 This video is part two of systems applications. One very common problem is to talk about money. So if I were to think about the difference between number of items and value of items, if I want to know the number of nickels, usually we don't know that when they start with a problem so I would call it in. And if I want to know the value of a nickel, the value of a nickel is 0.05, five cents. And then the total value would be the five cents times the number of coins that we had that were nickels. If I wanted to know dimes, I don't know how many dimes I have. Its value is 0.10 and the total value would be 0.10 times the number of dimes that I have. So if I wanted to write an equation, I actually would have totals down here would help me write my equations and I would total here, I would have the total number of coins and then I could just say n plus d is equal to the number of coins and here I would have the value or total value of the coins and then that if I just add these two, I would have that equal to the total value of my coins. Now we've got our money problem that we started without thinking about. Thomas says the collections of coins consisted of nickels and dimes. He has 50 coins total and they total 450. How many nickels and dimes are in his collection? Well, what are the two different things that we're discussing in here? We're discussing nickels and dimes. So that's what our items are. Okay, we put our nickels and dimes under the item because that's the two things we see that they're talking about and did they tell us how many of you either one we have? The only thing they told us was that we had 50 coins. So that would be the total amount of coins and they did tell us that the total value was 450. So that would be down here in our total row and we're gonna say that's 450 but we still have some things to fill in. We need to know how many that's what they're asking us to find. So those are our unknowns. We don't know how many nickels and we don't know how many dimes. But we do know their values. The value for a nickel again is 0.05 and a value for a dime is 0.10. So again, remember that its amount times the value is equal to the total. So we could say that we have 0.05 in and we have 0.10d and now we have two equations. And if we read those equations, remember we said that we could read them up and down or left and right. So n plus d would be equal to 50 and over here this equation is going to give us 0.05n plus 0.10d equal to 4.50 which is what I have here. So again we could do substitution, we could do elimination. Elimination is even the least likely of all the problems we've done. Elimination would not be the easiest of this problem because we've got all these decimals in here. So I would do substitution again. It doesn't really matter whether we substitute for n and we substitute for d. It might be a little easier to distribute the 0.1 but not necessarily. But I'm going to try it. I'm going to solve for d. So if I solve for d I'm going to subtract n from the side and that leaves me with d and subtract n from the 50 lead me with 50 minus n. So plugging into my bottom equation I have 0.05 and I have n because that's the variable I'm going to work with plus and then 0.1 remember I don't have to carry the zero along times the 50 let's put that in the 50 minus n and then that'll be equal to 4.5 zero if you really want the zero but again you don't have to have the zero. So 0.05n plus 0.1 remember you just move the decimal one place so instead of 50 it becomes 5 and then minus when I distribute to the negative n it's minus 0.1n equal to our 4.5. Combining my like terms I'm going to have 0.05 minus a 0.1 we have negative 0.05n plus 5 is equal to 4.5 and if we subtract 5 from both sides now we have negative 0.05n equal to negative 0.5 and if we divide by negative 0.05 divide by negative 0.05 that's a decimal in there let's see point my fault negative 0.5 divided by negative 0.05 gives us n being equal to 10 and remember that means 10 nickels and now know how many nickels I have to plug that into my top equation up here so D is equal to 50 minus that 10 so that tells us that we have D equal to 40 so I can't remember who he was but he has he has 10 nickels and 40 dimes at least so let's look at this one mission to a football game is $5 for students $8 for general admission if 700 tickets are sold and $5,000 is collected how many student tickets and general admission tickets are sold so first we have to decide what our variables are what are the two items that we're talking about and the items that we're talking about our student tickets and general admission tickets so we'll call this student and I'll call this gen add general admission and I'm gonna let it tells me that I have a total of 700 tickets collected but I don't know how many of each but I'm gonna plug in what I know I know there's 700 total and I know that the $5,000 is collected so that's a total amount of money or total value and now I just have to fill in everything else that I don't know so what's the amount of the student tickets it doesn't say it just tells me that they're sold for $5 so I know this this is a value so that's a $5 value and general admissions are $8 in value and now I'm ready to talk about how many student tickets they didn't tell me so I have to let that be a variable and how many general admission tickets I don't know they didn't tell me so I'll let it be a variable and now one more time amount times the value is equal to the total value so this would be 5s and 8g if I take the number of tickets that I sell for students and multiply that times 5 and add to that the number of general admission tickets times $8 for each one I should get a total of $5,000 and if I take those total number of student tickets plus the total number of general admission tickets there should be 700 of them so s plus g is 700 5s plus 8g is 5,000 which is what we have here so we've done substitution twice let's try elimination and I'm going to multiply the bottom equation because smaller numbers and I'm going to multiply it by negative five five is a little bit smaller might be a little bit easier to work with maybe maybe not but I'm going to say I want to multiply by negative five so my new system is the top equation stays the same and then I have to change the bottom equation distributing the five that becomes a negative 5s and minus 5g and 700 times negative 5 would be negative 3500 so my s is cancel out 8g plus a negative 5g will be 3g and 5,000 minus 3500 should be 1500 but just to be careful I'm going to say 5,000 here minus 3500 to double check that I'm right and I am so we divide by 3 to get to g and g it's going to be equal to 500 so 500 general admission tickets were sold how many student well we have to go back into one of our originals but this original right here has very nice coefficients of 1 so s plus my g that I now know is 500 is supposed to be equal to 700 and if I subtract my 500 from both sides then I find out that s is equal to 200 and we just say that there we can say 200 student and 500 general admission tickets were sold