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2D Linear Inequalities
The solutions of inequalities in x are segments of the real line.
2D Linear Inequalities
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
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
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.
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.
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.
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.
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.
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.
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.
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)
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)
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
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.
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.
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
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
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)
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)
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)
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)
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)
2D Linear Inequalities
To match the region that fits
2x + 3y > 12, sample (0, 0). 2x + 3y = 12 (0, 4)
(6, 0)
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)
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
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
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
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
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
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
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
Example C. Shade the system
2x – y < –2
x + y < 5
Find and label the tip of the region.
{
2D Linear Inequalities
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
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
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
For x + y < 5, graph x = y = 5
2D Linear Inequalities
For x + y < 5, graph x = y = 5
x y
0
0
2D Linear Inequalities
For x + y < 5, graph x = y = 5
x y
0 5
5 0
2D Linear Inequalities
For x + y < 5, graph x = y = 5
x y
0 5
5 0
2D Linear Inequalities
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
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
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
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
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
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
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
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
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
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
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.
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.
2D Linear Inequalities
2x – y = –2
x + y = 5+)
2D Linear Inequalities
2x – y = –2
x + y = 5+)
3x = 3
2D Linear Inequalities
2x – y = –2
x + y = 5+)
3x = 3
x = 1
2D Linear Inequalities
Set x = 1 in x + y = 5,
2x – y = –2
x + y = 5+)
3x = 3
x = 1
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
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{
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.

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3 6 2 d linear inequalities-x

  • 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.