The document contains a series of graphs illustrating linear equations and inequalities. It shows lines representing equations like y=x, y=2x, and y=3x, and uses points like (1,0) and (0,2) to demonstrate whether they satisfy the corresponding inequalities. The purpose is to teach about visualizing and verifying linear equations and inequalities using graphs.

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.