 ...(2)
From (2), 2x + 2 > 11 $\Rightarrow$ 2x > 9

If one number is 5, then the other is 7.
If one number is 7, then the other is 9.
Hence, the possible pairs are (5, 7), (7, 9).

Answer:

Let ‘x’ be the smaller of the two positive even numbers,
Then the other number is x + 2.
By the question,
x > 5 ...(1)
and x + (x + 2) < 23
i.e. 2x + 2 < 23 ...(2)
If the sum is less than 23, then it is also less than 24,
$\therefore$ 2x + 2 < 24 $\Rightarrow$ 2x < 22
$\Rightarrow$ x < 11.
Thus x may be 6, 8, 10.
Hence, the possible pairs are (6, 8), (8, 10) and (10, 12).

Answer:

Let ‘x’ be the shortest side.
Then longest side = 3x and third side = 3x – 2.
Now x + 3x + (3x – 2) $\ge$ 61
$\Rightarrow$ 7x $\ge$ 63
$\Rightarrow$ x $\ge$ 9.
Hence, the minimum length of the shortest side = 9 cm.

Answer:

Let ‘x’ be the length of the shortest board.
Then second length = x + 3
and third length = 2x.
By the question,
2x $\ge$ (x + 3) + 5
$\Rightarrow$ x $\ge$ 8 ...(1)
Also x + (x + 3) + 2x $\le$ 91
$\Rightarrow$ 4x $\le$ 88 $\Rightarrow$ x $\le$ 22 ...(2)
From (1) and (2),
8 $\le$ x $\le$ 22.
Hence, the shortest board is at least 8 cm long but not more than 22 cm long.

Answer:

The given inequality is x + y < 5 ... (1)
The corresponding equation is x + y = 5 ...(2)
This passes through the points A (5, 0) and B (0, 5).

(2) represents the line AB.
Putting x = 0, y = 0 in (1), we get :
0 + 0 < 5 i.e. 0 < 5,
which is true.
Hence, half-plane I is the solution region of the given inequality.

Answer:

The given inequality is 2x + y $\ge$ 6 ...(1)
The corresponding equation is 2x + y = 6 ...(2)
This passes through the points A (3, 0) and B (0, 6).
(2) represent the line AB.
Putting x = 0, y = 0 in (1) we get :

0 + 0 $\ge$ 6 i.e. 0 $\ge$ 6, which is not true.
Hence, half-Plane – II is the solution region of the given inequality.

Answer:

The given inequality is 3x + 4y $\le$ 12 ...(1)
The corresponding equation is 3x + 4y = 12 ...(2)
This passes through the points A (4, 0) and B (0, 3).
(2) represents the line AB.

Putting x = 0, y = 0 in (1), we get :
0 + 0 $\le$ 12, which is true.
Hence, half-plane I is the solution region of the given inequality.

Answer:

The given inequality is y + 8 $\ge$ 2x ...(1)
The corresponding equation is y + 8 = 2x ...(2)
This passes through the points A (4, 0) and B (0, – 8).
(2) represents the line AB.
Putting x = 0, y = 0 in (1), we get :
0 + 8 $\ge$ 0 i.e. 8 $\ge$ 0,
which is true.

Hence, half-plane I is the solution region of the given inequality.

Answer:

The given inequality is x – y $\le$ 2 ...(1)
The corresponding equation is x – y = 2 ...(2)
This passes through A (2, 0) and B (0, – 2)
(2) represents the line AB.

Putting x = 0, y = 0 in (1)
0 – 0 $\le$ 2 $\Rightarrow$ 0 < 2, which is true.
Hence, half-plane I is the solution region of the given inequality.

Answer:

The given inequality is 2x – 3y > 6 ...(1)
The corresponding equation is 2x – 3y = 6 ...(2)
This passes through A (3, 0) and B (0, – 2)
(2) represents the line AB.

Putting x = 0, y = 0 in (1), we get :
0 – 0 > 6 i.e. 0 > 6, which is not true.
Hence, half-plane II is the solution region of the given inequality.

Answer:

The given inequality is – 3x + 2y $\ge$ – 6 ...(1)
The corresponding equation is
– 3x + 2y = – 6
i.e. 3x – 2y = 6 ...(2)
This passes through A (2, 0) and B (0, – 3).
(2) represents the line AB.

Putting x = 0, y = 0 in (1) we get :
– 0 + 0 $\ge$ – 6 i.e. 0 $\ge$ – 6,
which is true.
Hence, half-plane I is the solution region of the given inequality.

Answer:

The given inequality is 3y – 5x < 30 ...(1)
The corresponding equation is 3y – 5x = 30 ...(2)
This passes through A (– 6, 0) and B (0, 10).
(2) represents the line AB.
Putting x = 0, y = 0 in (1), we get
0 – 0 < 30 i.e. 0 < 30, which is true.
Hence, half plane I is the solution region of the given inequality.

Answer:

The given inequality is y < – 2 ...(1)
The corresponding equation is y = – 2 ...(2)
This represents the line AB

Putting x = 0, y = 0 in (1), 0 < – 2, which is not true.
Hence, half-plane II is the solution region of the given inequality.

Answer:

The given inequality is x > – 3 ...(1)
The corresponding equation is x = – 3 ...(2)
This represents the line AB.
Putting x = 0, y = 0 in (1)
0 > – 3, which is true.

Hence, half-plane I is the solution region of the given inequality.

Answer:

x $\ge$ 3 represents the region on the right side of x = 3 including x = 3,
y $\ge$ 2 represents the region above y = 2 including y = 2.

Answer:

Let us draw the graphs of 3x + 2y = 12, x = 1, y = 2.
The st. line 3x + 2y = 12 passes through (4, 0) and (0, 6).
Let AB be this line.

Shaded portion is towards the origin.
[_ 0 $\le$ 12 is true]
x = 1, y = 2, are also shown.
Hence, the feasible region (shown shaded) is the solution set of the given system of inequations.

Answer:

The given inequalities are :
2x + y $\ge$ 6 ....(1)
and 3x + 4y $\le$ 12 ....(2)
The corresponding equations are :
2x + y = 6 ....(1)′
and 3x + 4y = 12 ....(2)′
(1)′ passes through A (3, 0) and B (0, 6).
$\therefore$ (1)′ represents the line AB.

(2)′ passes through C (4, 0) and D (0, 3).
$\therefore$ (2)′ represents the line CD.
Putting x = 0, y = 0 in (1),
0 + 0 $\ge$ 6 i.e. 0 $\ge$ 6, which is not true.
$\therefore$ Origin does not lie in this region.
$\therefore$ The region lying above AB and all points on AB represents (1).
Again putting x = 0, y = 0, in (2),
0 + 0 $\le$ 12 i.e. 0 $\le$ 12, which is true.
$\therefore$ The region lying below CD (towards origin) and all points lying on CD.
Hence, the common region (shown shaded) is the solution of (1) and (2).

Answer:

The given inequalities are
x + y > 4 ....(1)
and 2x – y > 0 ....(2)
The corresponding equations are
x + y = 4 ....(1)′
and 2x – y = 0 ....(2)′
(1)′ passes through A (4, 0) and B (0, 4)
$\therefore$ (1)′ represents the line AB.
Putting x = 0, y = 0 in (1) 0 + 0 > 4 i.e. 0 > 4, which is not true.
$\therefore$ Origin does not lie in the region
The line 2x – y = 0 passes thro’ O (0, 0) and P (1, 2).
This represents the line CD.
Putting x = 1, y = 0 in (2) 0 > 0, which is true.
$\therefore$ (1, 0) lies in the region.

The common region is shown as shaded in the figure.

Answer:

The given inequalities are :
2x – y > 1 ....(1)
and x – 2y < – 1 ....(2)
The corresponding equations are :
2x – y = 1 ...(1)′
and x – 2y = – 1 ...(2)′

$\therefore$ (1)′ represents the line AB.
Putting x = 0, y = 0 in (1), 0 – 0 > 1
i.e. 0 > 1, which is not true.
$\therefore$ Origin does not lie in the region.

$\therefore$ (2)′ represents the line CD.
Putting x = 0, y = 0 in (2), 0 – 0 < – 1
i.e. 0 < – 1, which is not true.
$\therefore$ Origin does not lie in the region.
The common region is as shown shaded in the figure.

Answer:

The given inequalities are :
x + y $\le$ 6 ....(1)
and x + y $\ge$ 4 ....(2)
The corresponding equations are :
x + y = 6 ....(1)′
and x + y = 4 ....(2)′
(1)′ passes thro’ A (6, 0) and B (0, 6).
$\therefore$ (1)′ represents the line AB.
Putting x = 0, y = 0 in (1), 0 + 0 $\le$ 6
i.e. 0 $\le$ 6, which is true.
$\therefore$ Origin lies in the region.
Again (2)′ passes thro’ C (4, 0) and D (0, 4).
$\therefore$ (2)′ represents the line CD.
Putting x = 0, y = 0 in (2), 0 + 0 $\ge$ 4
i.e. 0 $\ge$ 4, which is not true.
$\therefore$ Origin does not lie in the region.
The common region is shown as shaded in the figure.

Answer:

The given inequalities are
2x + y $\ge$ 8 ....(1)
and x + 2y $\ge$ 10 ....(2)
The corresponding equations are
2x + y = 8 ....(1)′
and x + 2y = 10 ....(2)′
(1)′ passes thro’ A (4, 0) and B (0, 8).
$\therefore$ (1)′ represents the line AB.
Putting x = 0, y = 0 in (1), 0 + 0 $\ge$ 8
i.e. 0 $\ge$ 8, which is not true.
$\therefore$ Origin does not lie in the region.
Again (2)′ passes thro’ C (10, 0) and D (0, 5).
$\therefore$ (2)′ represents the line CD.
Putting x = 0, y = 0 in (2), 0 + 0 $\ge$ 10 i.e. 0 $\ge$ 10,
which is not true.
$\therefore$ Origin does not lie in the region.
The common region is shown as shaded in the figure.

Answer:

The given inequalities are
x + y $\le$ 9 ....(1)
y > x ....(2)
and x $\ge$ 0 ....(3)
The corresponding equation of (1) is
x + y = 9 ....(1)′
(1)′ passes thro’ A (9, 0) and B (0, 9).
$\therefore$ (1)′ represents the line AB.
Putting x = 0, y = 0 in (1), 0 + 0 $\le$ 9
i.e. 0 $\le$ 9, which is true.
$\therefore$ Origin lies in the region.
Now corresponding equation of (2) is y = x.
This passes through (0, 0), (1, 1).
$\therefore$ (2)′ represents the line CD.
Putting x = 0, y = 1 in (2), y > x
i.e. y – x > 0,
1 – 0 > 0 i.e. 1 > 0, which is true.
$\therefore$ (0, 1) lies in the region
Lastly x > 0 represents the right of y-axis.
The common region is shown as shaded in the figure.

Answer:

Let us draw the graphs of 5x + 4y = 20, x = 1, y = 2.
The st. line 5x + 4y = 20 passes through (4, 0) and (0, 5).
Let AB be this line.

Shaded portion is towards the origin
[∵ 0 $\le$ 20 is true]
Region above y = 2 represents the inequality y $\ge$ 2 including the points on the line. Region on the right of x = 1
represents the inequality x $\ge$ 1, including the points on the line. The double shaded area bounded by ΔPQR is the required solution.

Answer:

The given inequalities are
3x + 4y $\le$ 60 ....(1)
x + 3y $\le$ 30 ....(2)
and x $\ge$ 0, y $\ge$ 0 ....(3)
The corresponding equations are
3x + 4y = 60 ....(1)′
x + 3y = 30 ....(2)′
and x = 0, y = 0 ....(3)′
(1)′ passes thro’ A (20, 0) and B (0, 15).
$\therefore$ (1)′ represents the line AB.
Putting x = 0, y = 0 in (1), 0 + 0 $\le$ 60
i.e. 0 $\le$ 60, which is true.
$\therefore$ Origin lies in the region.
Again (2)′ passes through C (30, 0) and D (0, 10).
$\therefore$ (2)′ represents the line CD.
Putting x = 0, y = 0 in (2), 0 + 0 $\le$ 30 i.e. 0 $\le$ 30, which is true.
$\therefore$ Origin lies in the region.
Now x $\ge$ 0 is the region on and right of y-axis.
And y $\ge$ 0 is the region on and above of x-axis
The common region is shown as shaded in the figure.

Answer:

The given inequalities are :
2x + y $\ge$ 4 ....(1)
x + y $\le$ 3 ....(2)
and 2x – 3y $\le$ 6 ....(3)
The corresponding equations are :
2x + y = 4 ....(1)′
x + y = 3 ....(2)′
and 2x – 3y = 6 ....(3)′
(1)′ passes thro’ A (2, 0) and B (0, 4).
$\therefore$ (1)′ represents the line AB.
Putting x = 0, y = 0 in (1), 0 + 0 $\ge$ 4 i.e. 0 $\ge$ 4, which is not true.
$\therefore$ Origin does not lie in the region.
(2)′ passes thro’ C (3, 0) and D (0, 3).
$\therefore$ (2)′ represents the line CD.
Putting x = 0, y = 0 in (2), 0 + 0 $\le$ 3 i.e. 0 $\le$ 3, which is true.
$\therefore$ Origin lies in the region.
(3)′ passes through C (3, 0) and E (0, – 2)
(3)′ represents the line CE.
Putting x = 0, y = 0 in (3) 0 – 0 $\le$ 6 i.e. 0 $\le$ 6, which is true.
$\therefore$ Origin lies in the region.
The common region is shown as shaded in the figure.

Answer:

The given inequalities are
x – 2y $\le$ 3 ...(1)
3x + 4y $\ge$ 12 ...(2)
and x $\ge$ 0, y $\ge$ 1 ...(3)
The corresponding equations are
x – 2y = 3 ...(1)′
3x + 4y = 12 ...(2)′
and x = 0, y = 1 ...(3)′

$\therefore$ (1)′ represents the line AB.
Putting x = 0, y = 0 in (1), 0 – 0 $\le$ 3 i.e. 0 $\le$ 3, which is true.
$\therefore$ Origin lies in the region.
(2)′ passes thro’ C (4, 0) and D (0, 3).
$\therefore$ (2)′ represents the line CD.
Putting x = 0, y = 0 in (2), 0 + 0 $\ge$ 12 i.e. 0 $\ge$ 12, which is not true.
$\therefore$ Origin does not lies in the region.

Now x $\ge$ 0 represents y-axis and right of y-axis.
And y $\ge$ 1 represents the line y = 1 and above it.
The common region is shown as shaded in the figure.

Answer:

The given inequalities are
4x + 3y $\le$ 60 ...(1)
y $\ge$ 2x ...(2)
x $\ge$ 3 ...(3)
and x, y $\ge$ 0 ...(4)
The corresponding equations are
4x + 3y = 60 ...(1)′
y = 2x ...(2)′
x = 3 ...(3)′
(1)′ passes thro’ A (15, 0) and B (0, 20).
$\therefore$ (1)′ represents the line AB.
Putting x = 0, y = 0 in (1), 0 + 0 $\le$ 60 i.e. 0 $\le$ 60, which is true.
$\therefore$ Origin lies in the region.
Now y = 2x passes thro’ O (0, 0) and (5, 10).
$\therefore$ (2)′ represents the line CD.
Putting x = 0, y = 5 in (2), 5 $\ge$ 0,
which is true.
$\therefore$ (0, 5) lies in the region.
x $\ge$ 3 represents the line x = 3 and right of it.
x, y $\ge$ 0 represents the region in first quadrant.

The common region is shown as shaded in the figure.

Answer:

The given inequalities are
3x + 2y $\le$ 150 ...(1)
x + 4y $\le$ 80 ...(2)
x $\le$ 15 ...(3)
y $\ge$ 0 ...(4)
The corresponding equations are
3x + 2y = 150 ...(1)′
x + 4y = 80 ...(2)′
x = 15 ...(3)′
and y = 0 ...(4)′
(1)′ passes thro’ A (50, 0) and B (0, 75).
$\therefore$ (1)′ represents the line AB.
Putting x = 0, y = 0 in (1), 0 + 0 $\le$ 150 i.e. 0 $\le$ 150, which is true.
$\therefore$ Origin lies in the region.
(2)′ passes thro’ C (80, 0) and D (0, 20).
$\therefore$ (2)′ represents the line CD.
Putting x = 0, y = 0 in (2), 0 + 0 $\le$ 80 i.e. 0 $\le$ 80, which is true.
$\therefore$ Origin lies in the region.
Now x $\le$ 15 represents the line x = 15 and left of it.
And y $\ge$ 0 represents x-axis and above it.

The common region is shown as shaded in the figure.

Answer:

The given inequalities are
x + 2y $\le$ 10 ....(1)
x + y $\ge$ 1 ....(2)
x – y $\le$ 0 ....(3)
x $\ge$ 0, y $\ge$ 0 ....(4)
The corresponding equations are
x + 2y = 10 ....(1)′
x + y = 1 ....(2)′
x – y = 0 ....(3)′
and x = 0, y = 0 ....(4)′
(1)′ passes thro’ A (10, 0) and B (0, 5)
$\therefore$ (1)′ represents the line AB.
Putting x = 0, y = 0, 0 + 0 $\le$ 10 i.e. 0 $\le$ 10, which is true.
$\therefore$ Origin lies in the region.
(2)′ passes thro’ C (1, 0) and D (0, 1)
$\therefore$ (2)′ represents the line CD.
Putting x = 0, y = 0, 0 + 0 $\ge$ 1 i.e. 0 $\ge$ 1, which is nottrue.
$\therefore$ Origin does not lie in the region.
(3)′ i.e. x – y = 0 passes thro’ (0, 0) and (1, 1).
$\therefore$ (3)′ represents the line EF.
Putting x = 0, y = 1, 0 – 1 $\le$ 0, i.e. – 1 $\le$ 0 which is true.
$\therefore$ (0, 1) lies in the region.
And x $\ge$ 0, y $\ge$ 0 represents Ist-quadrant.

The common region is shown as shaded in the figure.