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 regionsassociated 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 regionsassociated with y > x and y < x.

2D Linear Inequalities

y = xThe 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 regionsassociated with y > x and y < x.

2D Linear Inequalities

Points on the line fit the condition that y = x.

y = xThe 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 regionsassociated 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 = xThe 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 regionsassociated 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 regionsassociated 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 regionsassociated 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 regionsassociated 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 regionsassociated 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 regionsassociated 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 > ySpecifically, 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 < Cmeans 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 < Cmeans 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 < Cmeans 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 < Cmeans 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 y0

0

In general, to solve a linear inequalities Ax + By > C or Ax + By < Cmeans 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 y0 46 0

In general, to solve a linear inequalities Ax + By > C or Ax + By < Cmeans 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 y0 46 0

(0, 4)

(6, 0)

In general, to solve a linear inequalities Ax + By > C or Ax + By < Cmeans 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 ofthe solution. (Use a solid line for ≤ or ≥ )

x y0 46 0

2x + 3y = 12 (0, 4)

(6, 0)

In general, to solve a linear inequalities Ax + By > C or Ax + By < Cmeans 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 ofthe solution. (Use a solid line for ≤ or ≥ )

x y0 46 0

2x + 3y = 12 (0, 4)

(6, 0)

In general, to solve a linear inequalities Ax + By > C or Ax + By < Cmeans 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 ofthe solution. (Use a solid line for ≤ or ≥ )

x y0 46 0

2x + 3y = 12 (0, 4)

(6, 0)

In general, to solve a linear inequalities Ax + By > C or Ax + By < Cmeans 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 ofthe solution. (Use a solid line for ≤ or ≥ )

x y0 46 0

2x + 3y = 12 (0, 4)

(6, 0)

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

(6, 0)

2D Linear InequalitiesTo 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 InequalitiesTo 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 fits2x + 3y > 12 must be the other side of the line.

2x + 3y = 12 (0, 4)

(6, 0)

2x + 3y > 12

2D Linear InequalitiesTo 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 fits2x + 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 InequalitiesTo 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 fits2x + 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 InequalitiesTo 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 fits2x + 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 InequalitiesTo 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 fits2x + 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 InequalitiesTo 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 fits2x + 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 InequalitiesTo 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 fits2x + 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 < 5Find and label the tip of the region.{

2D Linear Inequalities

Example C. Shade the system 2x – y < –2 x + y < 5Find 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 < 5Find 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 < 5Find 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 y0

0

2D Linear Inequalities

Find the intercepts.

Example C. Shade the system 2x – y < –2 x + y < 5Find 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 y0 2–1 0

2D Linear Inequalities

Find the intercepts.

Example C. Shade the system 2x – y < –2 x + y < 5Find 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 y0 2–1 0

2D Linear Inequalities

Find the intercepts.

Example C. Shade the system 2x – y < –2 x + y < 5Find 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 y0 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 < 5Find 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 y0 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 < 5Find 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 y0 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 < 5Find 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 y0 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 < 5Find 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 y0 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 theinequality 2x – y < –2.

2D Linear Inequalities

Find the intercepts.

Example C. Shade the system 2x – y < –2 x + y < 5Find 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 y0 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 theinequality 2x – y < –2. Shade it.

2D Linear Inequalities

Find the intercepts.

Example C. Shade the system 2x – y < –2 x + y < 5Find 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 y0 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 theinequality 2x – y < –2. Shade it.

2D Linear Inequalities

Find the intercepts.

For x + y < 5, graph x = y = 52D Linear Inequalities

For x + y < 5, graph x = y = 5

x y0

0

2D Linear Inequalities

For x + y < 5, graph x = y = 5

x y0 55 0

2D Linear Inequalities

For x + y < 5, graph x = y = 5

x y0 55 0

2D Linear Inequalities

For x + y < 5, graph x = y = 5

x y0 55 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 y0 55 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 y0 55 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 y0 55 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 y0 55 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 y0 55 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 y0 55 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 InequalitiesFor x + y < 5, graph x = y = 5

x y0 55 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 InequalitiesFor x + y < 5, graph x = y = 5

x y0 55 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 InequalitiesFor x + y < 5, graph x = y = 5

x y0 55 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 InequalitiesFor x + y < 5, graph x = y = 5

x y0 55 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 y0 55 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 Inequalities2x – y = –2 x + y = 5+)

2D Linear Inequalities2x – y = –2 x + y = 5+)3x = 3

2D Linear Inequalities2x – 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 < –2x + 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 ≥ 04. 0 ≤ 8 – 2x 5. y < –x + 4 6. 2x/3 – 3 ≤ 6/57. 2x < 6 – 2y 8. 4y/5 – 12 ≥ 3x/4 9. 2x + 3y > 310. –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 ≥ –12x + 3y ≤ 5

20. {2x + 3y > –1 3x + 4y ≥ 2

21. {4x – 3y ≤ 3 3x – 2y > –4

{ x – y < 3

x – y ≤ –1

32

23

12

14

22. { x + y ≤ 1

x – y < –1

12

15

34

16

23.