 When working with linear functions, we often times have to solve linear equations, and we've seen some examples of that beforehand, like if you have something like y equals 2x minus 1, and you want to figure out when f of x equals 4, you just plug in 4 for the y-port and solve for it. It's not too complicated. Now, there are situations where something like the following might arise. This is sort of like a worst case scenario when it comes to solving linear equations. Take something like 1 half x plus 5 minus 4 equals 1 third 2x minus 1. Now, is solving such a thing just an algebraic chore, or is there actually situations where you might want to do something like this? Well, first of all, I want to mention that although it's not simplified, the left-hand side is a linear function. One of the benefits of not simplifying is we can see some of the transformations that have been done here. If you were to take the standard unmodified, untransformed linear function, you would actually get the identity function like this. That is, you get the graph y equals x. This is sort of like the untransformed linear function. When you look at things like this, what have we done to it? Well, we've shifted it to the right by 5, 1, 2, 3, 4, 5. We've stretched it vertically by a factor of 1 half. Changing the slope is changing the graph here, so it's kind of like now we lower the slope to get something like this. Oh, and then we also moved it down by 5. I guess I moved it to the right, so I should be getting something like this. Then I moved it down by 5 by 4, 1, 2, 3, 4. You get a graph that kind of looks like this. We've changed this line with transformations. Moved it to the right, moved it to the left by 5, compressed it vertically by 2, moved it down. We can do that. Same thing kind of going on over here. There's some type of horizontal stretch, vertical stretch. There's a horizontal shift. There's reasons why we might want to leave things unfactored, right? Because we can see transformations when they're not simplified right there. Another thing I should mention is that why would we want to set these two linear expressions equal to each other? Well, one thing that happens could be something like the following. Say we have two linear equations like this. Their graphs are going to be lines. Where is this common point of intersection? Where is this point, which we can see graphically right here? If we had something like f of x equals 1 half x plus 5 minus 4 and g of x equals 1 third 2x minus 1. If you were trying to figure out graphically where these two lines intersect each other, you're looking for some x-coordinate for which their y-coordinates are the same. Specifically, you'd be looking for when f of x equals g of x. Is there a common x-coordinate for which their y-coordinates f of x and g of x respectively are equal to each other? Well, solving that equation is exactly this equation right here. So when we start wondering where two graphs intersect each other, which is actually a very important question. When we start asking ourselves when two functions intersect each other, whether graphs intersect each other, that starts to mean that we get a whole linear expression on the left and a totally different linear expression on the right. We want to solve that. That comes down to this type of linear equation. So there's some stuff going on there. Whoops, I lost my screen. Here it is. There you are, you little dickens. So let's solve this equation. When it comes to solving linear equations, there's a basic template that one can go through and you can do it step by step. The first thing I do when I work with a linear equation, if I notice any fractions involved in it, like say you have one half and one third, we don't necessarily like doing arithmetic with fractions. You know, fractions are our friends, but it's kind of like that friend that's always borrowing your stuff. It's somewhat demanding. And so the first thing I want to do when it comes to an equation like this is this is sort of an optional step, but I like to clear the denominators. That is, I can actually remove all fractions from consideration. That way I can kind of postpone the difficulty of working with fractions to clear the denominators. What that means is we're going to first identify the denominators in presence. We have a one half, we have a one third. We want to identify the least common denominator. That is the least common multiple of the denominators present. And in this situation, we have two times three, which is equal to six. So what we're going to do is we are going to take our equation, one half x plus five minus four is equal to one third, two x minus one. And we're going to multiply both sides of the equation by the least common denominator. So we're going to multiply both sides of the equation by six, do six right here. Now on the left hand side, because we have actually two terms, we have a sum one half x plus five and a negative four. We're going to distribute the six onto both terms. Make sure you distribute. On the right hand side, we don't have a difference going on. We actually have a product we're not going to distribute. Now some people might think, well, why don't you distribute the six onto both pieces? And I can sort of counter that with the following. If I have something like two times three and I times that by four, do I distribute the four onto both pieces and end up with eight times 24? I'm sorry, eight times 12. And then continue on from there. No, that's not what you do. You simply just do two times three, which is six times four, which is that's where the 24 came from. So we don't multiply each factor in the product by the third factor. We only distribute when we have a plus or a difference, right? So if you have like two plus three times by four, that's the situation when you would distribute here. And so we're going to distribute the six on the left hand side right here. So this would give us a six over two X plus five. We're going to get a four times six. And then the right hand side will look like six over three and then two X minus one. When you distribute, when you multiply your six by a product, you're only going to multiply one of the factors. And the fact you want to multiply by is the fraction. That's what kind of initiated this whole process. Notice that six divided by two is a three. So we get three times X plus five minus 24. And then six divided by three is a two. You get two X minus one. Now, the fact that these factors are themselves sums and differences now means at this stage we can distribute the three. We can distribute the two and we end up with three X plus 15 minus 24. And then the right hand side, we're going to get four X minus two. So what we've done so far is we were able to clear up the denominators and we multiplied everything throughout. So now everything's been expanded out. This kind of leads into the second step when it comes to solving a linear equation. This is the step I refer to as combine like terms. And in fact, if I was going to separate these two steps, I would say that, you know, we'll draw this line to kind of separate here. Everything above this line was just the purpose was to clear the denominators. You'll notice that on this equation that has a star next to it, there's no more fractions. We can get rid of the fractions if we don't want to deal with them by multiplying by the LCD. Now we're in a situation where we want to combine like terms together. We want to add together the common factors of X and then we want to add the constants together. Oftentimes in order to get your variables together like the X is right here, you're going to first have to distribute because these variables are trapped in a cage, the cage of multiplication. The distributive property helps us get them out. So we get to a place like this one right here, 3x plus 15 minus 24. If you see like common terms on the same side of the equation, feel free to combine these together. 15 minus 24 is a negative 9. So we get 3x minus 9 equals 4x minus 2. That's the first part of combining like terms, but we also have to combine terms that are on opposite sides of the equation, right? We have sort of like this Berlin wall that separates the east and the west side of the equation. We want to reunite the X's. We want to reunite the constants. How do we do that? Well, the idea is what's good for the goose is good for the gander. As long as we do the same thing to both sides of the equation, equality will be preserved. So for example, if I subtract 3x from both sides, what you'll see here is that the 3x minus 3x on the left will cancel out. And on the right, the 4x combines with the negative 3x to give us negative 9 is equal to x plus 2. Was that a plus 2 or a minus 2? It's a minus 2. So we combine together the X's on the right-hand side. Why did I move things from the left to the right? It's mostly because I want to have a positive coefficient. It's just sort of an optimistic thing to do there. And now to solve for X, we have to get rid of the 2, so do its inverse operation plus 2 on the right. We add 2 to the left. And so we end up with X equals negative 7. And that is then our proposed solution to this equation. Now, good practice here would be to check our solution. We want to check to make sure that we have the right answer here. So coming back to the original equation, the original equation has been, you know, that's not filthy by the touch of man right here. We plug back the number into that situation. And so check the left-hand side. If you look at the left-hand side of this thing, you have one-half. X we think is negative 7 plus 5 minus 4 to that. Negative 7 plus 5 would be a negative 2. So we give one-half times negative 2 minus 4 there. One-half times negative 2 is going to be a negative 1 minus 4. That should give us a negative 5. That's what we think the left-hand side should be. We then want to compare that to the right-hand side. The right-hand side was just one-third, 2 times negative 7 minus 1. Negative 7 times 2 is a negative 14. Negative 14 minus 1, that gives us a negative 15. And the negative 15 over 3 gives us a negative 5. And so we can see that the left-hand side and the right-hand side agree with each other. So we did find the solution to this linear equation. When it comes to solving a linear equation, one only expects a single solution. There's only going to be one X intercept of a linear function. And so when you're looking for linear equations, no matter how complicated they're going to be, you only expect there to be one solution. We don't get any other situation when it comes to true linear functions.