 This video will talk about solving systems of linear equations. So a system of equations is a set of two or more, but we're going to look at two equations for which a common solution is sought. And that solution is going to be an ordered pair. And we're going to be looking at lines in this section. So our lines will usually have two different slopes. So we want to find out the point they share in common or where they intersect. Now we can use our calculator or we can draw these by hand. It doesn't matter. Either way, it's so much easier if we say that y is equal to, and in the first case in this problem, we already have it y equal. But then we also want to have the other one be y equal. And if I solve it, I have 5y is equal to 7 minus 4x. It's negative 5y is equal to 7 minus 4x. And then I divide by negative 5 and I can put it into my calculator just like this as long as I put parentheses around it. The numerator. So this is what I'm going to put in my calculator. Y is equal to parentheses, 7 minus 4x, close my parentheses, and then divide it by negative 5. I need that negative. And then my second equation, of course, is just 2x minus 5. And the standard window should work, so I'll do zoom 6 on this. And now I'm going to see my two lines. And if I want to let the calculator do it, I do second trace. And then we do intersect, which is 5. And since they're just two lines, we can say enter, enter, enter, and we find out that. So if I draw this for myself so that I have my work on my paper, it looks something like this, not exactly like this. But I see that I have this point here and I would say x equal 3, y equal 1. So now we can also use substitution. Substitutions must use when you have one variable equal to everything else with a coefficient of 1. So this is perfect. So what we're really going to say is this 2x minus 5 is going to replace this y. So I'm going to start by writing my problem 4x minus 5, but then I'm going to replace that y with something, and then it's equal to 7. So I replace it with the 2x minus 5, and now I just solve the equation. So I have 4x minus 10x, negative 5 times negative 5 is plus 25, equal to 7. And if I look at these two, I get negative 6x. And if I take this to the other side, it's going to be equal to negative 18. And divide by negative 6, x is going to be equal to positive 3. And then if I plug that back into my original equation, which was 2x minus 5 equal y, I have 2 times my x, which is 3, minus 5 is equal to y. So 6 minus 5 would be 1. And if you'll notice, this is the exact same problem that we had before. Same system. We just solved it two different ways and got the same answer of x equal 3, y equal 1. So the third way that we can solve this, and that's by elimination. This is best used when you have variables all over the place. I'll give you an example. Like if you had 2x plus 3y was equal to 7, and you had 5x plus 8y equal to negative 10, nothing looks really nice there. It's not easy to solve for y. It's not easy to solve for x. So we would use elimination. That what we want to do is get coefficient of 1. We have to multiply one or more equations to make opposite coefficients on a variable. That's the first step, and then the second step is to add the two equations. And this will eliminate one variable. And then you've got a one variable equation. It's very much like substitution at that point. So if I wanted to get rid of, let's say x here, I could multiply this top equation by a 5, and that would give me 10. And if I did that, I want opposite coefficients, so I'd have to multiply the bottom equation by negative 2. Let's just go ahead and set it up. That gives me 10x, distribute all the way through, plus 15y, all the way through, equal to 35. And negative 2 times 5 would be negative 10x, negative 2 times 8 would be minus 16y, and negative 2 times negative 10 would be positive 20, and then you would add them. So these would cancel out. This would give you negative y equal to 55. So we would know y would be negative 55, and then plug and chug. I'm afraid the numbers are going to be ugly. But we can try it. It goes to go back up to an original equation, and these are the smallest numbers, so that's where I'm going to go. And we're going to say that 2x plus 3 times my y, which I found out to be negative 55, is going to be equal to that 7. So 2x, this happens to be minus 165, equal to 7. 2x is going to be equal to, if I add 165 to 7, I'm going to have 172. And then if I divide by 2, I end up with 86. So we know by elimination that x would be 86 and y would be negative 55. Alright, so in application, generally they represent two different things. We're going to have two equations that represent two different things, and often you're going to be talking about value and quantity. And this is a typical kind of problem, a money kind of problem. So let's see what we have. Emma's doing odd jobs around the house trying to earn her money to buy a new guitar case. She saves all her quarters and dimes in her cedar box, where she places all nickels and pennies in a drawer to spend. So far she has 225 coins in her cedar box worth $45. How many quarters does Emma have? So we don't really care about nickels and pennies because she's going to spend those. But we do care about quarters and dimes. So let's see what we have. We want to know how many. That's what they're asking us for. So that means that when we represent our variables, it's got to be the number or how many quarters she has. And if she's also got dimes, I'll call that d, and that tells me the number of dimes that she has. And then the total number of coins that she has is, it said, 225. We could actually set up an equation now that says dimes plus quarters is equal to 225. The number of dimes plus the number of quarters is equal to 225. And now we also have this other thing, though, that says that she has a total of, a total value of all those coins is $45. Well, how did she get that? She took the value of her dimes and the value of her quarters. Well, we know that the value of a dime is .10, and then how many does she have? We don't know, so we call that d. And we know that the value of a quarter is 25 cents, but how many of those does she have? Because if I multiply 2 times 25, I know I have 50 cents. If she had 10, we'd know that she had $2.50. So it's a value of the coin times the number of coins. So we have a second equation then. So .10d plus .25q is equal to 45, and that's dollars, and this is number. So we need to solve this by any method that we can. We could graph it if we wanted to. We could do substitution. Elimination is probably the hardest method of all these. So I'm going to go ahead and do it by substitution. And I know that multiplying by .1 is a lot easier. So I'm going to solve for d, so I can plug in for d. So d is going to be 225 minus q. Plugging in, I have .10, but I'm going to replace the d plus .25q is equal to 45. And I'm replacing d with that 225 minus q. So I have to distribute, and that will give me 22.5, moving the decimal one place, minus .10 or .1q plus .25q is equal to 45. Well, if I subtract 22.5, I'm going to combine these two. That will give me positive .15q is equal to 22.5. And if I divide by .15, I'm going to find out that q is equal to 150. So go back and see if I've answered the question. How many quarters? That's all they asked me for. So Emma has 150 quarters.