Video Transcript: Multi-Step Inequalities

Inequalities with variables on both sides 

We're asked to solve for p. And we have the inequality here negative 3p minus 7 is less than  p plus 9. So what we really want to do is isolate the p on one side of this inequality. And  preferably the left-- that just makes it just a little easier to read. It doesn't have to be, but we  just want to isolate the p. So a good step to that is to get rid of this p on the right-hand side.  And the best way I can think of doing that is subtracting p from the right. But of course, if we  want to make sure that this inequality is always going to be true, if we do anything to the right, we also have to do that to the left. So we also have to subtract p from the left. And so the left hand side, negative 3p minus p-- that's negative 4p. And then we still have a minus 7 up  here-- is going to be less than p minus p. Those cancel out. It is less than 9. Now the next  thing I'm in the mood to do is get rid of this negative 7, or this minus 7 here, so that we can  better isolate the p on the left-hand side. So the best way I can think of to get rid of a negative 7 is to add 7 to it. Then it will just cancel out to 0. So let's add 7 to both sides of this inequality. Negative 7 plus 7 cancel out. All we're left with is negative 4p. On the right-hand side, we  have 9 plus 7 equals 16. And it's still less than. Now, the last step to isolate the p is to get rid  of this negative 4 coefficient. And the easiest way I can think of to get rid of this negative 4  coefficient is to divide both sides by negative 4. So if we divide this side by negative 4, these  guys are going to cancel out. We're just going to be left with p. We also have to do it to the  right-hand side. Now, there's one thing that you really have to remember, since this is an  inequality, not an equation. If you're dealing with an inequality and you multiply or divide both  sides of an equation by a negative number, you have to swap the inequality. So in this case,  the less than becomes greater than, since we're dividing by a negative number. And so  negative 4 divided by negative 4-- those cancel out. We have p is greater than 16 divided by  negative 4, which is negative 4. And we can plot this solution set right over here. And then we  can try out some values to help us feel good about the idea of it working. So let's say this is  negative 5, negative 4, negative 3, negative 2, negative 1, 0. Let me write that a little bit  neater. And then we can keep going to the right. And so our solution is p is not greater than or equal, so we have to exclude negative 4. p is greater than negative 4, so all the values above  that. So negative 3.9999999 will work. Negative 4 will not work. And let's just try some values  out to feel good that this is really the solution set. So first let's try out when p is equal to  negative 3. This should work. The way I've drawn it, this is in our solution set. p equals  negative 3 is greater than negative 4. So let's try that out. We have negative 3 times negative  3. The first negative 3 is this one, and then we're saying p is negative 3. Minus 7 should be  less than-- instead of a p, we're going to putting a negative 3. Should be less than negative 3  plus 9. Negative 3 times negative 3 is 9, minus 7 should be less than negative 3 plus 9 is 6. 9  minus 7 is 2. 2 should be less than 6, which, of course, it is. Now let's try a value that  definitely should not work. So let's try negative 5. Negative 5 is not in our solution set, so it  should not work. So we have negative 3 times negative 5 minus 7. Let's see whether it is less  than negative 5 plus 9. Negative 3 times negative 5 is 15, minus 7. It really should not be less  than negative 5 plus 9. So we're just seeing if p equals negative 5 works. 15 minus 7 is 8. And so we get 8 is less than 4, which is definitely not the case. So p equals negative 5 doesn't  work. And it shouldn't work, because that's not in our solution set. And now if we really want to feel good about it, we can actually try this boundary point. Negative 4 should not work, but it  should satisfy the related equation. When I talk about the related equation, negative 4 should  satisfy negative 3 minus 7 is equal to p plus 9. It'll satisfy this, but it won't satisfy this. Because when we get the same value on both sides, the same value is not less than the same value.  So let's try it out. Let's see whether negative 4 at least satisfies the related equation. So if we  get negative 3 times negative 4 minus 7, this should be equal to negative 4 plus 9. So this is  12 minus 7 should be equal to negative 4 plus 9. It should be equal to 5. And this, of course, 

is true. 5 is equal to 5. So it satisfies the related equation, but it should not satisfy this. If you  put negative 4 for p here-- and I encourage you to do so. Actually, we could do it over here.  Instead of an equals sign, if you put it into the original inequality-- let me delete all of that-- it  really just becomes this. The original inequality is this right over here. If you put negative 4,  you have less than. And then you get 5 is less than 5, which is not the case. And that's good,  because we did not include that in the solution set. We put an open circle. If negative 4 was  included, we would fill that in. But the only reason why we'd include negative 4 is if this was  greater than or equal. So it's good that this does not work, because negative 4 is not part of  our solution set. You can kind of view it as a boundary point. 

Inequalities with variables on both sides (with parentheses) 

Solve for x. And we have 5x plus 7 is greater than 3 times x plus 1. So let's just try to isolate "x"  on one side of this inequality. But before we do that, let's just simplify this righthand side. so we get 5x plus 7 is greater than - let's distribute this 3. So 3 times x plus 1 is the same thing as 3  times x plus 3 times 1 so it's going to be 3x plus 3 times 1 is 3. Now if we want to put our x's on  the lefthand side, we can subtract 3x from both sides. That will get rid of this 3x on the  righthand side. So let's do that. Let's subtract 3x from both sides, and we get on the lefthand  side: 5x minus 3x is 2x plus 7 is greater than - 3x minus 3x - those cancel out. That was the whole  point behind subtracting 3x from both sides - is greater than 3. Is greater than 3. No we can  subtract 7 from both sides to get rid of this positive 7 right over here. So, let's subtract, let's  subtract 7 from both sides. And we get on the lefthand side... 2x plus 7 minus 7 is just 2x. Is  greater than 3 minus 7 which is negative 4. And then let's see, we have 2x is greater than  negative 4. If we just want an x over here, we can just divide both sides by 2. Since 2 is a  positive number, we don't have to swap the inequality. So let's just divide both sides by 2, and  we get x is greater than negative 4 divide by 2 is negative 2. So the solution will look like this.  Draw the number line. I can draw a straighter number line than that. There we go. Still not that  great, but it will serve our purposes. Let's say that's -3, -2, -1, 0, 1, 2, 3. X is greater than  negative 2. It does not include negative 2. It is not greater than or equal to negative 2, so we  have to exclude negative 2. And we exclude negative 2 by drawing an open circle at negative 2,  but all the values greater than that are valid x's that would solve, that would satisfy this  inequality. So anything above it - anything above it will work. And let's just try, let's try just try  something that should work. and then let's try something that shouldn't work. So 0 should  work. It is greater than negative 2. It's right over here. So, let's verify that. 5 times 0 plus 7  should be greater than 3 times 0 plus 1. So this is 7 - 'cause this is just a 0 - 7 should be greater  than 3. Right. 3 times 1. So 7 should be greater than 3, and it definitely is. Now let's try  something that should not work. Let's try negative 3. So 5 times negative 3... 5 times negative 3 plus 7, let's see if it is greater than 3 times negative 3 plus 1. So this is negative 15 plus 7 is  negative 8 That is negative 8. Let's see if that is greater than negative 3 plus 1 is negative 2  times 3 is negative 6. Negative 8 is not - is not greater than negative 6. Negative 8 is more  negative than negative 6. It's less than. So, it is good that negative 3 didn't work 'cause we  didn't include that in our solution set. So we tried something that is in our solution set and it did work. And something that is not, and it didn't work. So we are feeling pretty good. 

Multi-step inequalities 

Let's do a few more problems that bring together the concepts that we learned in the last two 

videos. So let's say we have the inequality 4x plus 3 is less than negative 1. So let's find all of the  x's that satisfy this. So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from  both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's  cancel out. That just ends up with a zero. No reason to change the inequality just yet. We're just adding and subtracting from both sides, in this case, subtracting. That doesn't change the  inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is  negative 4. Negative 1 minus 3 is negative 4. And then we'll want to-- let's see, we can divide  both sides of this equation by 4. Once again, when you multiply or divide both sides of an  inequality by a positive number, it doesn't change the inequality. So the left-hand side is just x.  x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write  this in interval notation. All of the x's from negative infinity to negative 1, but not including  negative 1, so we put a parenthesis right there. Let's do a slightly harder one. Let's say we have  5x is greater than 8x plus 27. So let's get all our x's on the left-hand side, and the best way to do  that is subtract 8x from both sides. So you subtract 8x from both sides. The left-hand side  becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or  subtracting the same quantities on both sides. These 8x's cancel out and you're just left with a  27. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3,  we have to swap this inequality. It will go from being a greater than sign to a less than sign. And just as a bit of a way that I remember greater than is that the left-hand side just looks bigger.  This is greater than. If you just imagine this height, that height is greater than that height right  there, which is just a point. I don't know if that confuses you or not. This is less than. This little  point is less than the distance of that big opening. That's how I remember it. But anyway, 3x  over negative 3. So now that we divided both sides by a negative number, by negative 3, we  swapped the inequality from greater than to less than. And the left-hand side, the negative 3's  cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation,  it would be everything from negative infinity to negative 9, not including negative 9. If you  wanted to do it as a number line, it would look like this. This would be negative 9, maybe this  would be negative 8, maybe this would be negative 10. You would start at negative 9, not  included, because we don't have an equal sign here, and you go everything less than that, all  the way down, as we see, to negative infinity. Let's do a nice, hairy problem. So let's say we  have 8x minus 5 times 4x plus 1 is greater than or equal to negative 1 plus 2 times 4x minus 3.  Now, this might seem very daunting, but if we just simplify it step by step, you'll see it's no  harder than any of the other problems we've tackled. So let's just simplify this. You get 8x  minus-- let's distribute this negative 5. So let me say 8x, and then distribute the negative 5.  Negative 5 times 4x is negative 20x. Negative 5-- when I say negative 5, I'm talking about this  whole thing. Negative 5 times 1 is negative 5, and then that's going to be greater than or equal  to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. And now we can merge  these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to-- we can  merge these constant terms. Negative 1 minus 6, that's negative 7, and then we have this plus  8x left over. Now, I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's 

negative 20. Negative 20x minus 5. Once again, no reason to change the inequality just yet. All  we're doing is simplifying the sides, or adding and subtracting from them. The right-hand side  becomes-- this thing cancels out, 8x minus 8x, that's 0. So you're just left with a negative 7. And now I want to get rid of this negative 5. So let's add 5 to both sides of this equation. The left hand side, you're just left with a negative 20x. These 5's cancel out. No reason to change the  inequality just yet. Negative 7 plus 5, that's negative 2. Now, we're at an interesting point. We  have negative 20x is greater than or equal to negative 2. If this was an equation, or really any  type of an inequality, we want to divide both sides by negative 20. But we have to remember,  when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality. So let's remember that. So if we divide this side by negative 20 and we divide  this side by negative 20, all I did is took both of these sides divided by negative 20, we have to  swap the inequality. The greater than or equal to has to become a less than or equal sign. And,  of course, these cancel out, and you get x is less than or equal to-- the negatives cancel out--  2/20 is 1/10. If we were writing it in interval notation, the upper bound would be 1/10. Notice,  we're including it, because we have an equal sign, less than or equal, so we're including 1/10,  and we're going to go all the way down to negative infinity, everything less than or equal to  1/10. This is just another way of writing that. And just for fun, let's draw the number line. Let's  draw the number line right here. This is maybe 0, that is 1. 1/10 might be over here. Everything  less than or equal to 1/10. So we're going to include the 1/10 and everything less than that is  included in the solution set. And you could try out any value less than 1/10 and verify that it will  satisfy this inequality. 

Using inequalities to solve problems 

- [Instructor] We're told that Kayla wants to visit a friend who lives eight kilometers away. She'll  ride the subway as far as she can before walking the rest of the way. First, she needs to buy an  access pass that costs $5.50. There's also a fee of $1.25 per stop. This is an expensive subway.  Kayla doesn't want to spend more than $15 on the trip. So she wants to know the largest  number of stops she can afford. Let S represent the number of stops that Kayla buys. So first,  pause this video and see if you can write an inequality that describes how many stops, or that  describes the situation that describes that she wants to take as many stops as she can, but she  doesn't wanna spend more than $15. All right, now let's do this together. So first, let's just think about an expression for how much she spends. So no matter what, she's going to spend $5.50,  so we can write it like this, so $5.50, that's what she's going to spend, even if she doesn't take  any stops. And it's $1.25 per stop, and S is the number of stops. So the amount she's going to  spend just from the stops is going to be $1.25 times S. So it's going to be plus $1.25 S. This is the upfront she has to spend, and this is how much she's going to spend on stop. So this is an S  right over here and I wrote a five right next to it, they look kind of similar. And we know that  she doesn't want to spend more than $15. So she's willing to spend up to $15. So this total  amount that she spends has to be less than or equal to $15. Or if we didn't write it with the  dollar symbols, we would write 5.50 plus 1.25 S is going to be less than or equal to, or needs to  be less than or equal to 15. Now that we've written this inequality, what is the number of stops  that Kayla can afford? What's the largest number of stops that she can afford? Pause this video  and try to figure that out. Well, to figure that out, we just have to solve for S and then figure out

what the largest S is that satisfies the inequality once we've solved for S. So the first thing I  would do is subtract 5.50 from both sides. When we do that, we are left with 1.25 or $1.25 S is  less than or equal to 9.50. And then I would divide both sides by 1.25. And since I'm dividing  both sides by a positive value, it doesn't change the direction of the inequality, 1.25 and then  divide this by 1.25, 9.5 divide by 1.25 is equal to 7.6. So we get that S needs to be less than or  equal to 7.6. So we can't take a fractional number of stops. So the largest number of stops that  Kayla can take is going to be seven stops. She can't take eight, and she can't take 7 1/2 or 7.6.  So the largest number she can take is seven stops. So she can take as many as seven stops. And we are done.