Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Unions and intersections. Compound inequalities mean that we have two or more inequalities, and we're going to look at the solution in all the pieces. The Upside down U stands for intersection and if it's in the intersection it has to be in all sets. Solution must be in all sets if it's a union, IE the capital U must be in at least one set. So an example, if X is greater than 7, union X is less than or equal to 9. So I'm going to put 7 and I'm going to put 9 on my number line. X is greater than 7, so an open dot shaded to the right and then X is less than or equal to 9, so a solid dot shaded off to the left. Now this union says it has to be in at least one set. So if we read from left to right, we can see down here at negative Infinity, it's covered by this arrow here. When I get to 7, even though it's not covered by the bottom or the blue, it is covered by the top, which is the black. So 7 is defined in between 7:00 and 9:00, they're both OK. At 9 they're both OK again, and to the right it's that lower arrow or the blue one. So this solution is going to be all real numbers. Negative Infinity to Infinity is how we're going to write it in interval notation. If we wanted to do set notation, it gets a little trickier. It's X such that X is an element, funny little sideways E of the real numbers, and the real numbers we can write as a symbol capital R with two lines in the front. Now looking at the same problem, but this time doing an intersection. If we do an intersection, what we're wanting is we want where they're both true. So now when we look at the same problem, but the symbol has changed, I want where they're both true. They're both true only on 7:00 to 9:00, not including seven. So we're going to use a parenthesis including nine. So we're going to use a bracket that's only where both of them are true. Set notation would be X such that 7 is less than X is less than or equal to 9. Going to let you try a couple of those. If we look at the domain, the domain is all possible X values. So when we have a fraction, we know that the denominator can never equal 0. So 5 -, 10 X can never equal 0. We're going to go ahead and solve this. So we're going to take that five to the other side and divide by -10 so we can see that X is never equal to 1/2. So given myself some symbols at 1/2, we're going to have to have an open dot, but it's true anywhere else. So everywhere else it's true except for that 1/2 location. So our solution is going to be negative Infinity to 1/2, and we're going to use parenthesis because it doesn't include 1/2 union 1/2 to Infinity. Set notation gets a little trickier, X such that X is an element. But actually let's just do X such that X doesn't equal 1/2. We'll make the understanding that it's over the real numbers right now. This this next example, 8 -, 2 X. We know that the inside of a square root always has to be positive. So 8 -, 2 X is greater than or equal to 0. If we go ahead and solve this, we have -2 X greater than or equal to -8. If we divide or multiply by a negative, we have to flip that inequality sign so X is less than or equal to four. Solid dot at 4 shaded off to the left, so negative Infinity to four. We're going to use a bracket because it was a solid dot or X such that X less than or equal to four. The preference typically is the interval notation, which is the one with the parentheses and brackets. But I am showing both set and interval notation. Typically we use interval this one set couple for you to try. This next one gets a little more complex because we have multiple pieces. So we're going to just solve this all and then see where the solution is. So the first thing I'm going to do is multiply all three pieces by two. Six greater than or equal to X -, 4 greater than 12. Then I'm going to add 4, so 10 greater than or equal to X greater than 16. Well, if I look at 10 here, I want all of the X values that are less than or equal to 10. If I rewrote just this portion, I could see that 10 greater than or equal to X is the same thing as X less than or equal to 10. Then I also have this X greater than 16 portion 111213141516. Now because it was a compound inequality, this X is between. I need both of these to be true. Are both of these ever true? And the answer is no. I can't have both of them true at the same time. So this is actually a no solution. There's no place that both are true at the same time. So no solution or a circle with a slash through it is another way to write no solution or the empty set. We'll let you. Actually, let's go ahead and do this last one too. If I multiply 3 by a -3, we get 6 greater than or equal to four X + 14 greater than -3. Subtracting 14 from all three pieces, we get -8 greater than or equal to 4X greater than -17. If I divide by 4 negative 2 greater than or equal to X greater than -17 force. So we get this -2 greater than or equal to X&X greater than -17 force. So if I have -2 greater than or equal to XI could think of that as X less than or equal to -2. Here's 01/2, so X less than or equal to -2 X greater than -17 fourths. Well, four goes into 17 four times that's 16 and 1/4 leftover. So -2 negative three -4 and 1/4 right about here and it's going to be X greater than. So our solution here would be -4 and 1/4 to -2. Using the brackets and parentheses appropriately, we could write it also as X such that -4 and 1/4 less than X less than or equal to -2. When we write our set notation and our interval notation both, we should read from left to right so that -4 and one fourth came first and then the -2. Thank you and have a wonderful day.