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# Compound inequalities: OR

CCSS Math: HSA.REI.B.3

## Video transcript

Solve for z. 5z plus 7 is less than 27 or negative 3z is less than or equal to 18. So this is a compound inequality. 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 7, is 20. So we have 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 20/5. z is less than 4. Now, this was only one of the conditions. Let's [? look at ?] 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 or multiply both sides of an inequality by a negative number, you have to swap the inequality. So we could write 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 than or 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 interesting here. 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. We would put a circle around 4, since we're not including 4. And it'd be 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 in different color. It means you can include negative 6. I want to do that-- oh, here we go. It means you include negative 6. Let me do it in a more different color. Do it in orange. So z is greater than equal to negative 6. Means you can include negative 6. And it's everything greater than that, including 4. So it's everything greater than that. So what we see is 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 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. And you could just try it out with a bunch of numbers. 0 will work. 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. Remember, this is an or. So you just have to meet one of the constraints. So 4 meets this constraint. So even 4 works. So it's really the entire number line will satisfy either one or both of these constraints.