### Solving Systems of Inequalities By Graphing

- To solve a system, graph each inequality. The solution of the system is the intersection of the graphs of the inequalities.
- If the intersection is empty, the system has no solution.

### Solving Systems of Inequalities By Graphing

y > 4x + 3

y ≥ − 2x − 3

- Things to remember when graphing inequalities:
- a) Inequalities with " > " or " < " are dashed lines
- b) Inequalities with " ≥ " or " ≤ " are solid lines
- c) Solution is the common shaded area between the two inequalitites, if there is any.
- If the lines share no shading in common, it is said that the solution is the empty set, and there is no solution.
- Step 1: Graph both system of equations. Best method is to use y - intercept (b) and the slope (m)
- Equation 1: m = [rise/run] = [4/1] or [(−4)/(−1)]
- Equation 1: b = 3
- Equation 2: m = [rise/run] = [(−2)/1] or [2/(−1)]
- Equation 2: b = -3
- Step 2: Using a test point, check which way to shade for each graph. Always use the easiest point (0,0) if and only if is not on any of the lines
- Test Equation 1: (0,0)

y > 4x + 3

0 > 4(0) + 3

0 > 3

Not True; Shade away from (0,0) - Test Equation 2: (0,0)

y ≥ − 2x − 3

0 ≥ − 2(0) − 3

0 ≥ − 3

True; Shade Towards (0,0) - Step 3: Add a darker shade to the common shaded areas between the two lines.

y > 4x − 3

y > − x + 2

- Things to remember when graphing inequalities:
- a) Inequalities with " > " or " < " are dashed lines
- b) Inequalities with " ≥ " or " ≤ " are solid lines
- c) Solution is the common shaded area between the two inequalitites, if there is any.
- If the lines share no shading in common, it is said that the solution is the empty set, and there is no solution.
- Step 1: Graph both system of equations. Best method is to use y - intercept (b) and the slope (m)
- Equation 1: m = [rise/run] = [4/1] or [(−4)/(−1)]
- Equation 1: b = 3
- Equation 2: m = [rise/run] = [(−1)/1] or [1/(−1)]
- Equation 2: b = 2
- Step 2: Using a test point, check which way to shade for each graph. Always use the easiest point (0,0) if and only if is not on any of the lines
- Test Equation 1: (0,0)

y > 4x − 3

0 > 4(0) − 3

0 > − 3

True; Shade Towards (0,0) - Test Equation 2: (0,0)

y > − x + 2

0 > − (0) + 2

0 > 2

Not True; Shade Away From (0,0) - Step 3: Add a darker shade to the common shaded areas between the two lines.

x ≤ 3

y ≥ − [1/3]x − 2

- Things to remember when graphing inequalities:
- a) Inequalities with " > " or " < " are dashed lines
- b) Inequalities with " ≥ " or " ≤ " are solid lines
- c) Solution is the common shaded area between the two inequalitites, if there is any.
- If the lines share no shading in common, it is said that the solution is the empty set, and there is no solution.
- Step 1: Graph both system of equations. Best method is to use y - intercept (b) and the slope (m)
- Equation 1: m = Undefined
- Equation 1: b = none
- Equation 1: Solid Vertical Line Through x=3
- Equation 2: m = [rise/run] = [(−1)/3] or [1/(−3)]
- Equation 2: b = -2
- Step 2: Using a test point, check which way to shade for each graph. Always use the easiest point (0,0) if and only if is not on any of the lines
- Test Equation 1: (0,0)

x ≤ 3

0 ≤ 3

True; Shade Towards (0,0) - Test Equation 2: (0,0)

y ≥ − [1/3]x − 2

0 ≥ − [1/3](0) − 2

0 ≥ − 2

True; Shade Towards (0,0) - Step 3: Add a darker shade to the common shaded areas between the two lines.

y > 1

y < [3/2]x − 2

- Things to remember when graphing inequalities:
- a) Inequalities with " > " or " < " are dashed lines
- b) Inequalities with " ≥ " or " ≤ " are solid lines
- c) Solution is the common shaded area between the two inequalitites, if there is any.
- Step 1: Graph both system of equations. Best method is to use y - intercept (b) and the slope (m)
- Equation 1: m = [rise/run] = 0
- Equation 1: b = 1
- Equation 2: m = [rise/run] = [3/2] or [(−3)/(−2)]
- Equation 2: b = -2
- Test Equation 1: (0,0)

y > 1

0 > 1

Not True; Shade Away From (0,0) - Test Equation 2: (0,0)

y < [3/2]x − 2

0 < [3/2](0) − 2

0 < − 2

Not True; Shade Away From (0,0) - Step 3: Add a darker shade to the common shaded areas between the two lines.

y < [1/2]x − 1

y ≤ 2x + 2

- Things to remember when graphing inequalities:
- a) Inequalities with " > " or " < " are dashed lines
- b) Inequalities with " ≥ " or " ≤ " are solid lines
- c) Solution is the common shaded area between the two inequalitites, if there is any.
- Step 1: Graph both system of equations. Best method is to use y - intercept (b) and the slope (m)
- Equation 1: m = [rise/run] = [1/2] or [(−1)/(−2)]
- Equation 1: b = -1
- Equation 2: m = [rise/run] = [2/1] or [(−2)/(−1)]
- Equation 2: b = 2
- Test Equation 1: (0,0)

y < [1/2]x − 1

0 < [1/2](0) − 1

0 < − 1

Not True; Shade Away From (0,0) - Test Equation 2: (0,0)

y ≤ 2x + 2

0 ≤ 2(0) + 2

0 ≤ 2

True; Shade Towards (0,0) - Step 3: Add a darker shade to the common shaded areas between the two lines.

y > [2/3]x + 3

y < [2/3]x − 2

- Things to remember when graphing inequalities:
- a) Inequalities with " > " or " < " are dashed lines
- b) Inequalities with " ≥ " or " ≤ " are solid lines
- c) Solution is the common shaded area between the two inequalitites, if there is any.
- Step 1: Graph both system of equations. Best method is to use y - intercept (b) and the slope (m)
- Equation 1: m = [rise/run] = [2/3] or [(−2)/(−3)]
- Equation 1: b = 3
- Equation 2: m = [rise/run] = [2/3] or [(−2)/(−3)]
- Equation 2: b = -2
- Test Equation 1: (0,0)

y > [2/3]x + 3

0 > [1/2](0) + 3

0 > 3

Not True; Shade Away From (0,0) - Test Equation 2: (0,0)

y < [2/3]x − 2

0 < [2/3](0) − 2

0 < − 2

Not True; Shade Away From (0,0) - Step 3: Notice how this system of equations have no common shading in common.
- In this case, the solution is the Empty Set, or No Solution.

y < [3/2]x − 1

y < [1/2]x + 1

- Things to remember when graphing inequalities:
- a) Inequalities with " > " or " < " are dashed lines
- b) Inequalities with " ≥ " or " ≤ " are solid lines
- c) Solution is the common shaded area between the two inequalitites, if there is any.
- Step 1: Graph both system of equations. Best method is to use y - intercept (b) and the slope (m)
- Equation 1: m = [rise/run] = [3/2] or [(−3)/(−2)]
- Equation 1: b = -1
- Equation 2: m = [rise/run] = [1/2] or [(−1)/(−2)]
- Equation 2: b = 1
- Test Equation 1: (0,0)

y < [3/2]x − 1

0 < [3/2](0) − 1

0 < − 1

Not True; Shade Away From (0,0) - Test Equation 2: (0,0)

y < [1/2]x + 1

0 < [1/2](0) + 1

0 < 1

True; Shade Towards (0,0) - Step 3:Add a darker shade to the common shaded areas between the two lines.

2x + y < 3

3x − y ≥ 2

- Things to remember when graphing inequalities:
- a) Inequalities with " > " or " < " are dashed lines
- b) Inequalities with " ≥ " or " ≤ " are solid lines
- c) Solution is the common shaded area between the two inequalitites, if there is any.
- Step 1: Graph both system of equations. Best method is to find the x - and y - intercepts for both equations. You may replace inequalities with equal sign.
- Equation 1: x - Intercept

Set y to zero. Solve for x.

Point is (x,0)

2x + y = 3

2x + 0 = 3

2x = 3

x = [3/2]

( [3/2],0 ) - Equation 1: y - Intercept

Set x to zero. Solve for y.

Point is (0,y)

2x + y = 3

0 + y = 3

y = 3

(0,3)

Draw Dashed Line - Equation 2: x - Intercept

Set y to zero. Solve for x.

Point is (x,0)

3x − y = 2

3x − 0 = 2

3x = 2

x = [2/3]

([2/3],0) - Equation 2: y - Intercept

Set x to zero. Solve for y.

Point is (0,y)

3x − y = 2

0 − y = 2

− y = 2

y = − 2

(0, - 2)

Draw Solid Line - Test Equation 1: (0,0)

2x + y < 3

2(0) + 0 < 3

0 < 3

True; Shade Towards (0,0) - Test Equation 2: (0,0)

3x − y ≥ 2

3(0) − (0) ≥ 2

0 ≥ 2

Not True; Shade Away From (0,0) - Step 3:Add a darker shade to the common shaded areas between the two lines.

x + 2y ≤ − 4

3x + y ≤ 3

- Things to remember when graphing inequalities:
- a) Inequalities with " > " or " < " are dashed lines
- b) Inequalities with " ≥ " or " ≤ " are solid lines
- c) Solution is the common shaded area between the two inequalitites, if there is any.
- Step 1: Graph both system of equations. Best method is to find the x - and y - intercepts for both equations. You may replace inequalities with equal sign.
- Equation 1: x - Intercept

Set y to zero. Solve for x.

Point is (x,0)

x + 2y = − 4

x + 0 = − 4

x = − 4

( − 4,0) - Equation 1: y - Intercept

Set x to zero. Solve for y.

Point is (0,y)

x + 2y = − 4

0 + 2y = − 4

2y = − 4

y = − 2

(0, - 2)

Draw a Solid Line - Equation 2: x - Intercept

Set y to zero. Solve for x.

Point is (x,0)

3x + y = 3

3x + 0 = 3

3x = 3

x = 1

(1,0) - Equation 2: y - Intercept

Set x to zero. Solve for y.

Point is (0,y)

3x + y = 3

0 + y = 3

y = 3

(0,3)

Draw Solid Line - Test Equation 1: (0,0)

x + 2y ≤ − 4

0 + 2(0) ≤ − 4

0 ≤ − 4

Not True; Shade Away From (0,0) - Test Equation 2: (0,0)

3x + y ≤ 3

3(0) + (0) ≤ 3

0 ≤ 3

True; Shade Towards (0,0) - Step 3:Add a darker shade to the common shaded areas between the two lines.

2x − y < 3

4x + y < 3

- Things to remember when graphing inequalities:
- a) Inequalities with " > " or " < " are dashed lines
- b) Inequalities with " ≥ " or " ≤ " are solid lines
- c) Solution is the common shaded area between the two inequalitites, if there is any.
- Step 1: Graph both system of equations. Best method is to find the x - and y - intercepts for both equations. You may replace inequalities with equal sign.
- Equation 1: x - Intercept

Set y to zero. Solve for x.

Point is (x,0)

2x − y = 3

2x + 0 = 3

2x = 3

x = [3/2]

( [3/2],0) - Equation 1: y - Intercept

Set x to zero. Solve for y.

Point is (0,y)

2x − y = 3

0 − y = 3

− y = 3

y = − 3

(0, − 3)

Draw a Dashed Line - Equation 2: x - Intercept

Set y to zero. Solve for x.

Point is (x,0)

4x + y = 3

4x + 0 = 3

4x = 3

x = [3/4]

( [3/4],0) - Equation 2: y - Intercept

Set x to zero. Solve for y.

Point is (0,y)

4x + y = 3

0 + y = 3

y = 3

(0,3)

Draw Dashed Line - Test Equation 1: (0,0)

2x − y < 3

2(0) − 0 < 3

0 < 3

True; Shade Towards (0,0) - Test Equation 2: (0,0)

4x + y < 3

4(0) + (0) < 3

0 < 3

True; Shade Towards (0,0) - Step 3: Add a darker shade to the common shaded areas between the two lines.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Solving Systems of Inequalities By Graphing

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Solving by Graphing 0:08
- Graph Each Inequality
- Overlap
- Corresponding Linear Equations
- Test Point
- Example: System of Inequalities
- No Solution 7:06
- Empty Set
- Example: No Solution
- Example 1: Solve by Graphing 10:27
- Example 2: Solve by Graphing 13:30
- Example 3: Solve by Graphing 17:19
- Example 4: Solve by Graphing 23:23

### Algebra 2

### Transcription: Solving Systems of Inequalities By Graphing

*Welcome to Educator.com.*0000

*Today, we will be covering solving systems of inequalities by graphing.*0002

*In order to solve a system of inequalities by graphing, you need to graph the solution set of each inequality.*0009

*So, previously we discussed techniques; and you can review that lecture about how to find the solution set of an inequality by graphing.*0015

*Here, all you are going to do is graph each inequality.*0024

*And the solution of the system of the inequalities put together is the intersection of the solution sets of the inequalities.*0027

*So, what this is, really, is the overlap--the area of overlap--the points that are in common between each of the solution sets.*0035

*For example, if you are given a system of inequalities: y ≤ x - 3, and y > -2x + 1,*0044

*recall the techniques for graphing an inequality to determine the solution.*0056

*First, you want to graph the corresponding linear equation to find the boundary line of the solution set.*0063

*Then, use a test point to determine the half-plane containing the solution set.*0082

*Let's go ahead and do that, and then talk about how to find, then, the solution set.*0099

*This is just graphing each one--finding the solution set for each one; and then we have to talk about how to find the solution set for the system.*0105

*So, I am going to start with this one; and that is y ≤ x - 3; and the corresponding linear equation would be y = x - 3.*0111

*So, this is in slope-intercept form, y = mx + b, so I can easily graph this, because I know that the y-intercept is -3 and the slope is 1.*0126

*Since the slope is 1, increase y by 1; increase x by 1; increase y by 1; increase x by 1; and on.*0139

*Now, looking back here, this says y is less than or equal to -3; and what that tells me is that I am going*0148

*to use a solid line here for the boundary line, because the line is included in the solution set; so use a solid line to graph this.*0158

*The next thing is to determine which half-plane--the upper or lower half-plane--this solution set is in.*0174

*And I want to use an easy test point, and that is the origin (0,0), as my test point.*0179

*And it is well away from the boundary line, so I can use that.*0185

*If it were near or on the boundary, I would want to pick a different test point.*0189

*I am going to take my test point, (0,0), and substitute those values in to the inequality.*0194

*So, 0 is less than or equal to 0 minus 3; 0 is less than or equal to -3.*0200

*Well, that is not true; since that is not true, that means that the solution set is not in this upper half-plane.*0208

*(0,0) is not part of the solution set; therefore, the solution set is in this lower half-plane.*0216

*OK, that is the solution set for this inequality; now let's work on the other one.*0223

*Here, y is greater than -2x + 1; the corresponding linear equation is y = -2x + 1.*0229

*So, here I have the y-intercept at 1 and a slope of -2.*0240

*So, decrease y by 2...1, 2; increase x by 1; decrease y by 2; increase x by 1; and so on.*0246

*Now, since this is a strict inequality, I am going to use a dashed line, because the boundary line here is not part of the solution set.*0256

*OK, so here is my dashed line; next, I need to find a test point.*0267

*And again, I am going to use (0,0); I am going to use that test point.*0281

*Actually, let's use a different one, because that is a bit close; so let's go ahead and pick one*0293

*that is farther away, so we don't have any chance of causing confusion.*0297

*Let's use something up here...(2,1); (2,1) would be good for the test point.*0301

*So, my test point is going to be (2,1): I am going to insert (2,1) into that inequality.*0308

*1 is greater than -2; x is 2; plus 1; this is my test point; and 1 is greater than -4 plus 1; 1 is greater than -3.*0316

*And this is true; since this is true, that means that the test point is in the correct half-plane where the solution set lies.*0331

*So, for this line, this is the upper half-plane, and this is part of the solution set; so I am going to go ahead and shade that in.*0340

*OK, so far, I had just been going along, doing what I usually do when I would find the solution set for an inequality.*0356

*But remember, we are looking for the solution set for a system of inequalities.*0364

*So, this lower half-plane is the solution set for this inequality; this half-plane is the solution set for this inequality.*0369

*Now, I am looking for the intersection between the two solution sets, and that is right here, in this quadrant right here.*0376

*Darken that in even more; I am going to go ahead and use another color to emphasize that.*0382

*OK, so the area bounded by this line and including that line--that is the boundary line for the system of equations.*0396

*And then, the boundary line over here is the dashed line; so this line is not part of the solution set.*0406

*The technique: graph each of the inequalities; find their solution sets; and then find the area that is the intersection of the two solution sets.*0414

*You may come across a situation where there is no solution to a system of inequalities.*0427

*And this is because the two inequalities may not have any points in common.*0433

*Remember: the common points in each of the solution sets comprise the solution set for the system of inequalities.*0438

*If the two inequalities have no points in common, then the solution set of the system is the empty set.*0446

*For example, let's say I am given y < x - 4 and y ≥ x - 2.*0454

*So, I am going to go ahead and graph those, starting with y < x - 4.*0462

*I need to find the boundary line, so the corresponding linear equation is y = x - 4.*0467

*So, - 4 is the y-intercept, and the slope is 1.*0474

*So, increase x by 1, and increase y by 1, with each step.*0480

*Since this is a strict inequality, I am going to use a dashed line.*0487

*I have my boundary line; now I need to use a test point, and I am going to use this (the origin) as the test point.*0495

*And I am going to substitute (0,0) into the inequality and see what happens.*0503

*This tells me that 0 is less than -4; and that is not true.*0510

*Since that is not true, this is not part of the solution set; the solution set is actually below this line--the lower half-plane.*0516

*OK, that was my first inequality; now, my second inequality is y ≥ x - 2.*0525

*The corresponding linear equation is y = x - 2.*0532

*Here, I have a y-intercept of -2 and a slope of 1; so increase x by 1 and y by 1 each time.*0538

*Now, here I am going to use a solid line; this line is part of the solution set.*0551

*And a test point: I can use (0,0) again--that is well away from this boundary line.*0561

*Substituting these values into this inequality gives me 0 ≥ -2; and that is true: 0 is greater than -2.*0567

*So, this upper half-plane describes the solution set.*0578

*Now, you can see what happened here; and you may have already noticed that these two lines have the same slope.*0585

*Since they have the slope, and they are parallel lines, that means they are never going to intersect.*0592

*Since this solution set is above this line, and this one is below the line, there are not going to be any points in common.*0598

*These two lines will go along forever, and the points above and below them will never intersect.*0605

*So, this is a situation where there is no solution; we just say it is the empty set.*0611

*So here, we saw a situation where there is no solution, because there are no points in common.*0622

*OK, the first example is very straightforward: x ≥ 2; y > 3.*0628

*Starting with the x ≥ 2: the corresponding linear equation would be x = 2, and that just tells me that, if x equals 2,*0635

*no matter what the y-value is, it is just saying x is 2; so that is going to be a vertical line.*0650

*And since this is greater than or equal to, I am going to make this a solid line.*0657

*Now, you could certainly use a test point; or you could just look at this and say,*0674

*"Well, it is saying that x is greater than or equal to 2, which means that it is going to be values to the right."*0678

*You could certainly say, "OK, I want to do a test point at (0,0); 0 is greater than or equal to 2--not true, so this is not part of the solution set."*0685

*Or, like I said, you could just look back here and say, "OK, it is telling me that the values of x are greater than or equal to 2."*0697

*So, this half-plane contains the solution set.*0704

*The second inequality, y > 3, has a corresponding linear equation of y = 3.*0711

*So, if y equals 3, that is going to be a horizontal line right here.*0720

*And it is a strict inequality, so I am going to make that a dashed line, making it a different color so it shows up.*0725

*So, this blue line is that y = 3; OK.*0735

*Again, I could either just go back and say, "All right, this is y > 3, so those points would be up here";*0743

*or I can always use my test point, (0,0), and say 0 is greater than 3--that is not true, so I know this is not part of the solution set.*0750

*So, I need to go up here; OK.*0761

*So again, the solution set for the system of inequalities is the area of intersection of the two solution sets.*0774

*For this first inequality, the solution set is over here.*0782

*For the second inequality, the solution set is up here; and I can see the area of intersection is right up here.*0786

*And it is to the right of this solid vertical line, and it includes the points on the line.*0793

*And it is above the dashed blue line, and it does not include the points on the line.*0799

*We are just graphing each inequality and finding this area of intersection.*0805

*Here it is slightly more complicated, but the same technique.*0811

*Again, this is a system of inequalities, starting with the first one: y > x + 1.*0815

*The corresponding linear equation is y = x + 1.*0822

*The y-intercept is 1, and the slope is 1.*0828

*So, increase y by 1; increase x by 1; increase y by 1; increase x by 1.*0835

*And this is also going to be a dashed line, since it is a strict inequality; this boundary line is not part of the solution set.*0842

*Take a test point, (0,0); substitute back into that inequality to give 0 is greater than 0 + 1; so that says 0 is greater than 1.*0858

*Well, that is, of course, not true--which means that this half-plane where the test point is, is not the solution set.*0874

*Instead, it is the upper half-plane; so I am going to shade that in to indicate that this is the solution set for the first inequality.*0884

*The second inequality is y = -x + 2; well, the inequality is y > -x + 2; the corresponding linear equation is y = -x + 2.*0893

*So, that gives me a y-intercept of 2 and a slope of -1.*0907

*Decrease y by 1; increase x by 1; decrease y by 1; increase x by 1; and on down.*0914

*Again, it is a strict inequality, so we have another dashed line; I am making this a different color so it stands out as a separate line.*0923

*This blue line is the boundary line for the solution set of the second inequality.*0941

*OK, again, I am using (0,0) as my test point.*0949

*I am substituting into that inequality: 0 is greater than -0 + 2, so 0 is greater than 2.*0956

*And again, this is not true; so looking at this line, this point is not part of the solution set.*0966

*So, this lower half-plane is not the solution set.*0976

*I am going to go to the upper half-plane and shade that in.*0979

*OK, therefore, the solution set for the system of inequalities is right up here in this corner.*0996

*It is the points above the black line (but not including this line) and above the blue line (but not including that boundary line).*1004

*The technique, again: graph the first inequality (which was right here); we found the solution set in the upper half-plane.*1014

*We graphed the second inequality and found its solution set in the upper half-plane.*1022

*And then, we noted that this area up here is the overlap between the two inequalities; and that contains the solution set for the system.*1027

*OK, Example 3: 2x - y > 4 is my first inequality.*1041

*And one thing that I see right away is that this is not in the standard form we usually use.*1050

*So, I am going to work with this to put it in a standard form, where y is isolated on the left; and that is going to give me -y > -2x + 4.*1055

*Now, I have to divide by -1; and recall that when you divide an inequality by a negative number,*1067

*then you have to reverse the direction of the inequality symbol.*1073

*So, this is going to become y < 2x - 4; and it is absolutely crucial that, if you are multiplying*1079

*or dividing by a negative number, you immediately reverse the inequality symbol; or you won't end up with the correct solution set.*1089

*OK, so that is my first inequality; I am going to go ahead and do that same thing with the second inequality before I work with either one of these.*1096

*So, here I have 4x + 2y ≥ -2, so I am going to subtract 4x from both sides.*1107

*And then, I am going to divide both sides by 2; and since 2 is a positive number, I can just keep that inequality symbol as it was.*1118

*This is -4 divided by 2, so that gives me -2x - 1.*1125

*OK, looking at this first inequality: the corresponding linear equation, now in standard slope-intercept form, is y = 2x -4.*1138

*So, by putting it in that form, we made it much easier to graph.*1153

*The y-intercept is at -4, and the slope is 2.*1158

*So, increase y by 2; increase x by 1 for the slope.*1162

*Looking back, this is a strict inequality, so I need to use a dashed line, indicating that the boundary is not part of the solution set.*1169

*And then, a test point--I need to determine where the solution set lies--in the upper or lower half-plane.*1189

*I am using the origin as my test point, substituting (0,0) in for x and y.*1198

*This gives me 0 - 0 > 4, or 0 is greater than 4; and that is not true.*1208

*Since that is not valid, that is not part of the solution set; so this is not the correct half-plane.*1216

*I am going to shade the lower half-plane instead.*1221

*OK, over here, the corresponding linear equation is y = -2x - 1.*1228

*So, the y-intercept is -1; the slope is -2x.*1237

*So, when I decrease y by 2, I am going to increase x by 1; decrease y by 2; increase x by 1.*1243

*Again, I have to check here, and I see that I am going to use a solid line for this boundary line,*1258

*because the line is part of the solution set.*1264

*And then, I need a test point here; and I want to pick a test point that is not quite so close to this boundary line,*1278

*just in case my graphing wasn't perfect, so I am going to select (1,2) right up here as my test point, just to be safe.*1285

*Substitute those values into this inequality right here, or back up here--either way.*1296

*I am going to go ahead and use this one, because it is simpler: 2 ≥ -2(2) - 1.*1308

*2 is greater than or equal to -4 minus 1; so, 2 is greater than or equal to -5; and that is true.*1319

*So, my test point is in the half-plane containing the solution set.*1333

*I am going to shade this upper half-plane.*1340

*Once I have done that, I can see that the solution set for the set of inequalities, the system of inequalities, is right here.*1348

*It is the area bounded by this line and including the line, and then the area bounded by the dashed line,*1356

*right here, but not including it; so, this lower right section of the graph.*1363

*For this one, we had to take an extra step, just to make it easier--put this in standard form.*1372

*And then, we graphed the corresponding linear equation to find the boundary line.*1376

*We used the test point to find that the lower half-plane was the solution set for this first inequality.*1381

*The same technique for the other inequality: and the upper half-plane turned out to contain the solution set.*1388

*So, my area of intersection is right here; that is the solution set for the system.*1395

*OK, Example 4: y < 2x + 1; y ≥ 2x + 3.*1403

*We are going ahead and starting out with the first inequality, y < 2x + 1.*1412

*I am finding my boundary line with the corresponding linear equation, y = 2x + 1.*1424

*Here, the y-intercept is 1; the slope is 2; so increasing y by 2 means increasing x by 1, and continuing on.*1429

*I am checking back and seeing that I need a dashed line, because this is a strict inequality.*1446

*So, I am graphing out this line with a dashed line; so this line will not be part of the solution set--the points on the line.*1452

*The second inequality is y ≥ 2x + 3; the linear equation is y = 2x + 3.*1463

*I am going to go ahead and graph this out; and this tells me that the y-intercept here is 3.*1475

*And the slope is 2: increase y by 2, increase x by 1; or decrease y by 2, decrease x by 1; and so on.*1484

*So, this is going to give me another line, right next to this one.*1499

*But this time, I am actually going to use a solid line.*1504

*So, this is a solid line, because this is greater than or equal to.*1509

*Now, let's look at some test points for each.*1517

*For this first one, let's use a test point of (0,0) right here for this line and substitute in.*1520

*0 is less than 2 times 0 plus 1; that gives me 0 < 0 + 1, or 0 is less than 1.*1528

*And that is true; so I have this as part of my solution set right here, so it is the lower half-plane.*1537

*I found the solution set for this inequality.*1550

*For this one, I am also going to use (0,0); that is well away from that line--that is a good test point.*1552

*Substitute in: 0 is greater than or equal to 2 times 0, plus 3; 0 ≥ 0 + 3; 0 is greater than or equal to 3.*1560

*That is not true; so, for this line, (0,0) is not part of the solution set; so it is the upper half-plane.*1572

*Now, it is a little tough to draw: but if you look here, one thing that you will see is that these lines have the same slope.*1585

*Because of that, I know that these are parallel lines, so I know that these two lines are parallel and that they will never intersect.*1593

*Since the points are all below this parallel line in that half-plane, and above this line (they are parallel to it), these solution sets are never going to intersect.*1602

*In this case, the solution is the empty set; there are no points in common, since these are parallel lines.*1614

*That concludes this session on solving systems of inequalities by graphing at Educator.com; see you again!*1625

1 answer

Last reply by: Dr Carleen Eaton

Thu Jul 31, 2014 7:09 PM

Post by Philippe Tremblay on July 23, 2014

I draw a short dotted line (like that: ---) next to the equation if it's a strict inequality, and a short solid line when it's not. It takes no time and I find it useful.

1 answer

Last reply by: Dr Carleen Eaton

Tue Oct 9, 2012 11:45 PM

Post by Carroll Fields on October 1, 2012

In the latter part of Example III, 4x+2y> or = to -2. The test point is (1,2) , so the result should be 2 is greater than or equal to -2(1)-1; not 2 is > or = to -2(2)-1. Making the final answer 2> or = to -3, not 2> or = to -5 ( as listed )