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# Compound inequalities: OR

CCSS.Math:

## Video transcript

solve for Z 5z plus seven is less than 27 or negative 3z is less than or equal to 18 so this is a compound inequality we have two we have two conditions here so Z can satisfy this or Z can satisfy this over here so let's just solve each of these inequalities and just know that Z can satisfy either of them so let's just look at this so if we look at just this one over here we have 5z plus 7 is less than 27 let's isolate the Z's on the left-hand side so let's subtract 7 from both sides to get rid of this 7 on the left-hand side and so our left-hand side is just going to be 5z plus 7 minus 7 those cancel out 5z is less than 27 minus 20 sorry 27 minus 7 is 20 so we 5z is less than 20 now we can divide both sides of this inequality by 5 and we don't have to swap the inequality because we're dividing by a positive number and so we get Z is less than 20 over 5 Z is less than 4 now this was only one of the conditional script the other one over here we have negative 3z is less than or equal to 18 now to isolate the Z we could just divide both sides of this inequality by negative 3 but remember when you divide both or multiply both sides of an inequality by a negative number you have to swap the inequality so this is so we could write this negative 3z we're going to divide it by negative 3 and then you have 18 we're going to divide it by negative 3 but we're going to swap the inequality so the less Center equal will become greater than or equal to and so these guys cancel out negative 3 divided by negative 3 is 1 so we have Z is greater than or equal to 18 over negative 3 is negative 6 and remember it's this constraint or this constraint and this constraint right over here boils down to this and this one boils down to this so our solution set Z is less than 4 or Z is greater than or equal to negative 6 so let me make this clear let me rewrite it so Z could be less than 4 or Z is greater than or equal to negative 6 it can satisfy either one of these and this is kind of this is kind of interesting here because let's let's plot these well let's plot these so there's a number line right over there let's say that 0 is over here we have 1 2 3 4 is right over there and then negative 6 we have 1 2 3 4 5 6 that's negative 6 over there now let's think about Z being less than 4 z be Z being less than 4 we would put a circle around 4 since we're not including 4 and it would be everything everything less than 4 everything less than 4 now let's think about what Z being greater than or equal to negative 6 would mean that means you can include negative 6 and it's everything let me do that a different color it means you can include negative 6 I want to do that here we go it means you include negative 6 we do even a more different color do it in orange so Z is greater than or equal to negative 6 means you can include negative 6 and it's everything greater than that greater than that including 4 so it's everything greater than that so what we see as we've essentially shaded in the entire number line every number will meet either one of these constraints or both of them if we're over here we're going to meet both of the constraints if we're one of these number a number out here we're going to meet this constraint if we're a number down here we're going to meet this constraint you can just try it out with a bunch of numbers 0 will work 0 plus 0 plus 7 is 7 which is less than 27 and 3 times 0 is less than 18 so it meets both constraints if we put 4 here it should only meet one of the constraints negative 3 times 4 is negative 12 which is less than 18 so it meets this constraint but it won't meet this constraint because you do 5 times 4 plus 7 is 27 which is not less than 27 it's equal to 27 but remember this is an or so you just have to meet one of the constraints so 4 meets this constraint so even four works so it's really the entire number line will satisfy either one or both of these constraints