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CCSS.Math:

let's do some compound inequality problems and these are just inequality problems that have more than one set of constraints you're going to see what I'm talking about in a second so the first problem I have is negative 5 is less than or equal to X minus 4 which is also less than or equal to 13 so we have two sets of constraints on our on the set of X's that satisfy these equations X minus 4 has to be greater than or equal to negative 5 and X minus 4 has to be less than or equal to 13 so we could rewrite this compound inequality as negative 5 has to be less than or equal to X minus 4 and and X minus 4 and X minus 4 needs to be less than less than or equal to 13 and then we could solve each of these separately and then we have to remember this and there to think about the solution set because it has to be things that satisfy this equation and this equation so let's solve each of them individually so this one over here we can add 4 to both sides of the equation let's add 4 to both sides of the equation the left-hand side negative 5 plus 4 is negative 1 negative 1 is less than or equal to X right these fours just cancel out and you're just left with an X on this right-hand side so this the the left this part right here simplifies to X needs to be greater than or equal to negative 1 or negative 1 is less than or equal to X so we could also write it like this X needs to be greater than or equal to negative 1 these are equivalent I just swapped the sides now let's do this other condition here in green let's do this other condition over here in green let's add 4 to both sides of this equation let's add 4 to both sides of that equation the left-hand side we just get an X and then the right-hand side we get 13 plus 14 which is 17 so we get X is less than or equal to 17 so are two conditions X has to be greater than or equal to negative 1 and less than or equal to 17 so we could write this as again as a compound inequality if we want we can say that the solution set that X has to be less than or equal to 17 or or and greater than or equal to negative 1 has to satisfy both of these conditions so what would that look like on a number line on a number line so let's start a number line right there let's say that this is 17 maybe that's 18 you keep going down maybe this is zero I'm obviously skipping a bunch of stuff in between then we would have a negative 1 right there may be a negative 2 so X is greater than or equal to negative 1 so we would start at negative 1 we're going to circle it in because we have a greater than or equal to and then X is greater than that it's greater than that but it has to be less than or equal to 17 so it has to be less than or equal to 17 so it could be equal to 17 or less than 17 so this right here is a solution set everything that I've shaded in orange and if we wanted to write it in interval notation it would be X is between negative 1 and 17 and it it can also equal negative 1 so we put a bracket and can also equal 17 so this is the interval notation for this compound inequality right there let's do another one let's do another one let me let me get a good problem here let's say that let's say we have negative 12 I'm going to change the problem a little bit from the one that I've found here negative 12 is less than 2 minus 5x which is less than or equal to 7 I wanted to do a problem that has just a less than and a less than or equal to the problem in the book that I'm looking at has an equal sign here but I want to remove that intentionally because I want to show you when you have a hybrid situation when you have a little bit of both so first we can divide this into or separate this into two normal inequalities you have this inequality right there we know that negative twelve needs to be less than two minus five X that has to be satisfied and and let me do it another color this inequality also needs to be satisfied two minus five X has to be less than seven and greater than twelve less than or equal to seven and greater than negative 12 so and two minus five X has to be less than or equal to seven so let's just solve this the way we solve everything let's get this two on to the left-hand side here so let's subtract two from both sides of this equation so if you subtract two from both sides of this equation the left-hand side becomes negative 14 is less than these cancel out less than negative 5x now let's divide both sides by negative five and remember when you multiply or divide by a negative number the inequality swaps around so if you divide both sides by negative five you get a negative 14 over negative five and you have an X on the right hand side right if you divide that by negative 5 and this swaps from a less than sign to a greater than sign and these the negatives cancel out so you get 14 over 5 is greater than X or X is less than 14 over 5 which is what is this this is this is 2 and 4/5 X is less than 2 and 4/5 I just wrote this improper fraction as a mixed number now let's do the other constraint over here in magenta so let's subtract 2 from both sides of this equation just like we did before and actually you can do these simultaneously but it becomes kind of confusing so to avoid careless mistakes I encourage you to separate it out like this so if you subtract 2 from both sides of the equation the left-hand side becomes negative 5x the right-hand side you have less than or equal to the right-hand side becomes 7 minus 2 comes 5 now you divide both sides by negative 5 on the left hand side you get an X on the right hand side five divided by negative five is negative one and since we divided by a negative number we swap the inequality goes from less than or equal to to greater than or equal to so we have our two constraints X has to be less than 2 and 4/5 and it has to be greater than or equal to negative 1 so we could write it like this X has to be greater than or equal to negative 1 so that would be the lower bound on our interval and it has to be less than 2 and 4/5 and X has to be less than 2 and 4/5 and notice not less than or equal to that's why I wanted to show you you have the parentheses there because it can't be equal to 2 and 4/5 X has to be less than 2 and 4/5 or we could write it this way X has to be less than 2 and 4/5 that's just this inequality swapping the sides and it has to be greater than or equal to negative 1 so these two statements are equivalent and if I were to draw it on a number line if I were to draw it on a number line it would look like this so you have a negative 1 you have 2 and 4/5 over here 2 and 4/5 obviously you'll have stuff in between maybe you know 0 sitting there we have to be greater than or equal to negative 1 so we can be equal to negative 1 and we're going to be greater than negative 1 but we also have to be less than 2 and 4/5 so we can't include 2 and 4/5 there we can't be equal to 2 and 4/5 so we can only be less than so we put a empty circle around 2 and 4/5 and then we we fill in everything below that all the way down to negative 1 and we include negative 1 because we have this less than or equal sign so the last two problems I did are kind of and problems you have to meet both of these constraints now let's do an or problem let's do an or problem so let's say I have these inequalities let's say I'm given let's say that 4x minus 1 needs to be greater than or equal to 7 or or 9x over two needs to be less than three so now when we're saying or an X that would satisfy these or X's that satisfy either of these equations in the last few videos or the last few problems we have to find X's that satisfy both of these equations here this is much this is much more lenient we just have to satisfy one of these two so let's figure out the solution sets for both of these and then we figure out essentially their union their combination all the things that will satisfy either of these so on this one on the one on the Left we can add one to both sides you add one to both sides the left-hand side just becomes four X is greater than or equal to 7 plus 1 is 8 divide both sides by 4 you get X is greater than or equal to 2 or or let's do this one let's see if we multiply both sides of this equation by 2 ninths what do we get if we multiply both sides by two nines it's a positive number so we don't have to do anything to the inequality these cancel out and you get X X is less than 3 times 2 over 9 3 over 9 is the same thing as 1 over 3 so X needs to be less than two-thirds so or X is less than 2/3 so that's our solution set X needs to be greater than or equal to 2 or less than two-thirds so this is interesting let me plot the solution set on the number line let me plot it on the number line so that is our number line maybe maybe this is 0 this is 1 this is 2 3 maybe that is negative 1 so X can be greater than or equal to 2 so we could start let me do it in another color we can start at 2 here and be greater than or equal to 2 so include everything greater than or equal to 2 that's that condition right there or X could be less than 2/3 or X could be less than 2/3 so 2/3 is going to be right around here right that is 2/3 X could be less than 2/3 and this is interesting because you know if we pick one of these numbers it's going to satisfy this inequality if we pick one of these numbers it's going to satisfy that inequality if we had an end here there would have been no numbers that satisfy it because you can't be both greater than two and a less than two-thirds so the only way that there is any solution set here is because it's or you can satisfy one of the two inequalities anyway hopefully you found that fun