Hello everybody, welcome to Eck Math. I'm Mr. Eck, and today we're going to talk about section P.7, which is all about equations. This is a big section, but in contrast to the last section, which was both large and challenging, I think this one is just a large section that has a little bit of everything. So we're going to make this into three separate videos. I would encourage you to kind of bounce around and watch the pieces that you feel you need, maybe even try the problems and come back to these videos as you need them.
There is going to be stuff that you haven't seen in here before, but there's also probably going to be things that you have seen. And so I want you to be able to kind of customize your experience. We're going to go ahead and start with part one, which is all about linear equations with fractions and something else called rational equations.
Linear equations in general are things of the form, I guess, y equals mx plus b. But particularly what's important about linear equations is the exponent is degree one on them. And those are pretty easy to solve. You know how to solve them already. Here's an example of one.
You've been working on these since algebra 1. You probably don't need my help here. So what I'm really going to focus on is linear equations, I guess, that are in disguise. And here's an example.
So I have a linear equation that has a couple fractions involved. And I want to show you how to deal with the fractions. So when you have a linear equation and you have these fractions, Move is going to be to clear them out.
How are you going to do that? Well, you're going to multiply the entire equation. And what are you going to multiply it by? Well, let's multiply it by, in this case, 24. So this is a new tool. We've previously been dealing with expressions.
We are now dealing with an equation. So with an equation, it's something that's balanced. And that means you can multiply the left and right sides by the same thing. I'm writing it with a big bracket.
You can also think about saying, I want to multiply the right side by 24 and the left side by 24. But it's going to give you the same thing. Why did I choose 24? Well, I chose 24 because 3 times 8 is 24. So you're looking for the least common multiple.
of all the denominators, and in this case that is going to be 24. So what happens when you multiply by 24? Well, I'm remembering that 24 is 3 times 8. So when I multiply this side by 24, the 24 will distribute to both terms. And we get 24 times 5, which is...
should be 120. Does that sound right? then the 24 is going to distribute to x minus 2. What I'm really doing is multiplying by 8 times 3. When I multiply that, though, the 3s are going to reduce out to 1, and I really am just left with 8 times the quantity x minus 2. So don't actually distribute the 24 to both terms. It's not going to be very fruitful or productive. You want to go ahead and kind of reduce as you're going.
Here I'm going to multiply on the right side. Also by 24, you have to do the same number to be balanced. 24 is 8 times 3. So when I distribute 8 times 3 to this term, the 8s are going to reduce out and you really should be left with 3 times the quantity x plus 3. At this point, you're now in a linear equation.
This is the thing that we're really familiar with solving. So once you get rid of the fractions, it turns into that familiar context and everything is nice. Let's go ahead and solve it.
Take away 3x from both sides. If I'm grouping the x's on this side, so I'm going to get 5x on that side. I need to group the numbers on the other side. So I'll go ahead and add 16 and take away 120. I'm just going to use a calculator. Not like that.
I'm going to get negative 95. Then to solve I'll divide by 5. Both sides by 5, and x should be negative, did I say 19? And that's the solution for x. So it's a linear equation with some numbers and fractions, but you can easily clear out the fractions by choosing a number to multiply the entire equation by. What do you multiply by?
The least common multiple. Do we have another example? Yes we do.
So here's another one. Nothing super different. If you feel really confident, I'd say you could go ahead and skip this example, but we're going to do it anyway.
So I'm looking at the denominators and I see a 5, a 2, and a 3. So the least common multiple of 5, 2, and 3 is going to be 30. Is there another number that's smaller? No, because 5, 2, and 3 are all primes. So I'm going to go for the denominator of 30. Or I'm going to try to multiply everything by 30, which will cancel everything, should cancel out all the denominators.
So let's multiply this side by 30, and let's multiply this side by 30. When I bring 30 over to 3x over 5, 30 over 5 is really like 6, so this is like 6 times 3x minus, here 30 goes to this term, 30 divided by 2 is 15. multiplied by x minus 3 on this side. 30 times, 30 over 3 is 10, times x plus 2. And just like before, we'll go ahead and solve it. Did I get all the terms?
Yes. So something like that for an x value. Here's another flavor of problem.
This is a similar type of problem where you're gonna have, you know, fractions. It might turn out to be a linear equation in disguise, but the difference here is that there is an x term... there's an x term on the bottom. And that means that as soon as you have this x on the bottom, we have some restrictions on what the x variable is allowed to be. So I'm gonna make a note right now.
that x cannot equal anything that would make the denominator 0. What would make the denominator 0 in this case? Positive 2. So as soon as I see that in the equation and I have, you know, the same on both sides, I'd have that restriction. If I had a different number here, maybe this was a 5, then I would have two numbers that x couldn't be 2 or 5. It is a 2, so we're just going to leave that be. I have this written twice because I'm going to solve this in two different ways. The first way is going to be that kind of...
The first way might be is going to be not the least common multiple way. The first way is going to be kind of the blind way, and I'm doing that because I see I have a common denominator here. So one way to solve this is just to say, what about this last term? What if I took this 2, which is really 2 over 1, and gave it a common denominator of x minus 2 as well?
So I'm going to multiply that by x minus 2 to the top and bottom. x minus 2, x minus 2. That's the equal sign is right here. So on the one side of the equal sign, you'll have 2 over x minus 2 equals, and I'm going to combine these all into one fraction now.
We have an x minus 2 on the bottom still. We'll have an x minus 2x minus 2. So just carrying out the subtraction. Let's simplify that. So that'll be minus 2x plus x.
So that's going to be minus 1x altogether. And then minus 2 times minus 2 is going to be plus 4. So minus x plus 4 over x minus 2. Now I still have this equation, but I have a single term on each side. So now I guess I have to think about what I can multiply by.
I'm going to multiply by x. So this is a situation I want to go back. This is a situation where you might be looking at it, and you might be saying, I want to do that thing I learned in another class called cross multiplying, where you multiply each of those circle terms, and it cancels things out in the equation. We don't really like cross multiplying. It's more of a routine that people use, but not a, we'll say, actual mathematical thing with meaning.
So, Instead, what I want to think about is I want to multiply both sides by something. It can be anything I want, but it has to be something that will help me with the denominator. So I'm going to multiply by x minus 2 on both sides.
And if you do that, instead of like cross multiplying or whatever, you say, oh, wow, look at that. And look at that. and those will reduce out to 1 as long as x is not equal to 2. So we have 2 is equal to negative x plus 4. Let's see, let's go ahead and add x and take away 2, and so I get x equals positive 2. That's my answer.
Except, wait a minute, look at what I said right at the start. x can be anything except positive 2. So this is a situation where If there were a solution, it would be 2. But it turns out that x equals 2, which is what we call the candidate solution. Oh, wrong color. Candidate does not work. It's not in the domain.
So it turns out in this case, there are no real solutions. Which can happen. It's not common, but it's also not rare.
It happens sometimes, especially when you're doing problems written by a textbook that kind of, you know, are trying to test you on these things. I will say, you know, as a teacher, if you look at this and all you tell me is, oh, there's no real solutions, blah, I wouldn't consider that enough. I still need you to find me that solution. It's kind of an advanced technique, right? You need to find me the candidate solution, and then you need to tell me why that solution was not.
legal or did not work. Otherwise it's not really a complete problem. So we're looking for finding that answer and then explaining why that does not work.
Okay, so in this way, over on the right, I first did a denominator operation, common denominator. But if I'm thinking about what I did in the previous problems, where the first move was to multiply by the least common multiple of everything in the denominators, basically clearing out all fractions, That will work in this situation here, and it's actually going to work better. So that's going to be method 2, and it's my preferred method, which is to clear out fractions.
I look at all the denominators and I see that x minus 2 is the only thing in a denominator. So the LCM, least common multiple of all the denominators, is x minus 2. So I'm going to take this entire equation and kind of box the left sides, box the right sides, it's optional. I'm going to multiply by x minus 2 right here. On the right side, it's easy.
Those will reduce out and you will just get 2. On the, I said right, I meant left. On the real right side, the x minus 2 is a binomial. It has to distribute to both terms here because they're separated by a negative sign. So there is a distribution. When you distribute the x minus 2 to the first term, Those will reduce out and you'll just get the numerator x.
When you distribute the x minus 2 to the second term you get minus because minus 2 x minus 2. From here it's easy to solve because it's a linear equation. You get the again the candidate solution x equals 2. But x can't be 2, so no solutions. Real solutions.
Same answer as before. I really like that clear-out fraction method. It'll work all the time, even if you don't have the same thing in the numerator and the denominator.
That's a really helpful thing to do. So that's a linear equation with numbers on the bottom. And it looks like we've got just one more example here.
with some different things on the top and bottom. So let's do that, it's nothing new. Let's just see what happens. First thing that you wanna do in a situation like this is factor anything that's not yet factored. So x squared minus two x minus three, I need to factor it.
I wanna give you a problem solving hint. You can always look at the other side and say, hmm, I wonder, are those gonna be the factors of this? Just like, is our problem writer really crafty in trying to make things match up?
It won't always be true. That's something else to be aware of. But if it is, of course, your life is going to be easy. So let's check. It needs to be an x here and an x there.
Minus 3 and minus 2, so would minus 3 and plus 1 work? Absolutely. So those are the factors. So now I'm looking at all the denominators and I see a denominator of x minus 3 and x plus 1. So what do I need to multiply by?
Well, I'm going to group the left side. I'm going to group the right side and I'm going to multiply by both of these terms. So I'm going to multiply by x minus 3. and x plus 1. I'm going to do it at the same time. So it looks like the right side is easier.
Those will all cancel out or reduce out, and you should just get 8. That was pretty nice. On the other side, we have to do it in two steps. When we distribute to the 1, the x minus 3 cancels out, and you're left with 1 times what I didn't multiply by. or what didn't reduce out, x plus 1. Minus sign carries down. When you distribute this to the second term the x plus 1s will reduce out and we'll have a minus 2 times x minus 3. Again, simple linear equation.
All we have to do is solve. x equals negative 1 is our candidate solution, but wait, I want to check something. Can I plug in negative 1 into the original? Well, let's see.
Let's check it in there. You can maybe already see what's going on. 1 over negative 1 minus 3 is negative 1 fourth, that's okay.
Minus 2 over negative 1 plus 1, uh-oh, that's minus 2 over 0. So the candidate answer actually causes me to divide by zero, which means although this is the candidate answer, it's not a solution. It's not in the domain of the original, so no real solutions. And of course I could have noticed that right at the start by saying, oh anything in the denominator is going to give me a restriction on x.
x cannot be positive 3 or negative 1. So I could have listed that out as excluded values from the very start. And that's how you solve linear equations when there are fractions involved. If you have fractions on, if you have numbers on the bottom, just get rid of the fractions. If you have x values on the bottom, You also just get rid of the fractions.
You just have to do some work with polynomials and be sure you watch out for excluded values. Usually you will have solutions. I actually think it's just a pure coincidence that the two examples I chose had no solutions, but it is something to watch out for.
So sometimes you'll get a mix. You might get two or three possible solutions and only one of them will not be in the domain. Depends on the problem. All right folks, that's gonna be the end of the first video. Stay tuned, we're going to talk about quadratics and all kinds of other situations as well.
We'll see you next time.