The power point explains nicely the graph of inequalities of XI CBSE

1. POST GRAGUATETEACHER(MATHEMATICS)
KENDRIYA VIDYALAYA,FORT WILLIAM,KOLKATA

2. . . . -3 -2 .-1 . 1. 2. 3. 4 . 5 .6 .7
. . . . . . . . . . . . . . . . . .
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
. . -6 -5 -4

3. X=0

4. 7
6
x=0
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

5. 7
6
5
x 0
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

6. 7
6
x 0
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

7. 7
6
x=3
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

8. 7
6
5
x 3
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

9. 7
6
5
x 3
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

10. 7
6
5
x 3
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

11. 7
6
5
x 3
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

12. 7
6
x=-3
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

13. 7
6
5
x 3
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

14. 7
6
5
x 3
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

15. 7
6
y=0
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

16. 7
6
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
y 0
-3
-6 -5 -4

17. 7
6
y 0
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

18. 7
6
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
y 0 -3
x 0
-6 -5 -4

19. 7
6
y=x
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

20. 7
Let us verify the in-equation
6
With the point(1,0). y
0 x
1
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
(1,0) does not belongs to shaded region
as equation is not true when we put 1
-3
for x and 0 for y.
-6 -5 -4

21. 7
(1,0) belongs to shaded region
6
as equation is true when we put 1
5
for x and 0 for y.
4
Let us verify the in-equation y x1
0
3
With the point(1,0).
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

22. 7
6
y =2 x
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

23. 7
6
y 2 x0
2
5
2
4
3
Let us verify the in-equation
2
1
With the point(0,2).
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
(0,2) belongs to shaded
Region as equation is true
-3
when we put 0 for x and 2 for y.
-6 -5 -4

24. 7
y 2x
2 0
6
5
4
(0,2) does not belongs to shaded region
3
as equation is not true when we put 0
2
for x and 2 for y.
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
Let us verify the in-equation
-6 -5 -4
With the point(0,2).

25. 7
6
y =3 x
5
4
3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

26. 7
6
3x+2y=6
5
4
x y
1
3
2 3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

27. 7
Let us verify the in-equation
6
With the point(0,0).
5
x y
4
3x + 2y = 66
3x 2 y 1
3
2 3
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
(0,0) does not belongs to shaded region
as equation is not true when we put 0
-3
for x and 0 for y.
-6 -5 -4

28. 7
6
2x+3y=6
5
4
x y
1
3
3 2
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

29. 7
6
5
2 2 0 33 y0 6
x
2x+3y=6 6
4
3
x y
2
1
1
3 2
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
(0,0) does not belongs to shaded region
as equation is true when we put 0
-3
for x and 0 for y.
-6 -5 -4

30. 7
6
2x+3y=6 6
2x+3y
5
0+0 6
4
x y
3
1
3 2
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
(0,0) belongs to shaded region
as equation is true when we put 0
-3
for x and 0 for y.
-6 -5 -4

31. 7
6
2x-3y=6
5
4
x y
1
3
3 2
2
1
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2 -1
-3
-6 -5 -4

32. I acknowledge my son Avinash
Nayak(IIT,KGP) for the
presentation.