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Question-1

Solve the inequation:  3x – 7 > x + 3

Solution:
3x – 7 > x + 3
3x – 7 + 7 > x + 3 + 7
3x > x + 10
3x – x > x + 10 - x
2x > 10
x > 5

Question-2

Solve the inequation:   x + 12 < 4x – 2

Solution:
x + 12 < 4x – 2
x + 12 - 12< 4x – 2 – 12
x < 4x – 14
x – 4x < 4x – 14 – 4x
- 3x < -14
- 3x/-3 > -14/-3
x > 14/3

Question-3

Solve the inequation:  4x – 7 < 3 – x

Solution:
4x – 7 < 3 – x
4x – 7 + 7 < 3 – x + 7
4x < 10 – x
4x + x < 10 – x + x
5x < 10
x < 2

Question-4

Solve the inequation:  3x + 17 2(1 - x)

Solution:
3x + 17 2(1 - x)
3x
- 2x - 15
3x + 2x - 2x – 15 + 2x
5x
– 15
5x/5
– 15/5
x
– 3

Question-5

Solve the inequation:   –2x + 6 5x – 4

Solution:
–2x + 6 5x – 4
–2x 5x – 10
–2x – 5x 5x – 10 – 5x
–7x – 10
–7x/-7 – 10/-7
x 10/7

Question-6

Solve the inequation:   –(x - 3) + 4 > - 2x + 5

Solution:
–(x - 3) + 4 > - 2x + 5
–x + 7 > - 2x + 5
–x + 7 – 7 > - 2x + 5 – 7
–x > - 2x - 2
x < 2x + 2
x – 2x < 2x + 2 – 2x
- x < 2
  x > -2

Question-7

Solve the inequation:   2(2x + 3) – 10 < 6(x - 2)

Solution:
2(2x + 3) – 10 < 6(x - 2)
4x + 6 – 10 < 6x - 12
4x – 4 < 6x - 12
4x – 4 + 4 < 6x – 12 + 4
4x < 6x - 8
4x – 6x < 6x – 8 – 6x
- 2x < – 8
- 2x/-2 > -8 /-2
x > 4

Question-8

Solve the inequation:  2-3x 2(x + 6)

Solution:
2 - 3x 2x + 12
2 - 3x – 2
2x + 12 – 2
- 3x
2x + 10
- 3x – 2x
2x + 10 – 2x
- 5x
10
- 5x/-5
10/-5
x
-2

Question-9

Solve the inequation:  37 – (3x + 5) 9x – 8(x - 3)

Solution:
37 – 3x - 5 9x – 8x + 24
32 – 3x
x + 24
32 – 3x - 32
x + 24 – 32
– 3x
x - 8
– 3x – x
x – 8 – x
- 4x
– 8
- 4x/-4
– 8/-4
x
2

Question-10

Solve the inequation:   +

Solution:
+
10x + 3x 39
13x 39
13x/13 39/13
x 3

Question-11

Solve the inequation: - 3

Solution:
- 3
2(4 + 2x)
3x - 18 (Multiplying by 6 both sides)
8 + 4x
3x – 18
8 + 4x – 8 3x – 18 – 8
4x
3x – 26
4x – 3x
3x – 26 – 3x
x
– 26

Question-12

Solve the inequation:

Solution:

9(x – 2)
25(2 - x) (Multiplying by 15 both sides)
9x – 18 50 - 25x
9x – 18 + 18
50 - 25x + 18
9x
68 - 25x
9x + 25x 68 - 25x + 25x
34x
68
34x/34
68/34
x
2

Question-13

Solve the inequation:<

Solution:
<
15x < 20(5x - 2) – 12(7x - 3) (Multiplying by 60 both sides)

15x < 100x - 40 – 84x + 36
15x < 16x - 4
15x – 16x < 16x – 4 – 16x
– x < – 4
x > 4

Question-14

Solve the inequation:

Solution:

2(5 – 2x) x – 30 (Multiplying by 6 both sides)
10 – 4x
x – 30
10 – 4x – 10 x – 30 – 10
– 4x
x – 40
– 4x –x
x – 40 –x
– 5x
– 40
– 5x/ -5
– 40/ -5
x
8

Question-15

Solve the inequation:

Solution:

(Multiplying by 6 both sides)
3(3x + 20) 10(x - 6) (Multiplying by 5 both sides)
9x + 60
10x - 60
9x + 60 – 60
10x - 60 – 60
9x 10x – 120
9x – 10x
10x – 120 - 10x
-x
– 120
  x
120

Question-16

Solve the following system of inequations: x –2 > 0, 3x < 18

Solution:
x –2 > 0 .
3x < 18  
x –2 > 0
x > 2...................(1)

3x < 18
x < 6 ..................(2)

From (1) and (2), solutions of the given system are, therefore, given by 2 < x < 6
Hence the solution of the system is 2 < x < 6.

Question-17

Solve the following system of inequations:5x + 1 > -24, 5x – 1 < 24

Solution:
5x + 1 > -24                
5x – 1 < 24
5x + 1 > -24
5x > - 25
x > - 5 .....................(1)

5x – 1 < 24
5x < 25
x < 5.......................(2)

From (1) and (2), solutions of the given system are, therefore, given by -5 < x < 5
Hence the solution of the system of is –5 < x < 5.

Question-18

Solve the following system of inequations:x + 2 5, 3x – 4 > - 2 + x

Solution:
x + 2 5                   
3x – 4 > - 2 + x    
x + 2
5
x
3 .......................(1)

3x – 4 > - 2 + x
3x > 2 + x
2x > 2
x > 1 .....................(2)

From (1) and (2), solutions of the given system are, therefore, given by 1 < x
3.
Hence the solution of the system of is 1 < x
3.

Question-19

Solve the following system of inequations: 4x + 5 > 3x, -(x + 3) + 4 –2x + 5

Solution:
4x + 5 > 3x                                      
-(x + 3) + 4
–2x + 5   
4x + 5 > 3x
4x > 3x - 5
x > - 5 ....................(1)

-(x + 3) + 4
–2x + 5
-x - 3 + 4
–2x + 5
-x + 1
–2x + 5
-x
–2x + 4
x
4 ......................(2)

From (1) and (2), solutions of the given system are, therefore, given by -5 < x
4
Hence the solution of the system of is -5 < x
4.

Question-20

Solve the following system of inequations: ,

Solution:
                         
                    


16x – 27 < 12x + 9             (Multiplying by 12 both sides)
16x < 12x + 36
4x< 36
x< 9 .........................(1)


2(7x - 1) – (7x + 2) > 6x     (Multiplying by 6 both sides)
14x – 2 – 7x – 2 > 6x
7x – 4 > 6x
7x > 6x + 4
x > 4 .......................(2)

From (3) and (4), solutions of the given system are, therefore, given by 4 < x < 9
Hence the solution of the system of is 4 < x < 9.

Question-21

Solve the following system of inequations: 2(x + 1) < x + 5, 3(x + 2) > 2 – x.

Solution:
2(x + 1) < x + 5                 
3(x + 2) > 2 – x
2(x + 1) < x + 5
2x + 2 < x + 5
2x < x + 3
x < 3........................ (1)

3(x + 2) > 2 – x
3x + 6 > 2 – x
3x > -4 – x
4x > -4
x > -1 ......................(2)

From (1) and (2), solutions of the given system are, therefore, given by -1 < x < 3
Hence the solution of the system of is -1 < x < 3.

Question-22

Solve the following system of inequations: 3x – 1 5, x + 2 > -1

Solution:
3x – 1 5                  
x + 2 > -1                 
3x – 1
5
3x
5 + 1
3x 6
x
2 ...................(1)

x + 2 > -1
x > -1 – 2
x > -3 .................(2)

From (1) and (2), solutions of the given system are, therefore, given by x
2
Hence the solution of the system of is x
2.

Question-23

Solve the following system of inequations: 3x – 7 > 2(x - 6), 6 – x > 11 – 2x

Solution:
3x – 7 > 2(x - 6)                 
6 – x > 11 – 2x     
3x – 7 > 2x - 12

3x – 7 + 7 > 2x - 12 + 7
3x > 2x - 5
3x – 2x > 2x – 5 – 2x
x > – 5  ............................(1)

6 – x > 11 – 2x
6 – x – 6 > 11 – 2x – 6
- x > 5 – 2x
– x + 2x > 5 – 2x + 2x
x > 5 ...............................(2)

From (1) and (2), solutions of the given system are, therefore, given by x > 5
Hence the solution of the system is x > 5.

Question-24

Solve the following system of inequations: –2 - , 3 – x < 4(x - 3)

Solution:
–2 -                    
3 – x < 4(x - 3)                 
–2 -

–24 – 3x
4(1 + x)      (Multiplying by 12 both sides)
–24 – 3x
4 + 4x
– 3x
4x + 28
– 7x
28
x
-4 .......................... (1)

3 – x < 4x - 12
- x < 4x – 15
- 5x < - 15
x > 3  ........................(2)

From (1) and (2), solutions of the given system are, therefore, given by x > 3
Hence the solution of the system is x > 3.

Question-25

Solve the following system of inequations:
5(2x - 7) – 3(2x + 3) 0, 2x + 19 6x + 47

Solution:
5(2x - 7) – 3(2x + 3) 0                 
2x + 19
6x + 47                           
5(2x - 7) – 3(2x + 3)
0
10x - 35 – 6x - 9
0
4x - 44
0
4x
44
x
11 ........................... (3)

2x + 19 6x + 47
2x
6x + 28
-4x
28
x
-7 ............................ (4)

From (1) and (2), solutions of the given system are, therefore, given by -7
x 11
Hence the solution of the system is -7
x 11.

Question-26

Solve the following system of inequations: 2x – 7 <11, 3x + 4 < - 5

Solution:
2x – 7 <11                   
3x + 4 < - 5
2x – 7 <11
2x <18
x < 9 ...........................(1)

3x + 4 < - 5
3x < - 9
x < - 3  .......................(2)

From (1) and (2), solutions of the given system are, therefore, given by x < - 3
Hence the solution of the system is x < - 3.

Question-27

Solve the following system of inequations: 4 – 5x > -11, 4x + 11 -13

Solution:
4 – 5x > -11                  
4x + 11
-13                
4 – 5x > -11
– 5x > -15
x < 3 ............................ (1)

4x + 11
-13
4x
- 24
x
- 6 .......................... (2)

From (1) and (2), solutions of the given system are, therefore, given by x
- 6.
Hence the solution of the system is x
- 6

Question-28

Solve the following system of inequations: 4x – 5 < 11, -3x – 4 8.

Solution:
4x – 5 < 11                 
-3x – 4
8                   
4x – 5 < 11
4x < 16
x < 4 ...............................(1)

-3x – 4
8
-3x
12
x
-4................................(2)

 From (1) and (2), solutions of the given system are, therefore, given by x -4.
Hence the solution of the system is x
-4.

Question-29

Solve the following system of inequations: 5x – 7 < 3 (x + 3), 1 – x - 4

Solution:
5x – 7 < 3 (x + 3)                 
1 –
x – 4                      
5x – 7 < 3 (x + 3)
5x – 7 < 3x + 9
5x < 3x + 16
2x < 16
x < 8.............................(1)

 
1 – x – 4            (Multiplying both sides by 2)
2 – 3x
2x – 8
- 3x
2x – 10
- 5x
– 10
x
2 ............................(2)

From (1) and (2), solutions of the given system are, therefore, given by x
2.
Hence the solution of the system is x
2.

Question-30

Solve the following system of inequations:2(2x + 3) – 10 < 6(x - 2),

Solution:
2(2x + 3) – 10 < 6(x - 2) 
                      
2(2x + 3) – 10 < 6(x - 2)
4x + 6 – 10 < 6x - 12
4x – 4 < 6x – 12
4x < 6x – 8
- 2x < – 8
x > 4 .................................. (1)



3(2x - 3) + 72 24 + 16x           (Multiplying both sides by 12)
6x - 9 + 72
24 + 16x
6x + 63
24 + 16x
6x
16x – 39
-10x
- 39
x
39/10............................(2)

From (1) and (2), the system has no solution.

Question-31

Represent the following inequation graphically in two dimensional plane and hence solve them:  x – 2y + 4 0

Solution:
We draw the graph of the equation x – 2y + 4 = 0
 

x

0

-4

y

2

0


Put x = 0

Then 0 – 2y + 4 0

or -2y - 4

or y 2

Put x = 0, y = 0

Then 0 – 2(0) + 4 0

or 4 0, which is false.

Hence, half plane II is not the solution of the given inequation.

Therefore, the shaded half plane I is the solution region of the inequation including points on the line

x – 2y + 4 = 0.

 Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

Question-32

Represent the following inequation graphically in two dimensional plane and hence solve them: 2x + y > 3

Solution:
We draw the graph of the equation 2x + y = 3
 

X

0

3/2

Y

3

0


Put x = 0

Then 2(0) + y > 3
or y > 3

Put x = 0, y = 0

Then 2(0) + 0 > 3

or 0 > 3, which is false.

Hence, half plane I is not the solution of the given inequation.

Therefore, the shaded half plane II is the solution region of the inequation excluding points on the line

2x + y = 3.

Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

Question-33

Represent the following inequation graphically in two dimensional plane and hence solve them:  3x – 4y < 12

Solution:
We draw the graph of the equation 3x – 4y = 12

x

0

4

y

-3

0


Put x = 0

Then 3(0) – 4y < 12

or - y < 3

or y > -3

Put x = 0, y = 0

Then 3(0) – 4(0) < 12

or 0 < 12, which is true.

Hence, half plane II is not the solution of the given inequation.

Therefore, the shaded half plane I is the solution region of the inequation excluding points on the line

3x – 4y = 12.

 Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

Question-34

Represent the following inequation graphically in two dimensional plane and hence solve them:  y + 8 2x

Solution:
We draw the graph of the equation y + 8 = 2x

x

2

4

y

-4

0


Put x = 0

Then y + 8 2x

or y + 8 0

Put x = 0, y = 0

Then 0 + 8 2(0)

or 8 0, which is true.

Hence, half plane I is not the solution of the given inequation.

Therefore, the shaded half plane II is the solution region of
the inequation including points on the line

y + 8 = 2x.

Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

Question-35

Represent the following inequation graphically in two dimensional plane and hence solve them:  2x 6 – 3y

Solution:
We draw the graph of the equation 2x = 6 – 3y
 

x

0

3

y

2

0


Put x = 0

Then 2(0) 6 – 3y
or y 2

Put x = 0, y = 0

Then 2(0) 6 – 3(0)
or 0 6, which is true.

Hence, half plane I is not the solution of the given inequation.

Therefore, the shaded half plane II is the solution region of the inequation including points on the line 2x = 6 – 3y

Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

Question-36

Represent the following inequation graphically in two dimensional plane and hence solve them:   0 2x – 5y + 10

Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation 2x – 5y + 10 = 0

x

0

-5

y

2

0


Put x = 0

Then 0 2(0) – 5y + 10

or -10 – 5y

Put x = 0, y = 0

Then 0 2(0) – 5(0) + 10

or 0 10, which is false.

Hence, half plane I is not the solution of the given inequation.

Therefore, the shaded half plane II is the solution region of the inequation including points on the line

2x – 5y + 10 = 0.

Question-37

Represent the following inequation graphically in two dimensional plane and hence solve them:   2x – 3y < 6

Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
 

We draw the graph of the equation 2x – 3y = 6

x

0

3

y

-2

0


Put x = 0

Then 2(0) – 3y < 6
or y < -2
Put x = 0, y = 0

Then 2(0) – 3(0) < 6
or 0 < 6, which is true.

Hence, half plane II is not the solution of the given inequation.

Therefore, the shaded half plane I is the solution region of the inequation excluding points on the line

2x – 3y = 6

Question-38

Represent the following inequation graphically in two dimensional plane and hence solve them:   x > - 2

Solution:
1cm = 1unit along x-axis
1cm = 1unit along y-axis

The graph of the equation x = - 2 is vertical line parallel to y – axis.

Put x = 0

Then 0 > -2, which is true.

Hence, the solution region is the shaded region on the right hand side of the line x = -2 containing the origin. 

Hence every point on the shaded region is the solution of the given inequation.

Question-39

Represent the following inequation graphically in two dimensional plane and hence solve them:    y < -2

Solution:
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation y = -2

Put y = 0

Then 0 < -2, which is false.

Hence, the solution region is the shaded region below line y = -2 containing the origin. 

Hence every point below the shaded region is the solution of the given inequation.

Question-40

Represent the following inequation graphically in two dimensional plane and hence solve them: x 8 – 4y

Solution:
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation x = 8 – 4y

x

0

8

y

2

0

Put x = 0

Then 0 8 – 4y
or y 2

Put x = 0, y = 0

Then (0) 8 – 4(0)

or 0 8, which is true.
Hence, half plane I is not the solution of the given inequation.

Therefore, the shaded half plane II is the solution region of the inequation including points on the line x = 8 – 4y.

Question-41

Solve the following system of inequations graphically.
x + 2y 20, 3x + y 15.

Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation x + 2y = 20

x

0

20

y

10

0

Put x = 0

Then 0 + 2y 20
or y 10

Put x = 0, y = 0

Then 0 + 2(0) 20
or 0 20, which is false.

3x + y 15

We draw the graph of the equation 3x + y = 15

x

0

5

y

15

0

Put x = 0

Then 3x + y 15
or y 15

Put x = 0, y = 0

Then 3(0) + (0) 15
or 0 15, which is true.           

Every point in the common shaded region represents a solution of the given system of inequations.

Question-42

Solve the following system of inequations graphically.
4x + 3y 12, 4x – 5y -20.

Solution:
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation 4x + 3y = 12

x

0

3

y

4

0

Put x = 0

Then 4(0) + 3y 12
or y 4

Put x = 0, y = 0

Then 4(0) + 3(0) 12
or 0 12, which is false.
3x + y 15

We draw the graph of the equation 4x – 5y = -20

x

0

-5

y

4

0

Put x = 0

Then 4(0) – 5y -20
or y 4

Put x = 0, y = 0

Then 4(0) – 5(0) -20

or 0 -20, which is true. 

Every point in the common shaded region represents a solution of the given system of inequations.

Question-43

Solve the following system of inequations graphically.
2x + y 8, x + 2y 10.

Solution:
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation 2x + y = 8

x

0

4

y

8

0

Put x = 0

Then 2(0) + y 8
or y 8

Put x = 0, y = 0

Then 2(0) + (0) 8
or 0 8, which is false.

We draw the graph of the equation x + 2y =10

x

0

10

y

5

0

Put x = 0

Then 0 + 2y 10
or y 5

Put x = 0, y = 0

Then 0 + 2(0) 10
or 0 10, which is false. 

Every point in the common shaded region represents a solution of the given system of inequations.

Question-44

Solve the following system of inequations graphically.
y 4, x 1

Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation y = 4

Put y = 0

Then 0 4, which is true.

We draw the graph of the equation x = 1

Put x = 0

Then 0 1, which is false.

Question-45

Solve the following system of inequations graphically.
5x + 6y 30, x 0, y 0

Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation 5x + 6y = 30.

x

0

6

y

5

0

Put x = 0
Then 5(0) + 6y 30
or y 5

Put x = 0, y = 0

Then 5(0) + 6(0) 30
or 0 30, which is false.

We draw the graph of the equation x 0.

Put x = 0

Then 0 0, which is false.

We draw the graph of the equation y = 0.

Put y = 0

Then 0 0, which is false. 

Every point in the common shaded region represents a solution of the given system of inequations.

Question-46

Solve the following system of inequations graphically.
x + y 9, y > x, x 1.

Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation x + y = 9

x

0

9

y

9

0

Put x = 0

Then 0 + y 9
or y 9

Put x = 0, y = 0

Then 0 + 0 9
or 0 9, which is true.

We draw the graph of the equation y = x.

Put x = 0

Then y > 0

Put x = 0, y = 0

Then 0 > 0, which is false.

We draw the graph of the equation x = 1.

Put x = 0

Then 0 1, which is false.           

Every point in the common shaded region represents a solution of the given system of inequations.

Question-47

Solve the following system of inequations graphically.
x + 3y 12, 3x + y 12, x 0, y 0.

Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation x + 3y = 12

x

0

12

y

4

0

Put x = 0

Then 0 + 3y 12
or y 4

Put x = 0, y = 0

Then 0 + 3(0) 12
or 0 12, which is true.

We draw the graph of the
equation 3x + y = 12

X

0

4

Y

12

0

Put x = 0

Then 0 + y 12
or y 12

Put x = 0, y = 0

Then 3(0) + 0 12
or 0 12, which is true.

We draw the graph of the equation x = 0.

Put x = 0

Then 0 0, which is false.

We draw the graph of the equation y = 0.

Put y = 0

Then 0 0, which is false. 

Every point in the common shaded region represents a solution of the given system of inequations.

Question-48

Solve the following system of inequations graphically.
2x + y 4, x + y 3, 2x – 3y 6.

Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation 2x + y = 4

x

0

2

y

4

0

Put x = 0

Then 2(0) + y 4
or y 4

Put x = 0, y = 0

Then 2(0) + (0) 4
or 0 4, which is false.

We draw the graph of the equation x + y = 3

x

0

3

y

3

0

Put x = 0

Then 0 + y 3
or y 3

Put x = 0, y = 0

Then 0 + 0 3
or 0 3, which is true.

We draw the graph of the equation 2x – 3y = 6

x

0

3

y

-2

0

Put x = 0

Then 2(0) – 3y 6
or y -2

Put x = 0, y = 0

Then 2(0) – 3(0) 6
or 0 6, which is true.  

Every point in the common shaded region represents a solution of the given system of inequations.

Question-49

Solve the following system of inequations graphically.
x + y < 6, 7x + 4y 28, x 0, y 0.

Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation x + y = 6

x

0

6

y

6

0

Put x = 0

Then 0 + y < 6
or y < 6

Put x = 0, y = 0

Then 0 + 0 < 6
or 0 < 6, which is true.

We draw the graph of the equation 7x + 4y = 28

x

0

7

y

4

0

Put x = 0

Then 7(0) + 4y 28
or y 4

Put x = 0, y = 0

Then 7(0) + 4(0) 28
or 0 28, which is true.

We draw the graph of the equation x = 0.

Put x = 0

Then 0 0, which is false.

We draw the graph of the equation y = 0.

Put y = 0

Then 0 0, which is false. 

Every point in the common shaded region represents a solution of the given system of inequations.

Question-50

Solve the following system of inequations graphically.
3x + 2y 24, x + 2y 16, x + y 10, x 0, y 0.

Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis

We draw the graph of the equation 3x + 2y = 24

x

0

8

y

12

0

Put x = 0

Then 3(0) + 2y 24
or y 12

Put x = 0, y = 0

Then 3(0) + 2(0) 24
or 0 24, which is true.

We draw the graph of the equation x + 2y = 16

x

0

16

y

8

0

Put x = 0

Then 0 + 2y 16
or y 8

Put x = 0, y = 0

Then 0 + 2(0) 16
or 0 16, which is false.

We draw the graph of the equation x + y =10

x

0

10

y

10

0

Put x = 0

Then x + y 10
or y 10

Put x = 0, y = 0

Then 0 + 0 10
or 0 10, which is true.

We draw the graph of the equation x = 0.

Put x = 0

Then 0 0, which is false.

We draw the graph of the equation y = 0.

Put y = 0

Then 0 0, which is false.

Every point in the common shaded region represents a solution of the given system of inequations.





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