 Let's take a look at solving some more linear inequalities. Remember, we might have two or more inequalities linked with either AND or OR. For example, negative 5 less than or equal to 3x plus 7 AND 3x plus 7 less than 18. Another type of compound inequality involves OR, 3x plus 7 less than negative 5 OR, 2x minus 8 greater than negative 2. The important things to remember, if inequalities are ANDed, a solution must solve all of them. Meanwhile, if inequalities are ORed, a solution must solve at least one of them. For example, let's consider the compound inequality 3x plus 7 less than negative 5 OR, 2x minus 8 greater than negative 2. So we have two inequalities, and so there are two corresponding equations to solve. This gives us two critical values, so we need to check our critical values. If x equals negative 4, 3x plus 7 less than negative 5 is false. But because this is an OR inequality, only one of the inequalities needs to be true. So maybe our other inequality is true. Well, let's check it out. So if x equals negative 4, 2x minus 8 greater than negative 2 is also false. So x equals negative 4 fails both inequalities, and so we exclude x equals negative 4. Let's go ahead and graph this. Since x equals negative 4 is a critical value, but it's excluded, we'll put an open circle at negative 4. We'll check the other critical value. If x equals 3, 2x minus 8 greater than negative 2 is false. Again, because this is an OR inequality, only one needs to be true, so we'll check the other one. If x equals 3, 3x plus 7 less than negative 5 is also false. So we exclude x equals 3 from our solution. And again, since x equals 3 is a critical value, but it's excluded, we'll put an open circle at 3. Now because we have two critical values, we have three intervals to check. So let's go big or go home in our first interval. If x equals minus 1 million, 3x plus 7 less than 15 is true. And since this is an OR inequality, only one of our inequalities needs to be true. So we found an inequality that's true for x equals minus 1 million, so we include the interval that contains it. Now in our middle interval, we can try x equals 0 as a test point. If x equals 0, 3x plus 7 less than 15 is false. But since this is an OR inequality, we need to check the other because if that's true, we do get to include this interval. So if x equals 0, 2x minus 8 greater than negative 2 is also false. So again, remember that if inequalities are ORed, a solution must solve at least one of them. And since x equals 0 solves neither, then x equals 0 doesn't solve the inequality. And so we exclude this middle interval. On the right side, if x equals 1 million, 3x plus 7 less than 15 is false. But since this is an OR inequality, we need to check the other one. If x equals 1 million, 2x minus 8 greater than negative 2 is true. And because this is an OR inequality, we only need one of the inequalities to be true. Well, we found one, and so we include the interval that contains x equals 1 million. Now that we have the graph of the solution, let's go ahead and write our solution in interval notation. There's two intervals that will be unioned together. This first interval goes from minus infinity up to negative 4 excluding negative 4. And this second interval goes from 3 to infinity excluding both endpoints. What if we have an AND inequality? So again, we'll solve the corresponding equations. This gives us critical values x equals 1 and x equals 29 fifths. And remember, a good math student and a good human being remembers that the inequality exists. So we do have to check our critical values. Because this is an AND inequality, both inequalities must be satisfied by any solution. So if x equals 1, 8 minus 3x greater than or equal to 5 is true. But 5x minus 4 greater than or equal to 25 is false. So x equals 1 is not part of the solution. As before, we'll graph our solution. So x equals 1 is a critical value, but it's excluded. So we'll put an open circle at x equals 1. If x equals 29 fifths, 5x minus 4 greater than or equal to 25 is true. But 8 minus 3x greater than or equal to 5 is false. So again, x equals 29 fifths is not part of the solution. So again, we want to graph our solution. We'll put an open circle at x equals 29 fifths. So now we have two critical values in three intervals. So we'll check a test point in each interval. In our first interval, we try x equals negative 1 million. If x equals negative 1 million, then 5x minus 24 greater than or equal to 25 is false. Since this is an AND inequality, we need both inequalities to be satisfied. And since this first one isn't, there's no need to check the other. The interval containing x equals minus 1 million is not part of the solution. Let's take a look at our middle interval. If x equals 2, then 5x minus 4 greater than or equal to 25 is false. And again, both inequalities have to be satisfied. And since this one isn't, there's no need to check the other. The interval containing x equals 2 is not part of the solution. If x equals 1 million, then 5x minus 4 greater than or equal to 25 is true. But don't celebrate just yet. Since this is an AND inequality, we need both inequalities to be satisfied. So we have to check the other one. If x equals 1 million, then 8 minus 3x greater than or equal to 5 is false. Since one of the inequalities isn't true, the interval containing x equals 1 million is not part of the solution. And notice that we've eliminated every real number, and so that tells us that this inequality has no solution.