INSTRUCTORS Carleen Eaton Grant Fraser

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 1 answerLast reply by: Dr Carleen EatonThu Jul 31, 2014 7:09 PMPost by Philippe Tremblay on July 23, 2014I 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 answerLast reply by: Dr Carleen EatonTue Oct 9, 2012 11:45 PMPost by Carroll Fields on October 1, 2012In 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 )

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

Solve 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
Solve by graphing:
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
• Test Equation 2: (0,0)
y >− x + 2
0 >− (0) + 2
0 > 2
Not True; Shade Away From (0,0)
Solve by graphing:
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
• Test Equation 2: (0,0)
y ≥ − [1/3]x − 2
0 ≥ − [1/3](0) − 2
0 ≥ − 2
Solve by graphing:
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.
• 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] = 0
• Equation 1: b = 1
• Equation 2: m = [rise/run] = [3/2] or [(−3)/(−2)]
• 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 > 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)
Solve by graphing:
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.
• 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] = [1/2] or [(−1)/(−2)]
• Equation 1: b = -1
• Equation 2: m = [rise/run] = [2/1] or [(−2)/(−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 < [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
Solve by graphing:
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.
• 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] = [2/3] or [(−2)/(−3)]
• Equation 1: b = 3
• Equation 2: m = [rise/run] = [2/3] or [(−2)/(−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)
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.
Solve by graphing:
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.
• 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] = [3/2] or [(−3)/(−2)]
• Equation 1: b = -1
• Equation 2: m = [rise/run] = [1/2] or [(−1)/(−2)]
• Equation 2: b = 1
• 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 < [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
Solve by graphing:
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.
• 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 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
• 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)
2x + y < 3
2(0) + 0 < 3
0 < 3
• Test Equation 2: (0,0)
3x − y ≥ 2
3(0) − (0) ≥ 2
0 ≥ 2
Not True; Shade Away From (0,0)
Solve by graphing:
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.
• 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 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
• 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 + 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
Solve by graphing:
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.
• 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 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
• 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)
2x − y < 3
2(0) − 0 < 3
0 < 3
• Test Equation 2: (0,0)
4x + y < 3
4(0) + (0) < 3
0 < 3

*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.

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

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 going0148

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 one0293

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 multiplying1079

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

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