1. 2D Linear Inequalities

2. The solutions of inequalities in x are segments of the real line.
2D Linear Inequalities

3. The solutions of inequalities in x are segments of the real line.
The solutions of inequalities in x and y are regions of the plane.
2D Linear Inequalities

4. The solutions of inequalities in x are segments of the real line.
The solutions of inequalities in x and y are regions of the plane.
Example A. Use the graph of y = x to identify the regions
associated with y > x and y < x.
2D Linear Inequalities

5. The solutions of inequalities in x are segments of the real line.
The solutions of inequalities in x and y are regions of the plane.
Example A. Use the graph of y = x to identify the regions
associated with y > x and y < x.
2D Linear Inequalities
y = x
The graph of y = x is the diagonal line.

6. The solutions of inequalities in x are segments of the real line.
The solutions of inequalities in x and y are regions of the plane.
Example A. Use the graph of y = x to identify the regions
associated with y > x and y < x.
2D Linear Inequalities
Points on the line fit the condition
that y = x.
y = x
The graph of y = x is the diagonal line.

7. The solutions of inequalities in x are segments of the real line.
The solutions of inequalities in x and y are regions of the plane.
Example A. Use the graph of y = x to identify the regions
associated with y > x and y < x.
2D Linear Inequalities
Points on the line fit the condition
that y = x. Points not on the line
fit the condition y = x.
y = x
The graph of y = x is the diagonal line.

8. The solutions of inequalities in x are segments of the real line.
The solutions of inequalities in x and y are regions of the plane.
Example A. Use the graph of y = x to identify the regions
associated with y > x and y < x.
2D Linear Inequalities
Points on the line fit the condition
that y = x. Points not on the line
fit the condition y = x.
y = x
Specifically, the line y = x divides
the plane into two half-planes.
The graph of y = x is the diagonal line.

9. The solutions of inequalities in x are segments of the real line.
The solutions of inequalities in x and y are regions of the plane.
Example A. Use the graph of y = x to identify the regions
associated with y > x and y < x.
2D Linear Inequalities
Points on the line fit the condition
that y = x. Points not on the line
fit the condition y = x.
y = x
Specifically, the line y = x divides
the plane into two half-planes.
One of them fits the relation that
y < x and the other fits x < y.
The graph of y = x is the diagonal line.

10. The solutions of inequalities in x are segments of the real line.
The solutions of inequalities in x and y are regions of the plane.
Example A. Use the graph of y = x to identify the regions
associated with y > x and y < x.
2D Linear Inequalities
Points on the line fit the condition
that y = x. Points not on the line
fit the condition y = x.
y = x
Specifically, the line y = x divides
the plane into two half-planes.
One of them fits the relation that
y < x and the other fits x < y.
To identify which half-plane matches which inequality,
sample any point in the half planes.
The graph of y = x is the diagonal line.

11. The solutions of inequalities in x are segments of the real line.
The solutions of inequalities in x and y are regions of the plane.
Example A. Use the graph of y = x to identify the regions
associated with y > x and y < x.
2D Linear Inequalities
Points on the line fit the condition
that y = x. Points not on the line
fit the condition y = x.
y = x
Specifically, the line y = x divides
the plane into two half-planes.
One of them fits the relation that
y < x and the other fits x < y.
To identify which half-plane matches which inequality,
sample any point in the half planes. For example, let's select
(0, 5) to test the inequalities.
(0, 5)The graph of y = x is the diagonal line.

12. The solutions of inequalities in x are segments of the real line.
The solutions of inequalities in x and y are regions of the plane.
Example A. Use the graph of y = x to identify the regions
associated with y > x and y < x.
2D Linear Inequalities
Points on the line fit the condition
that y = x. Points not on the line
fit the condition y = x.
y = x
Specifically, the line y = x divides
the plane into two half-planes.
One of them fits the relation that
y < x and the other fits x < y.
To identify which half-plane matches which inequality,
sample any point in the half planes. For example, let's select
(0, 5) to test the inequalities. Since 0 < 5 so the side that
contains (0, 5) is x < y.
The graph of y = x is the diagonal line.
x < y
(0, 5)

13. The solutions of inequalities in x are segments of the real line.
The solutions of inequalities in x and y are regions of the plane.
Example A. Use the graph of y = x to identify the regions
associated with y > x and y < x.
2D Linear Inequalities
Points on the line fit the condition
that y = x. Points not on the line
fit the condition y = x.
y = x
x > y
Specifically, the line y = x divides
the plane into two half-planes.
One of them fits the relation that
y < x and the other fits x < y.
To identify which half-plane matches which inequality,
sample any point in the half planes. For example, let's select
(0, 5) to test the inequalities. Since 0 < 5 so the side that
contains (0, 5) is x < y. It follows that the other side is x > y.
The graph of y = x is the diagonal line.
x < y
(0, 5)

14. In general, to solve a linear inequalities
Ax + By > C or Ax + By < C
means to identify which side of the line Ax + By = C
is the half-plane that is the solution of the inequality in question.
2D Linear Inequalities

15. In general, to solve a linear inequalities
Ax + By > C or Ax + By < C
means to identify which side of the line Ax + By = C
is the half-plane that is the solution of the inequality in question.
2D Linear Inequalities
Example B. Shade 2x + 3y > 12.

16. In general, to solve a linear inequalities
Ax + By > C or Ax + By < C
means to identify which side of the line Ax + By = C
is the half-plane that is the solution of the inequality in question.
To accomplish this, after graphing the line Ax + By = C,
sample any point not on the line.
2D Linear Inequalities
Example B. Shade 2x + 3y > 12.

17. In general, to solve a linear inequalities
Ax + By > C or Ax + By < C
means to identify which side of the line Ax + By = C
is the half-plane that is the solution of the inequality in question.
To accomplish this, after graphing the line Ax + By = C,
sample any point not on the line.
2D Linear Inequalities
Example B. Shade 2x + 3y > 12.
Use the intercepts method
to graph 2x + 3y = 12.
x y
0
0

18. In general, to solve a linear inequalities
Ax + By > C or Ax + By < C
means to identify which side of the line Ax + By = C
is the half-plane that is the solution of the inequality in question.
To accomplish this, after graphing the line Ax + By = C,
sample any point not on the line.
2D Linear Inequalities
Example B. Shade 2x + 3y > 12.
Use the intercepts method
to graph 2x + 3y = 12.
x y
0 4
6 0

19. In general, to solve a linear inequalities
Ax + By > C or Ax + By < C
means to identify which side of the line Ax + By = C
is the half-plane that is the solution of the inequality in question.
To accomplish this, after graphing the line Ax + By = C,
sample any point not on the line.
2D Linear Inequalities
Example B. Shade 2x + 3y > 12.
Use the intercepts method
to graph 2x + 3y = 12.
x y
0 4
6 0
(0, 4)
(6, 0)

20. In general, to solve a linear inequalities
Ax + By > C or Ax + By < C
means to identify which side of the line Ax + By = C
is the half-plane that is the solution of the inequality in question.
To accomplish this, after graphing the line Ax + By = C,
sample any point not on the line.
2D Linear Inequalities
Example B. Shade 2x + 3y > 12.
Use the intercepts method
to graph 2x + 3y = 12.
Draw a dotted line because
2x + 3y = 12 is not part of
the solution.
(Use a solid line for ≤ or ≥ )
x y
0 4
6 0
2x + 3y = 12 (0, 4)
(6, 0)

21. In general, to solve a linear inequalities
Ax + By > C or Ax + By < C
means to identify which side of the line Ax + By = C
is the half-plane that is the solution of the inequality in question.
To accomplish this, after graphing the line Ax + By = C,
sample any point not on the line.
2D Linear Inequalities
Trick: Sample a point on the axes. Use (0, 0) if it’s possible.
Example B. Shade 2x + 3y > 12.
Use the intercepts method
to graph 2x + 3y = 12.
Draw a dotted line because
2x + 3y = 12 is not part of
the solution.
(Use a solid line for ≤ or ≥ )
x y
0 4
6 0
2x + 3y = 12 (0, 4)
(6, 0)

22. In general, to solve a linear inequalities
Ax + By > C or Ax + By < C
means to identify which side of the line Ax + By = C
is the half-plane that is the solution of the inequality in question.
To accomplish this, after graphing the line Ax + By = C,
sample any point not on the line. If the selected point fits
inequality then the half-plane containing the test point is the
solution.
2D Linear Inequalities
Trick: Sample a point on the axes. Use (0, 0) if it’s possible.
Example B. Shade 2x + 3y > 12.
Use the intercepts method
to graph 2x + 3y = 12.
Draw a dotted line because
2x + 3y = 12 is not part of
the solution.
(Use a solid line for ≤ or ≥ )
x y
0 4
6 0
2x + 3y = 12 (0, 4)
(6, 0)

23. In general, to solve a linear inequalities
Ax + By > C or Ax + By < C
means to identify which side of the line Ax + By = C
is the half-plane that is the solution of the inequality in question.
To accomplish this, after graphing the line Ax + By = C,
sample any point not on the line. If the selected point fits
inequality then the half-plane containing the test point is the
solution. Otherwise, the other side is the solution the inequality.
2D Linear Inequalities
Trick: Sample a point on the axes. Use (0, 0) if it’s possible.
Example B. Shade 2x + 3y > 12.
Use the intercepts method
to graph 2x + 3y = 12.
Draw a dotted line because
2x + 3y = 12 is not part of
the solution.
(Use a solid line for ≤ or ≥ )
x y
0 4
6 0
2x + 3y = 12 (0, 4)
(6, 0)

24. 2D Linear Inequalities
To match the region that fits
2x + 3y > 12, sample (0, 0). 2x + 3y = 12 (0, 4)
(6, 0)

25. 2D Linear Inequalities
To match the region that fits
2x + 3y > 12, sample (0, 0).
Plug in (0, 0) into the inequality,
we’ve 0 + 0 > 12 which is false.
2x + 3y = 12 (0, 4)
(6, 0)

26. 2D Linear Inequalities
To match the region that fits
2x + 3y > 12, sample (0, 0).
Plug in (0, 0) into the inequality,
we’ve 0 + 0 > 12 which is false.
Hence the half-plane that fits
2x + 3y > 12 must be the other
side of the line.
2x + 3y = 12 (0, 4)
(6, 0)
2x + 3y > 12

27. 2D Linear Inequalities
To match the region that fits
2x + 3y > 12, sample (0, 0).
Plug in (0, 0) into the inequality,
we’ve 0 + 0 > 12 which is false.
Hence the half-plane that fits
2x + 3y > 12 must be the other
side of the line.
If the inequality is > or < , then the boundary is also part
of the solution and we draw a solid line to express that.
2x + 3y = 12 (0, 4)
(6, 0)
2x + 3y > 12

28. 2D Linear Inequalities
To match the region that fits
2x + 3y > 12, sample (0, 0).
Plug in (0, 0) into the inequality,
we’ve 0 + 0 > 12 which is false.
To find the region that fits a system of two x&y linear
inequalities, graph the equations first.
Hence the half-plane that fits
2x + 3y > 12 must be the other
side of the line.
If the inequality is > or < , then the boundary is also part
of the solution and we draw a solid line to express that.
2x + 3y = 12 (0, 4)
(6, 0)
2x + 3y > 12

29. 2D Linear Inequalities
To match the region that fits
2x + 3y > 12, sample (0, 0).
Plug in (0, 0) into the inequality,
we’ve 0 + 0 > 12 which is false.
To find the region that fits a system of two x&y linear
inequalities, graph the equations first. In general, we get two
intersecting lines that divide the plane into 4 regions.
Hence the half-plane that fits
2x + 3y > 12 must be the other
side of the line.
If the inequality is > or < , then the boundary is also part
of the solution and we draw a solid line to express that.
2x + 3y = 12 (0, 4)
(6, 0)
2x + 3y > 12

30. 2D Linear Inequalities
To match the region that fits
2x + 3y > 12, sample (0, 0).
Plug in (0, 0) into the inequality,
we’ve 0 + 0 > 12 which is false.
To find the region that fits a system of two x&y linear
inequalities, graph the equations first. In general, we get two
intersecting lines that divide the plane into 4 regions.
Then we sample to determine the two half-planes that fit the
two inequalities.
Hence the half-plane that fits
2x + 3y > 12 must be the other
side of the line.
If the inequality is > or < , then the boundary is also part
of the solution and we draw a solid line to express that.
2x + 3y = 12 (0, 4)
(6, 0)
2x + 3y > 12

31. 2D Linear Inequalities
To match the region that fits
2x + 3y > 12, sample (0, 0).
Plug in (0, 0) into the inequality,
we’ve 0 + 0 > 12 which is false.
To find the region that fits a system of two x&y linear
inequalities, graph the equations first. In general, we get two
intersecting lines that divide the plane into 4 regions.
Then we sample to determine the two half-planes that fit the
two inequalities. The overlapped region of the two half-planes
is the region that fits the system.
Hence the half-plane that fits
2x + 3y > 12 must be the other
side of the line.
If the inequality is > or < , then the boundary is also part
of the solution and we draw a solid line to express that.
2x + 3y = 12 (0, 4)
(6, 0)
2x + 3y > 12

32. 2D Linear Inequalities
To match the region that fits
2x + 3y > 12, sample (0, 0).
Plug in (0, 0) into the inequality,
we’ve 0 + 0 > 12 which is false.
To find the region that fits a system of two x&y linear
inequalities, graph the equations first. In general, we get two
intersecting lines that divide the plane into 4 regions.
Then we sample to determine the two half-planes that fit the
two inequalities. The overlapped region of the two half-planes
is the region that fits the system. To give the complete solution,
we need to locate the tip of the region by solving the system.
Hence the half-plane that fits
2x + 3y > 12 must be the other
side of the line.
If the inequality is > or < , then the boundary is also part
of the solution and we draw a solid line to express that.
2x + 3y = 12 (0, 4)
(6, 0)
2x + 3y > 12

33. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
2D Linear Inequalities

34. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
We are looking for the region that satisfies both inequalities.
2D Linear Inequalities

35. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
We are looking for the region that satisfies both inequalities.
For 2x – y < –2, graph the equation 2x – y = –2.
2D Linear Inequalities

36. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
We are looking for the region that satisfies both inequalities.
For 2x – y < –2, graph the equation 2x – y = –2.
x y
0
0
2D Linear Inequalities
Find the intercepts.

37. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
We are looking for the region that satisfies both inequalities.
For 2x – y < –2, graph the equation 2x – y = –2.
x y
0 2
–1 0
2D Linear Inequalities
Find the intercepts.

38. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
We are looking for the region that satisfies both inequalities.
For 2x – y < –2, graph the equation 2x – y = –2.
x y
0 2
–1 0
2D Linear Inequalities
Find the intercepts.

39. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
We are looking for the region that satisfies both inequalities.
For 2x – y < –2, graph the equation 2x – y = –2.
x y
0 2
–1 0
Since the inequality
includes “ = ”, use a solid
line for the graph.
2D Linear Inequalities
Find the intercepts.

40. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
We are looking for the region that satisfies both inequalities.
For 2x – y < –2, graph the equation 2x – y = –2.
x y
0 2
–1 0
Since the inequality
includes “ = ”, use a solid
line for the graph.
2D Linear Inequalities
Find the intercepts.

41. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
We are looking for the region that satisfies both inequalities.
For 2x – y < –2, graph the equation 2x – y = –2.
x y
0 2
–1 0
Since the inequality
includes “ = ”, use a solid
line for the graph.
Test (0, 0),
2D Linear Inequalities
Find the intercepts.

42. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
We are looking for the region that satisfies both inequalities.
For 2x – y < –2, graph the equation 2x – y = –2.
x y
0 2
–1 0
Since the inequality
includes “ = ”, use a solid
line for the graph.
Test (0, 0), it does not fit.
2D Linear Inequalities
Find the intercepts.

43. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
We are looking for the region that satisfies both inequalities.
For 2x – y < –2, graph the equation 2x – y = –2.
x y
0 2
–1 0
Since the inequality
includes “ = ”, use a solid
line for the graph.
Test (0, 0), it does not fit.
Hence, the other side fits the
inequality 2x – y < –2.
2D Linear Inequalities
Find the intercepts.

44. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
We are looking for the region that satisfies both inequalities.
For 2x – y < –2, graph the equation 2x – y = –2.
x y
0 2
–1 0
Since the inequality
includes “ = ”, use a solid
line for the graph.
Test (0, 0), it does not fit.
Hence, the other side fits the
inequality 2x – y < –2.
Shade it.
2D Linear Inequalities
Find the intercepts.

45. Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
We are looking for the region that satisfies both inequalities.
For 2x – y < –2, graph the equation 2x – y = –2.
x y
0 2
–1 0
Since the inequality
includes “ = ”, use a solid
line for the graph.
Test (0, 0), it does not fit.
Hence, the other side fits the
inequality 2x – y < –2.
Shade it.
2D Linear Inequalities
Find the intercepts.

46. For x + y < 5, graph x = y = 5
2D Linear Inequalities

47. For x + y < 5, graph x = y = 5
x y
0
0
2D Linear Inequalities

48. For x + y < 5, graph x = y = 5
x y
0 5
5 0
2D Linear Inequalities

49. For x + y < 5, graph x = y = 5
x y
0 5
5 0
2D Linear Inequalities

50. For x + y < 5, graph x = y = 5
x y
0 5
5 0
Since the inequality
is strict, use a dotted line
for the graph.
2D Linear Inequalities

51. For x + y < 5, graph x = y = 5
x y
0 5
5 0
Since the inequality
is strict, use a dotted line
for the graph.
2D Linear Inequalities

52. For x + y < 5, graph x = y = 5
x y
0 5
5 0
Since the inequality
is strict, use a dotted line
for the graph.
Test (0, 0),
2D Linear Inequalities

53. For x + y < 5, graph x = y = 5
x y
0 5
5 0
Since the inequality
is strict, use a dotted line
for the graph.
Test (0, 0), it fits x + y < 5.
2D Linear Inequalities

54. For x + y < 5, graph x = y = 5
x y
0 5
5 0
Since the inequality
is strict, use a dotted line
for the graph.
Test (0, 0), it fits x + y < 5.
Hence, this side fits x + y < 5.
2D Linear Inequalities

55. For x + y < 5, graph x = y = 5
x y
0 5
5 0
Since the inequality
is strict, use a dotted line
for the graph.
Test (0, 0), it fits x + y < 5.
Hence, this side fits x + y < 5.
Shade them.
2D Linear Inequalities

56. x y
0 5
5 0
Since the inequality
is strict, use a dotted line
for the graph.
Test (0, 0), it fits x + y < 5.
Hence, this side fits x + y < 5.
Shade them.
2D Linear Inequalities
For x + y < 5, graph x = y = 5

57. x y
0 5
5 0
Since the inequality
is strict, use a dotted line
for the graph.
Test (0, 0), it fits x + y < 5.
Hence, this side fits x + y < 5.
Shade them.
The region that fits the system is
the region has both shading.
2D Linear Inequalities
For x + y < 5, graph x = y = 5

58. x y
0 5
5 0
Since the inequality
is strict, use a dotted line
for the graph.
Test (0, 0), it fits x + y < 5.
Hence, this side fits x + y < 5.
Shade them.
2D Linear Inequalities
For x + y < 5, graph x = y = 5

59. x y
0 5
5 0
Since the inequality
is strict, use a dotted line
for the graph.
Test (0, 0), it fits x + y < 5.
Hence, this side fits x + y < 5.
Shade them.
The region that fits the system is
the region has both shading.
2D Linear Inequalities
For x + y < 5, graph x = y = 5

60. x y
0 5
5 0
Since the inequality
is strict, use a dotted line
for the graph.
Test (0, 0), it fits x + y < 5.
Hence, this side fits x + y < 5.
Shade them.
The region that fits the system is
the region has both shading.
2D Linear Inequalities
2x – y = –2
x + y = 5{
For x + y < 5, graph x = y = 5
To find the tip of the region,
we solve the system of equations
for their point of intersection.

61. x y
0 5
5 0
Since the inequality
is strict, use a dotted line
for the graph.
Test (0, 0), it fits x + y < 5.
Hence, this side fits x + y < 5.
Shade them.
The region that fits the system is
the region has both shading.
2D Linear Inequalities
2x – y = –2
x + y = 5{
Add these equations to remove
the y.
For x + y < 5, graph x = y = 5
To find the tip of the region,
we solve the system of equations
for their point of intersection.

62. 2D Linear Inequalities
2x – y = –2
x + y = 5+)

63. 2D Linear Inequalities
2x – y = –2
x + y = 5+)
3x = 3

64. 2D Linear Inequalities
2x – y = –2
x + y = 5+)
3x = 3
x = 1

65. 2D Linear Inequalities
Set x = 1 in x + y = 5,
2x – y = –2
x + y = 5+)
3x = 3
x = 1

66. 2D Linear Inequalities
Set x = 1 in x + y = 5, we get 1 + y = 5  y = 4.
2x – y = –2
x + y = 5+)
3x = 3
x = 1

67. 2D Linear Inequalities
Set x = 1 in x + y = 5, we get 1 + y = 5  y = 4.
2x – y = –2
x + y = 5+)
3x = 3
x = 1
Hence the tip of the region is (1, 4).
(1, 4)
2x – y < –2
x + y < 5{

68. Exercise A. Shade the following inequalities in the x and y
coordinate system.
1. x – y > 3 2. 2x ≤ 6 3. –y – 7 ≥ 0
4. 0 ≤ 8 – 2x 5. y < –x + 4 6. 2x/3 – 3 ≤ 6/5
7. 2x < 6 – 2y 8. 4y/5 – 12 ≥ 3x/4 9. 2x + 3y > 3
10. –6 ≤ 3x – 2y 11. 3x + 2 > 4y + 3x 12. 5x/4 + 2y/3 ≤ 2
2D Linear Inequalities
16.{–x + 2y ≥ –12
2x + y ≤ 4
Exercise B. Shade the following regions. Label the tip.
13.{x + y ≥ 3
2x + y < 4
14. 15.{x + 2y ≥ 3
2x – y > 6
{x + y ≤ 3
2x – y > 6
17. {3x + 4y ≥ 3
x – 2y < 6
18. { x + 3y ≥ 3
2x – 9y ≥ –4
19.{–3x + 2y ≥ –1
2x + 3y ≤ 5
20. {2x + 3y > –1
3x + 4y ≥ 2 21. {4x – 3y ≤ 3
3x – 2y > –4
{
x – y < 3
x – y ≤ –1
3
2
2
3
1
2
1
4
22. {
x + y ≤ 1
x – y < –1
1
2
1
5
3
4
1
6
23.