One-step inequalities examples 

In this video, I want to tackle some inequalities that involve multiplying and dividing by positive and negative numbers, and you'll see that it's a little bit more tricky than just the adding and  subtracting numbers that we saw in the last video. I also want to introduce you to some other  types of notations for describing the solution set of an inequality. So let's do a couple of  examples. So let's say I had negative 0.5x is less than or equal to 7.5. Now, if this was an  equality, your natural impulse is to say, hey, let's divide both sides by the coefficient on the x  term, and that is a completely legitimate thing to do: divide both sides by negative 0.5. The  important thing you need to realize, though, when you do it with an inequality is that when you multiply or divide both sides of the equation by a negative number, you swap the inequality.  Think of it this way. I'll do a simple example here. If I were to tell you that 1 is less than 2, I think  you would agree with that. 1 is definitely less than 2. Now, what happens if I multiply both sides of this by negative 1? Negative 1 versus negative 2? Well, all of a sudden, negative 2 is more  negative than negative 1. So here, negative 2 is actually less than negative 1. Now, this isn't a  proof, but I think it'll give you comfort on why you're swapping the sign. If something is larger,  when you take the negative of both of it, it'll be more negative, or vice versa. So that's why, if  we're going to multiply both sides of this equation or divide both sides of the equation by a  negative number, we need to swap the sign. So let's multiply both sides of this equation.  Dividing by 0.5 is the same thing as multiplying by 2. Our whole goal here is to have a 1  coefficient there. So let's multiply both sides of this equation by negative 2. So we have  negative 2 times negative 0.5. And you might say, hey, how did Sal get this 2 here? My brain is  just thinking what can I multiply negative 0.5 by to get 1? And negative 0.5 is the same thing as  negative 1/2. The inverse of that is negative 2. So I'm multiplying negative 2 times both sides of  this equation. And I have the 7.5 on the other side. I'm going to multiply that by negative 2 as  well. And remember, when you multiply or divide both sides of an inequality by a negative, you  swap the inequality. You had less than or equal? Now it'll be greater than or equal. So the left hand side, negative 2 times negative 0.5 is just 1. You get x is greater than or equal to 7.5 times  negative 2. That's negative 15, which is our solution set. All x's larger than negative 15 will  satisfy this equation. I challenge you to try it. For example, 0 will work. 0 is greater than  negative 15. But try something like-- try negative 16. Negative 16 will not work. Negative 16  times negative 0.5 is 8, which is not less than 7.5. So the solution set is all of the x's-- let me  draw a number line here-- greater than negative 15. So that is negative 15 there, maybe that's  negative 16, that's negative 14. Greater than or equal to negative 15 is the solution. Now, you  might also see solution sets to inequalities written in interval notation. And interval notation, it  just takes a little getting used to. We want to include negative 15, so our lower bound to our  interval is negative 15. And putting in this bracket here means that we're going to include  negative 15. The set includes the bottom boundary. It includes negative 15. And we're going to  go all the way to infinity. And we put a parentheses here. Parentheses normally means that  you're not including the upper bound. You also do it for infinity, because infinity really isn't a  normal number, so to speak. You can't just say, oh, I'm at infinity. You're never at infinity. So  that's why you put that parentheses. But the parentheses tends to mean that you don't include  that boundary, but you also use it with infinity. So this and this are the exact same thing.  Sometimes you might also see set notations, where the solution of that, they might say x is a 

real number such that-- that little line, that vertical line thing, just means such that-- x is  greater than or equal to negative 15. These curly brackets mean the set of all real numbers, or  the set of all numbers, where x is a real number, such that x is greater than or equal to negative  15. All of this, this, and this are all equivalent. Let's keep that in mind and do a couple of more  examples. So let's say we had 75x is greater than or equal to 125. So here we can just divide  both sides by 75. And since 75 is a positive number, you don't have to change the inequality. So  you get x is greater than or equal to 125/75. And if you divide the numerator and denominator  by 25, this is 5/3. So x is greater than or equal to 5/3. Or we could write the solution set being  from including 5/3 to infinity. And once again, if you were to graph it on a number line, 5/3 is  what? That's 1 and 2/3. So you have 0, 1, 2, and 1 and 2/3 will be right around there. We're going  to include it. That right there is 5/3. And everything greater than or equal to that will be  included in our solution set. Let's do another one. Let's say we have x over negative 3 is greater  than negative 10/9. So we want to just isolate the x on the left-hand side. So let's multiply both  sides by negative 3, right? The coefficient, you could imagine, is negative 1/3, so we want to  multiply by the inverse, which should be negative 3. So if you multiply both sides by negative 3,  you get negative 3 times-- this you could rewrite it as negative 1/3x, and on this side, you have  negative 10/9 times negative 3. And the inequality will switch, because we are multiplying or  dividing by a negative number. So the inequality will switch. It'll go from greater than to less  than. So the left-hand side of the equation just becomes an x. That was the whole point. That  cancels out with that. The negatives cancel out. x is less than. And then you have a negative  times a negative. That will make it a positive. Then if you divide the numerator and the  denominator by 3, you get a 1 and a 3, so x is less than 10/3. So if we were to write this in  interval notation, the solution set will-- the upper bound will be 10/3 and it won't include 10/3.  This isn't less than or equal to, so we're going to put a parentheses here. Notice, here it  included 5/3. We put a bracket. Here, we're not including 10/3. We put a parentheses. It'll go  from 10/3, all the way down to negative infinity. Everything less than 10/3 is in our solution set.  And let's draw that. Let's draw the solution set. So 10/3, so we might have 0, 1, 2, 3, 4. 10/3 is 3  and 1/3, so it might sit-- let me do it in a different color. It might be over here. We're not going  to include that. It's less than 10/3. 10/3 is not in the solution set. That is 10/3 right there, and  everything less than that, but not including 10/3, is in our solution set. Let's do one more. Say  we have x over negative 15 is less than 8. So once again, let's multiply both sides of this  equation by negative 15. So negative 15 times x over negative 15. Then you have an 8 times a  negative 15. And when you multiply both sides of an inequality by a negative number or divide  both sides by a negative number, you swap the inequality. It's less than, you change it to  greater than. And now, this left-hand side just becomes an x, because these guys cancel out. x  is greater than 8 times 15 is 80 plus 40 is 120, so negative 120. Is that right? 80 plus 40. Yep,  negative 120. Or we could write the solution set as starting at negative 120-- but we're not  including negative 120. We don't have an equal sign here-- going all the way up to infinity. And  if we were to graph it, let me draw the number line here. I'll do a real quick one. Let's say that  that is negative 120. Maybe zero is sitting up here. This would be negative 121. This would be  negative 119. We are not going to include negative 120, because we don't have an equal sign  there, but it's going to be everything greater than negative 120. All of these things that I'm  shading in green would satisfy the inequality. And you can even try it out. Does zero work? 

0/15? Yeah, that's zero. That's definitely less than 8. I mean, that doesn't prove it to you, but you could try any of these numbers and they should work. Anyway, hopefully, you found that  helpful. I'll see you in the next video. 

One-step inequalities: -5c< or = 15 

Solve for c and graph the solution. We have negative 5c is less than or equal to 15. So negative  5c is less than or equal to 15. I just rewrote it a little bit bigger. So if we want to solve for c, we  just want to isolate the c right over here, maybe on the left-hand side. It's right now being  multiplied by negative 5. So the best way to just have a c on the left-hand side is we can  multiply both sides of this inequality by the inverse of negative 5, or by negative 1/5. So we  want to multiply negative 1/5 times negative 5c. And we also want to multiply 15 times  negative 1/5. I'm just multiplying both sides of the inequality by the inverse of negative 5,  because this will cancel out with the negative 5 and leave me just with c. Now I didn't draw the  inequality here, because we have to remember, if we multiply or divide both sides of an  inequality by a negative number, you have to flip the inequality. And we are doing that. We are  multiplying both sides by negative 1/5, which is the equivalent of dividing both sides by  negative 5. So we need to turn this from a less than or equal to a greater than or equal. And  now we can proceed solving for c. So negative 1/5 times negative 5 is 1. So the left-hand side is  just going to be c is greater than or equal to 15 times negative 1/5. That's the same thing as 15  divided by negative 5. And so that is negative 3. So our solution is c is greater than or equal to  negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative 1, negative  2, negative 3. And then I could go above, 1, 2. And so c is greater than or equal to negative 3. So  it can be equal to negative 3. So I'll fill that in right over there. Let me do it in a different color.  So I'll fill it in right over there. And then it's greater than as well. So it's all of these values I am  filling in in green. And you can verify that it works in the original inequality. Pick something that should work. Well, 0 should work. 0 is one of the numbers that we filled in. Negative 5 times 0 is 0, which is less than or equal to 15. It's less than 15. Now let's try a number that's outside of it.  And I haven't drawn it here. I could continue with the number line in this direction. We would  have a negative 4 here. Negative 4 should not be included. And let's verify that negative 4  doesn't work. Negative 4 times negative 5 is positive 20. And positive 20 is not less than 15, so  it's good that we did not include negative 4. So this is our solution. And this is that solution  graphed. And I wanted to do that in that other green color. Here you go. That's what it looks  like. 

One-step inequality word problem 

A contractor is purchasing some stone tiles for a new patio. Each tile costs $3, and he wants to  spend less than $1,000. And it's less than $1,000, not less than or equal to $1,000. The size of  each tile is one square foot. Write an inequality that represents the number of tiles he can  purchase with a $1,000 limit. And then figure out how large the stone patio can be. So let x be  equal to the number of tiles purchased. And so the cost of purchasing x tiles, they're going to  be $3 each, so it's going to be 3x. So 3x is going to be the total cost of purchasing the tiles. And  he wants to spend less than $1,000. 3x is how much he spends if he buys x tiles. It has to be less  than $1,000, we say it right there. If it was less than or equal to, we'd have a little equal sign  right there. So if we want to solve for x, how many tiles can he buy? We can divide both sides of 

this inequality by 3. And because we're dividing or multiplying-- you could imagine we're  multiplying by 1/3 or dividing by 3 -- because this is a positive number, we do not have to swap  the inequality sign. So we are left with x is less than 1,000 over three, which is 333 and 1/3. So he has to buy less than 333 and 1/3 tiles, that's how many tiles, and each tile is one square foot. So  if he can buy less than 333 and 1/3 tiles, then the patio also has to be less than 333 and 1/3  square feet. Feet squared, we could say square feet. And we're done.



Last modified: Tuesday, March 8, 2022, 11:06 AM