Beg Algebra: Solving Absolute Value Equations

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SOLVING ABSOLUTE-VALUE EQUATIONS
So, let’s take a look at solving an equation that actually has an absolute value sign in it. Now, how do you do this? This is actually no big deal at all. The only thing you have to remember, the key to this whole business, is what the absolute value means. So, let me just recap this, just two seconds. If you want a more extended recap, you can just click somewhere around me and you can actually watch a lecture just on absolute values. But just to sort of refresh your memory, let me tell you that if I write A, what does that equal? Well, there are two possibilities. It either equals A, if A is positive, or it could be -A, if A were negative. So, if I say to you that A is some value, let me call it a question mark, then what do I know?
I know the two possibilities. Either A = ?, or A = -?. Do you see why? Because when I take absolute values I’m going to get just the question mark. So, if you just remember this basic principle this will lead you through all these questions. Because basically what you have to do is if you have an equality, which has an absolute value, that actually means there are two different equalities that you need to solve. One that looks like A equals the thing, and the other one is A equals negative the thing. That’s all there is to it. So, when you have an absolute value––let me give you a really simple example and then we’ll do a real one.
Suppose I tell you x = 3, what’s the answer? Well, there’s two answers. Do you see it? x equals either 3 or -3. So, either x = 3 or x = -3. So, always with absolute values we’re going to have these two equations that require to be solved. One where the thing is just with the number, inside there with the number and the other one where we have the inside there with negative the number there. Let’s now do this with an actual real world, more complicated situation.
Suppose I have 3x - 1 is equal to 2. Let’s find all the solutions to that. So, what do I do? I have to set up two equalities. One is, first of all, if the inside equals 2 and the other possibility is if the inside equals -2. Because notice, when I take absolute values, in both cases the absolute value will be 2. Since the inside is blind to negative signs, we have to consider two cases. 3x - 1 = 2, and the other possibility, we have 3x - 1 = -2. Always set them up the same way. We have either absolute equals something. The thing equals this or the thing equals negative that. Now you just solve away.
So, I bring the -1 over and I see 3x = 3. So, x has to equal 1. There’s one answer. If I bring this over here, the -1 comes over as a plus 1. I see 3 x = -1. I take -2 and I add 1. So, therefore, x = -1/3. There are two answers to this. You can check to see if they’re right. If I plug in a 1 here look what happens. If I plug in a 1 for x, I see 3 - 1. That’s 2. Absolute value of 2? 2 checks. What about if I put in a -1/3? If I put in a -1/3, I see -1/3 times 3 is -1, -1 -1 is -2. But I’m taking absolute values, so I still get 2. You see how I get those two answers with the absolute value? I have the thing with the number itself and then I have the thing again, with negative the number. That’s all there is to absolute values.
Let’s try another example. Look at this one. 5 over x - 3 absolute equals 10. How would you solve this? Again, the same idea. If I have the absolute value of something, that something either equals this or that something equals negative this. So, I set up two equations: 5 over x - 3 = 10 and the other one is 5 over x - 3 = -10. Okay, so now we solve these things. So, how would you solve this? Well, there are a variety of ways. One thing you can do, for example, is cross multiply. If you realize that this is actually 10 over 1, I could cross multiply and here I would see that 10 times x - 3, I take this product here. That will equal this product here.
Another thing you can do is multiply everything through by the quantity x - 3 and you get the exact same answer if you cancel here. Then what would we have? We’d have 10x - 30 = 5. Then how would I solve this? I’d bring the 30 over and I’d see 10x = 35 and so I’d see that x = 35/10. I can reduce that a little teeny bit by dividing top and bottom by 5 and see 7 over 2. So, there’s one solution. What’s the other solution? Well, I have to solve this now. So, I can cross multiply again, if you want. If I multiply through again by x - 3 and I would see -10 times x - 3 = 5. So, I would see -10, I’m distributing now, -10x + 30 = 5. If I bring this plus 30 over it becomes a -30. So, I’d see -10x = 5 - 30, is -25. If I divide both sides by -10, I would see x equals just 25 over 10, because those negative signs cancel each other out. If I cut through by the 5, I see 5 over 2.
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So, I see two answers. I see 7 halves and I see 5 halves. Once again, you can check, plug in 7 halves for x and see that in fact this will actually produce a 10, which in absolute value is equal to 10. If you put in 5 halves here, you can check that this would actually produce -10. But since I’m taking absolute values I’m still a 10. So, there are always two solutions here when I have an absolute value of this kind. So, that’s sort of the basics of looking at an equation with absolute values. Actually, up next what I’ll do are some slightly more exotic, slightly more complicated--no new ideas, but just to get a chance for you to see and maybe try some other ones together. I’ll see you there.