In this video, we discuss equations or equalities when things are equal, and inequalities, a complementary notion when things are not equal. Often, we're trying to isolate, evaluate, or estimate some unknown quantity typically denoted by X. It turns out that essentially the same algebraic tools apply for both equalities and inequalities, though a little bit more care is needed for inequalities. Let's begin with a simple equation involving an unknown X. Two X plus one is equal to seven, and as usual X is an unknown and we want to find out what it is. The idea is that you try and isolate X. So, X is embedded in this expression two X plus one. We want to somehow separate it out so that we get it standing on its own. Okay. So, we just get X in two steps. We'll take one away from both sides, and we get two X is equal to six. Then we divide by two, and we end up with X is equal to three. Of course, when you do a calculation like this, you can always check that your answer works. Right, two times three plus one is equal to six plus one, which is equal to seven. So, the left-hand side and the right-hand side of the original equations agree when X is equal to three. Again now, let's make things a little bit more elaborate. Again, I'm going to write down an equation, but X will appear more than once. In fact, X appears three times and is sort of mixed up in, with fractions. Again, the aim is to isolate X. So, we'll start off by simplifying the fraction, the expressions involving fractions on the right-hand side, by getting a common denominator of 10, and then the numerator simplifies to seven X minus nine, and then we multiply both sides by 10, and you can simplify the left-hand side a little bit, and then multiply both sides by two, add 18 to both sides, subtract five X from both sides, and we get 18 is equal to nine X. Divide through by nine, and then put X first, and we get X is equal to two. That was rather long process, but can you see that in principle it was no more difficult than the previous exercise. There's a little bit more rearranging and X appeared more than once. It was like gathering apples or oranges. We just collect together everything involving X until we just got one x. Then simplify the expressions until we get X on its own. Of course, you can check that the answer is correct by feeding back into the original equation and checking that the left-hand sides and the right-hand sides are in fact equal. But, what I want to do next is something that's qualitatively quite different. It involves a completely new idea to just simply rearranging information. Let's consider the following equation. Again, X appears more than once. It appears twice. This equation is very special because zero appears on the right-hand side. The left-hand side is a product of two numbers, X minus one and X minus two. Whenever you have a factorization of zero, you're able to make progress almost immediately. A consequence of this equation is that at least one of the factors has to be equal to zero, and why is that? Well, that's a property of multiplication of real numbers. If I take a rectangle side lengths A and B, then the result of multiplying A and B which we call AB, geometrically, is the area of this rectangle. Okay. Now, let's suppose the product of A and B is equal to zero. What does that imply about the geometry of the rectangle? That means the area has to be zero. So, how can that be the case? It means the rectangle becomes infinitely thin. There's at least two possible ways. The rectangle could become infinitely thin of length B. So, A turns out to be zero. Another possibility is that the area could become zero by becoming an infinitely thin line of length A, and now B has become equal to zero because it is in fact the third possibility, and that's the whole thing contracts to a single point with zero area, in which case both A and B are equal to zero. But the upshot of this is that if AB is equal to zero, if the area of the rectangle is equal to zero, then either A is equal to zero or B is equal to zero. It is this fact that was used, if we go back to the equation to deduce that either X minus one is equal to zero, or X minus two is equal to zero where X minus one has the role of A and X minus two the role of B. So, what we have is that our original equation is split up into two equations where in each case X appears only once. So, by simple rearrangements of the information in each case, we isolate X as either one or two. I would like to revisit this equation using a technique that I saw a student use years ago, and the student had very little mathematical experience and background. So, he was using his own ingenuity, and he ignored this property of zero that we've used just now to solve this equation. He just started off by expanding the left-hand side. Okay. So, if you multiply out the brackets, you get X squared minus three X plus two is equal to zero. His idea was to again try to isolate X, and the first step will be perhaps to strip away the two from the left-hand side, take it away, and you get X squared minus three X equals minus two. Then he did something really really clever. He added nine divided by four to both sides. So, why on Earth nine on four? Well, let's just say that the right-hand side simplifies to a quarter. Now I'll just explain why nine over four. Notice that 9 on 4 is actually minus 3 on 2 , all squared, and the minus 3 on 2 is a half of the coefficient of x. Now, where is all these leading? By choosing minus 3 on 2 squared, the left-hand side can be rewritten as x minus 3 on 2 all squared. What the student was doing, in fact, is an application of the following algebraic identity: a plus b squared is a squared plus 2 ab plus b squared, where a is equal to x and b is equal to minus 3 on 2. This trick of adding 9 on 4 to both sides a few steps back produces this perfect square effect. This technique is called completing the square. Again, it's a very important trick that you're going to see formalized in the next module when we discuss the quadratic formula. What's the whole point of this? Well. Notice the last line, we've got an expression on the left-hand side with only one instance of x. So, now to isolate x is relatively straightforward. Reduce the problem to something similar to what we've seen before. You just strip away the expression until you get x on its own. So we get x minus 3 on 2. If we take square roots on both sides, on the left-hand side, we get x minus 3 on 2. On the right-hand side, we get plus or minus the square root of a quarter, which is plus or minus a half. By adding 3 on 2 to both sides, we isolate x as 3 on 2 plus or minus a half. 3 on 2 plus a half is 2. 3 on 2 minus a half is 1. So, we get the two answers that we saw previously. I thought I'd mention that because, firstly, it's a precursor to something that we see in more detail in the next module, the quadratic formula, but also I just want to highlight the fact that even though the student had very little mathematical background, just thinking about imaginatively about the numbers and expressions involved, he was able to do something really original and creative. I want to emphasize this fact because a lot of you possibly don't have much knowledge of calculus, but I want you to feel like you're on a journey of discovery. Even if you discover something that's well-known or even possibly very old or ancient, I want you to feel like you're discovering it for the first time. So far, we've just discussed equations. I want to move on to discuss inequalities where things aren't necessarily equal. But in order to formalize that'll make sense of that, we need an order in what's called an ordering of the real number line. Let's take a real number line. There are two points, a and b, and a appears to the left of b, then we say that a is less than b. We write that symbolically with a less than symbol. This here is a less than symbol. Notice that the pointy bit, the symbol points to the left-hand side where it sort of opens up, it points to the right-hand side. If b is to the right of a, we say a is less than b. We can rewrite this also using a greater than symbol. To say a is less than b is equivalent to saying b is greater than a. The greater than symbol means that the thing that is greater is to the right of the thing that's less. There are two other symbols. We say a is less than or equal to b using this symbol if a is less than b to the left of b or possibly equal to b. We also write a is greater than or equal to b if a is greater than b to the right of b or possibly equal to b. So, for example, here is a real number line and, say, 0 in the middle. Here's a few numbers. So because I've pointed out negative 20 on the line, that's less than negative 15, which is less than minus 10 and so on, less than 0, less than say 5, less 10, less than 15. Now notice that minus 20 is less than 5. Minus 20, you might think of as a sort of large number and size and indeed, it has a large also called absolute value or magnitude, something I'll talk about in a future video. But in terms of the ordering on the number line, we still say negative 20 is less than 5. It might sound a little bit unintuitive, but when we say less than, we just mean to the left of on the number line. Now we've got these symbols less than, greater than, less than or equals and greater than or equals. Let's explore some relationships between numbers that use some of these symbols. Let's revisit the very first example we discussed, 2x plus 1 equals 7, but instead of equals, let's turn it into an inequality. Let's have less than. Now to solve this inequality, it's exactly the same process that you saw when it was equals. You subtract 1 from both sides and the inequality is retained, and you isolate x finally by dividing by 2. So the answer is x is less than 3. Let's do something a little bit more elaborate. Let's solve this inequality. 1 minus x is less than 3x plus 6. Notice that x appears more than once, so there's a little bit of work to do to isolate x. Now it's really very similar to the first couple of examples, but a little bit of care is needed when you're dealing with inequalities. I'll show you what I mean. I could, for example, add x to both sides which becomes 4x plus 6 on the right-hand side. Adding x doesn't change the inequality. Then I could subtract 6 from both sides, which also doesn't change the inequality. 1 minus 6 is minus 5. We usually like to have the x on the left-hand side. So, now what I'll do is move this around. If minus 5 is less than 4x then 4x is greater than minus 5. See how the inequality has changed? I divide by 4, then we finally get x is greater than minus 5 over 4 and that's the solution. Let's approach this problem in a slightly different way just to illustrate the sorts of variations in algebraic manipulation. Starting off with 1 minus x is less than 3x plus 6. What I'm going to do is multiply both sides by minus 1. I'm going to take the negative of both sides and when you do that, when you multiply an inequality through by a negative number, it flips over. The reason for that is that when you multiply by a negative number, the relative left right positions of numbers on the number line get inverted. It's like looking in a mirror. Left becomes right and right becomes left. That's a geometric explanation, but I have also given an algebraic explanation in the accompanying notes. Anyway, the rule is if you multiply by a negative number or just take the negatives, then the inequality flips over. So, less than becomes greater than, greater than becomes less than and so on. So, now what's the negative of the left-hand side? That's just x minus 1. The negative of the right-hand side is minus 3x minus 6. We can gather together the x's, add 3x to both sides. Left-hand side, we get 4x minus 1. On the right-hand side, minus 6. Add 1 to both sides. Divide both sides by 4, and we've got the same answer that we had before: x is greater than minus 5 over 4. So, most of the manipulations we did, except all but 1, maintained the inequality. You just have to be careful that when you multiply by a negative number, you change the inequality. This was just to give you a taste of manipulations with equations and inequalities using a few contrasting examples. In the next video, we'll introduce the notion of solution sets, which apply generally in all situations and provide a very convenient way of capturing important information about real numbers using interval notation. Please read and digest the notes accompanying this video and then try the exercises. Thank you for watching and we look forward to seeing you again soon.
