WEBVTT 00:21.520 --> 00:27.962 In this video, we're going to discuss compound inequalities. Compound inequalities 00:28.000 --> 00:38.205 are two inequalities that are joined by the word "and" or "or." A value is a solution of a compound 00:38.240 --> 00:48.182 inequality containing "and" if it is a solution of both inequalities. Let's consider the inequality x 00:48.240 --> 00:57.158 is less than or equal to five and x is greater than or equal to three. To find the solution set 00:57.200 --> 01:05.065 of this compound inequality, let's first consider the inequality x is less than or equal to five. 01:05.120 --> 01:15.709 This would contain five and those points to the left of five on the number line. For the 01:15.760 --> 01:25.219 inequality x is greater than or equal to three, that solution set would include the value three 01:25.280 --> 01:33.060 and all of the numbers to the right of three on the number line. 01:33.120 --> 01:40.267 The solution set to the compound inequality x is less than or equal to five and x is greater 01:40.320 --> 01:47.374 than or equal to three is the intersection of these two solution sets, meaning that the 01:47.440 --> 01:53.314 solution set to the compound inequality must include the values that are in both 01:53.360 --> 02:04.758 solution sets. So that would include the values from three to five, including three and five. 02:04.800 --> 02:15.069 The places where these solution sets overlap is the solution set to this compound inequality. 02:15.102 --> 02:26.180 We can also write this compound inequality in the following way: three is less than or equal to x is 02:26.213 --> 02:36.557 less than or equal to five. Now let's look at some examples of compound inequalities. 02:36.590 --> 02:42.730 We will solve the compound inequality and graph the solution set on the number line for each 02:42.763 --> 02:52.339 of these examples. Our first example is negative two is less than two x minus eight is less than six. 02:52.400 --> 02:59.780 We'll begin by using the Addition Property of Inequalities and adding eight to all three 02:59.840 --> 03:08.122 sides of this inequality. When I add a number or subtract a number to an inequality, it does not 03:08.160 --> 03:19.700 change the direction of the inequality sign. So we have six is less than two x is less than fourteen. 03:19.760 --> 03:27.307 Next, we'll use the Multiplication Property of Inequalities, and we'll divide all three sides by 03:27.360 --> 03:34.581 two. This is a positive value two, so it does not change the direction of our inequality symbols. 03:34.640 --> 03:44.925 Dividing all three pieces, or sides, by two gives us three is less than x is less than 03:44.960 --> 03:54.735 seven. Now, let's graph the solution set on our number line. 03:54.800 --> 04:00.908 Since my end points here of this inequality contain three and seven those are the values 04:00.941 --> 04:10.250 that I'll include on my number line, three and seven. Notice that there are no equal to components 04:10.284 --> 04:17.224 on these inequalities. They are simply less than signs. Therefore, three is not a solution of this 04:17.257 --> 04:25.165 compound inequality, nor is seven a solution of this compound inequality. So we'll put open circles at 04:25.199 --> 04:33.006 three and seven, and the solution set of this inequality contains all of the values that are both 04:33.040 --> 04:43.317 less than seven and greater than three which are all the values in between three and seven. 04:43.360 --> 04:51.458 In our next example, we have negative one is less than or equal to two-thirds x plus three is 04:51.520 --> 05:01.802 less than four. Let's begin by subtracting three from all three sides of this linear inequality. 05:01.840 --> 05:09.309 Subtracting three does not change the direction of our inequality symbols. Negative one 05:09.360 --> 05:17.885 minus three is negative four. The symbol stays the same less than or equal to 05:17.920 --> 05:27.728 two-thirds x. And again, the symbol stays the same less than four minus three, which is one. 05:27.761 --> 05:36.603 And now, we'll multiply by the reciprocal of two-thirds which is three-halves to all three sides. 05:36.637 --> 05:43.243 Since three-halves is a positive number, this does not change the direction of any of the inequality symbols. 05:43.277 --> 05:56.657 Multiplying by three-halves here gives us three times negative two, which is negative six. 05:56.690 --> 06:12.973 Multiplying by three-halves here gives us x. And multiplying by three-halves here gives us three-halves. 06:13.006 --> 06:19.513 Again, notice that the inequality symbols did not change directions. Now let's graph the 06:19.546 --> 06:31.224 solution set on a number line. We have negative six and a positive three-halves, which is one and a half. 06:31.280 --> 06:37.864 Those are the endpoints of this solution set, so I'll put those numbers on my number line. 06:37.920 --> 06:46.106 Negative six is part of the solution set because of this inequality that has an equal to component. 06:46.160 --> 06:52.112 Negative six is less than or equal to x, so we'll put a closed circle at negative six. And we'll 06:52.160 --> 06:57.617 have an open circle at positive three-halves since three-halves is not part of the solution 06:57.680 --> 07:06.393 set. And the values for x which are both less than three-halves and greater than or equal to 07:06.426 --> 07:19.306 negative six include the values between those two circles. This is the solution set of the inequality. 07:19.360 --> 07:27.314 Our third example is zero is less than or equal to four minus x divided by five is less than or 07:27.360 --> 07:34.821 equal to two. Let's begin by multiplying all three pieces of this inequality by positive 07:34.880 --> 07:42.596 five. Multiplying by positive five does not change the direction of the sign of these inequalities. 07:42.640 --> 07:50.604 Zero times five is zero. Multiplying by five on the inside portion of this inequality 07:50.640 --> 08:01.715 leaves us with four minus x. And multiplying by five on the right side gives us ten. Now we'll use 08:01.760 --> 08:09.155 the Addition Property of Inequality to subtract four from all three sides of this inequality. 08:09.200 --> 08:20.600 Zero minus four is negative four. Four minus x minus four is negative x. And ten minus four 08:20.640 --> 08:30.577 is six. Now I have a negative x in the center of this compound inequality. So to undo this negative 08:30.640 --> 08:38.285 in front of the x, I'm going to either multiply or divide all three pieces by negative one. 08:38.320 --> 08:44.357 When you multiply or divide by a negative in an inequality, it does change the direction of these 08:44.400 --> 08:52.899 symbols. Multiplying negative four by negative one gives us positive four, and we'll flip the sign 08:52.960 --> 09:00.674 to a greater than or equal to symbol. Negative x times negative one is positive x. We'll flip 09:00.720 --> 09:09.015 this sign to a greater than or equal to symbol. And six times negative one is negative six. 09:09.049 --> 09:17.157 Now while this is a true inequality or a correct inequality equivalent to the given statement, 09:17.200 --> 09:21.695 the standard format here would be to put the smallest number 09:21.760 --> 09:28.501 on the left side of the inequality. So I will rewrite this inequality as negative six is less 09:28.560 --> 09:39.145 than or equal to x is less than or equal to four. And now let's graph the solution set. 09:39.200 --> 09:46.353 The endpoints of the solution set are negative six and four. And because there's an equal to 09:46.400 --> 09:53.326 component on both of these inequality symbols, we'll put a closed circle at each of those points. 09:53.360 --> 09:57.797 And the values of x that are both less than or equal to four 09:57.840 --> 10:07.240 and greater than or equal to negative six are the values here between these two closed circles. 10:07.280 --> 10:18.418 This is the graph of the solution set of this inequality. Now let's talk about another type of 10:18.451 --> 10:23.223 compound inequality. A value is a solution of a compound 10:23.256 --> 10:32.365 inequality containing "or" if it is a solution of either inequality. Let's consider the example 10:32.400 --> 10:39.072 x is less than or equal to one or x is greater than or equal to three. 10:39.120 --> 10:45.945 First, we'll begin by graphing the solution set to x is less than or equal to one. So on our number 10:46.000 --> 10:55.221 line we'll have the numbers one two and three. And we'll begin by graphing x is less than or 10:55.280 --> 11:02.662 equal to one by putting a closed circle at one and shading those values to the left of one 11:02.720 --> 11:10.603 because they are less than one. So this is the solution set to x is less than or equal to 11:10.640 --> 11:21.481 one. Now let's graph the solution set to x is greater than or equal to three. 11:21.520 --> 11:27.787 We'll begin with a closed circle at three because three is included in the solution set due to the 11:27.840 --> 11:34.360 equal to component of that inequality symbol. And we'll shade those numbers to the right of three 11:34.400 --> 11:43.636 on our number line since those are greater than three. So this is the graph of the solution set 11:43.680 --> 11:51.878 of x is greater than or equal to three. For the inequality that contains x is less than or equal 11:51.920 --> 12:01.955 to one or x is greater than or equal to three, this graph is the union of these two solution sets, 12:02.000 --> 12:13.633 meaning the graph should contain all of the solutions from each of the two solution sets. 12:13.680 --> 12:24.344 So it will contain all of the values here less than or equal to one as well as the values here 12:24.400 --> 12:32.752 greater than or equal to three. This is the graph of the compound inequality 12:32.800 --> 12:40.827 x is less than or equal to one or x is greater than or equal to three. 12:40.880 --> 12:46.599 Let's take a look at these next two examples. We're going to solve the compound inequality for 12:46.640 --> 12:53.873 each and then graph the solution set on a number line. Our first compound inequality is five x 12:53.920 --> 13:01.147 minus three is less than or equal to ten or x plus one is greater than or equal to five. 13:01.200 --> 13:07.787 Recall that a compound inequality has two inequalities combined by an "and" or an "or" 13:07.840 --> 13:15.695 statement. We'll begin by solving this first piece of this compound inequality. Using the Addition 13:15.760 --> 13:25.471 Property of Inequality, we'll add three to both sides, giving us five x is less than or equal to 13:25.520 --> 13:31.944 thirteen. And now we'll use the Multiplication Property of Inequality and divide both sides by 13:32.000 --> 13:39.318 five. Since five is a positive number, that does not change the direction of the inequality sign. 13:39.360 --> 13:48.828 This gives us x is less than or equal to thirteen-fifths. Our compound inequality 13:48.880 --> 13:55.535 contained the word "or." We will continue to write that throughout this process of solving. 13:55.600 --> 14:02.508 To solve this piece, we'll subtract one from both sides, giving us x is greater than or equal to 14:02.560 --> 14:09.882 four. So our complete solution here to this compound inequality will be x is less than or 14:09.920 --> 14:23.863 equal to thirteen-fifths or x is greater than or equal to four. Now let's graph this solution set. 14:23.920 --> 14:36.509 Thirteen-fifths is two and three-fifths, so this will be between the numbers two and three. 14:36.560 --> 14:42.582 And I also have the number four that's relevant in this solution set as well. So 14:42.640 --> 14:53.726 two and three-fifths will be close to the center here. This will be the value thirteen-fifths. 14:53.760 --> 15:01.467 We'll have a closed circle at thirteen-fifths, and we'll shade all the values that are less than 15:01.520 --> 15:07.873 thirteen-fifths or to the left of that number on the number line. 15:07.920 --> 15:12.445 In addition, we'll have all of the values that are greater than or equal to four. 15:12.480 --> 15:20.519 So we'll have a closed circle at four, and we'll shade all of the values to the right of four. 15:20.560 --> 15:25.157 Recall that the solution set to a compound inequality containing "or" 15:25.200 --> 15:35.001 must be a solution to either of these two or the union of these two solution sets. 15:35.040 --> 15:44.577 Now let's take a look at this compound inequality. Five times x minus two is less than five 15:44.640 --> 15:53.019 or x plus nine is greater than ten. We'll begin by solving this left piece of the inequality. 15:53.052 --> 15:58.024 I'll divide both sides by five, which is a positive number thus not changing 15:58.080 --> 16:05.064 the direction of our inequality sign. And leaving us with x minus two is less than one. 16:05.120 --> 16:15.007 Then adding two to both sides gives us x is less than three. We'll continue to write "or" 16:15.041 --> 16:21.840 throughout our solution process. And solving this piece of the inequality, we'll subtract nine from 16:21.847 --> 16:30.022 both sides, giving us x is greater than one. So the solution to this inequality will contain 16:30.080 --> 16:43.002 x is less than three or x is greater than one. Now let's graph the solution set on a number line. 16:43.040 --> 16:50.042 I'll put the numbers one, two, and three on this number line. Let's start 16:50.080 --> 16:56.182 with x is less than three. That would have an open circle at three and 16:56.240 --> 17:06.258 all of the values left of three, or smaller than three, would be part of the solution set. 17:06.320 --> 17:14.200 In addition, we have to include the solution set for x is greater than one. So at one we would have 17:14.240 --> 17:22.041 an open circle, but notice that at x equals one we already have this as part of the solution set. 17:22.080 --> 17:29.081 And now we will shade all the values to the right of one which are greater than one. So 17:29.120 --> 17:35.087 that's going to include this number three and the numbers all the way to the right 17:35.120 --> 17:43.462 on this number line. Notice we have shaded the entire number line. So every number on this number 17:43.520 --> 17:59.840 line is a solution to this compound inequality, meaning that the solution set is all real numbers.