WEBVTT mathematics/algebra-1/smith
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Welcome back to www.educator.com.
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In this lesson we are going to look at applications of systems of equations.
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Think about word problems involving the systems of equations that we saw earlier.
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Our main goal will be looking at these word problems and figuring out how we can interpret them and build the system from there.
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Here is a part of this that we will be actually looking at our solutions and be able to figure out how it fits in the context of that word problem.
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A good way to think of how to start picking these apart is to take every little bit of information in there.
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Say 2 different situations and create an equation for each of those situations.
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You will notice some of these problems look similar to ones we did in the earlier applications of linear equations.
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In that lesson we tried to write everything in terms of one variable.
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Now that we know more about systems, we can use 2 variables to help us out and create an equation for each of the situations present.
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It will actually make things a little bit neater.
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Once we have built our system of equations, we are getting down to being able to solve that system.
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We will use our techniques for solving a system such as elimination method or the substitution method.
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We can also use tables like we did earlier to help you organize this information
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but I will mainly focus on just the methods of substitution and elimination to tackle the systems of equations.
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One very important part to remember is that all of these are going to be word problems.
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Even though we get a solution like x = 3 and y = 5, we have to take those and interpret them and say what is x and y.
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Always interpret these in the context of the problem.
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One thing that will make this very easy and not so bad, is to go ahead and write down what your variables are at the very beginning.
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In that way, once you are all done with the problem you can reference that and say okay my x represents this and my y represents this.
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That way you can easily say here is how it fits in the context of the problem.
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Let us jump into our examples and see how we can actually start creating a system of equations.
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In this first one, we have a restaurant that needs to have 2 seat tables and 3 seat tables.
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According to local fire code, it looks like the restaurants maximum occupancy is 46 costumers.
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If the owners have hardly enough servers to handle 17 tables of customers, how many of each kind of table should they purchase?
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Notice that we are under a few different constraints.
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One has to do with simply the number of people that you can fit in this restaurant.
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We want to make sure that we have no more than 46 customers.
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Another restriction that we are under is the number of tables that our servers can handle and provide good service and we can only do 17 tables.
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From the owner's perspective, we do not want to go less than these numbers.
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After all, if we have less than 46 customers, we are not making as much money.
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If you have less than 17 tables then we have servers that are not doing a lot of work.
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We are going to try and aim to get these exact when we set up our equations.
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Let us hunt down some unknowns and first write those down.
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I'm going to say let x that will be the number of 2 seat tables.
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Let y be the number of 3 seat tables.
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We can use these unknowns to start connecting our information gathered.
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In the first situation, we are going to focus on the number of people that we want to fit in this restaurant.
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We are looking to put 46 people in this restaurant and it will come from seating them either at these 2 seat table or at the 3 seat table.
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There will be 2 people for every 2 seat table so we can represent that by saying 2 × X.
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There will be 3 people at every 3 seat table or 3y.
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That expression right there just represents a number of people and we want it equal to 46 customers.
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That is dealing with our people, maybe I will even label that people.
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The other restriction is what we can do with our servers in handling all of these tables.
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We want the total number of tables and that is our 2 seat tables and our 3 seat tables to equals 17.
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You can say we have encapsulated all the information to one equation or the other.
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We have actually set up the system and it is just a matter of going through and solving it.
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You could use the elimination method or you could use the substitution method, both of them should get you to the same answer.
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I’m going to go through the elimination method.
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I will do this by taking the second equation here and we will multiply that one by -2.
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Let us see what the result will be.
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The first equation will remain unchanged and everything in the second one will be multiplied by -2.
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We have done that, we can go ahead and combine both of those equations together.
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You will notice that the x’s are canceling out since one is 2 and one is -2, that is exactly what we want with the elimination method.
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3y + - 2y will give us 1y, and now we have 46 - 834 and that will give us 12.
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One of the great thing is about writing down our variables at the very beginning is not only do I know that y = 12,
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but I can interpret this as the number of 3 seat tables since I have it right here that y is the number of 3 seat tables.
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We are going to continue on, and I will figure out what x is so we can figure out the number for both of the tables.
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It is not so bad, just always remember that the total number of tables must be equal to 17.
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I will borrow one of our original equations down here and just substitute in the 12 for y.
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To solve that one we will subtract 12 from both sides and get that x = 5.
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Now we can fully answer this problem.
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The number of 3 seat tables is 12 and the number of 2 seat tables that will be 5.
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We will keep that as our final answer.
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Set down your unknowns, set up your system, solve it and interpret it in the context of the problem.
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Let us try another problem, it has a lot of the same feel to it.
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This one says that at a concession stand, if you buy 5 hotdogs and 2 hamburgers the cost is $9.50, 2 hotdogs and 5 hamburgers cost $13.25.
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We are interested in finding the total cost of just one hamburger.
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First, we need to go ahead and label our unknown.
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We have the number of hotdogs and we have our hamburgers.
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Let us work on that.
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I will say that we will use and see d for hotdogs and let us use h for our hamburgers.
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Both of these situations here are involving costs.
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In the first one, we are looking at so many hotdogs, 5 of them and 2 hamburgers all equal to $9.50.
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In the next situation, we have a different combination of hotdogs and hamburgers to equal the $13.25.
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We will take each of these and package them into their own equation.
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5 hotdogs 2 hamburgers equals to $9.50.
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2 hotdogs 5 hamburgers equals to $13.25.
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There is our system of equations.
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When we are done solving we should be able to get the price for just one hotdog and a price for just one hamburger.
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In this system, you could solve it using elimination or substitution.
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I think I'm going to attack this using our elimination method.
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I will have to multiply both equations by something to get something to cancel out.
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Let us multiply the first one here or multiply everything in there by -2.
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We will take our second equation and I will multiply everything there by 5.
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Let us see what the result is, -10 - 4h -19.
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Then we will take our second equation 10 d + 25h (5 × 13.25).
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Adding these 2 equations together, let us see what our result is.
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The 10 d and -10 d both will cancel each other out and we will have 25h - 4h = 21h.
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We can add together the 66 + 25 -19.
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In this new equation, we only have h to worry about and we can get that all by itself just by dividing both sides by 21.
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This will give me that h = $2.25.
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By looking at what we have identified earlier, I know that this is the cost of one hamburger $2.25.
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Even though the problem is not asking for it, let us go ahead and figure out how much a hotdog is
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by taking this amount and substituting it back into one of our original equations.
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You can see that I have put this into the first equation, now I need to multiply 2 by $2.25 that will $4.50.
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Subtract $4.50 from both sides and divide both sides by 5.
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Now we have our entire solution.
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I know for sure that the hamburgers will cost $2.25.
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I also know that one hotdog will cost exactly $1.
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Try and pick out each situation there and interpret it into its own equation.
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Let us try this one out.
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In the U.S. Senate, there are 100 representatives, if there are exactly 10 more democrats than republicans, how many of each are in the senate?
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One assumption that we are going to use here is that all representatives are either democrats or republicans.
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We are not even going to consider any third parties here.
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Let us see if I can break this down but first let us identify some variables.
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Let us say r is the number of republicans and we will let d be the number of democrats.
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We need to set up an equation for each of these and one of the first big bits of information I get is that the total number of representatives should be 100.
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We can interpret that as the number of republicans + the number of democrats that should be equal to 100.
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We need to interpret that there are exactly 10 more democrats than republicans.
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To look at that, let us compare the 2.
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If I was to take republicans and democrats and put them on each side of the equal sign,
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they would not exactly be equal because I have 10 more democrats than republicans.
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The way you want to interpret is that right now this side got a lot more, it has 10 more than the republican side.
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If I want to balance out this equation, I need to take something away from the heavier side.
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Since it is exactly 10 more, I will take away 10 and that should balance it just fine.
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My second equation is r = d -10.
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I will get that relationship that there are exactly 10 more democrats than republicans.
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When we are done setting up the system, notice how it set up a good for substitution.
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The reason is if you notice here our r is already solved.
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Let us use that substitution method to help us out.
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We will take the d – 10 and substitute it into the first equation.
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In this new equation, we only have these to worry about.
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Let us combine the like terms, 1d and another d will give us a total of 2d -10 =100 and we can add 10 to both sides.
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One last step, let us go ahead and divide both sides by 2.
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This will give me that the number of democrats should be 55.
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Since I also know that there are exactly 10 more democrats than republicans,
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I can take this number then subtract 10 and that will give me my republicans.
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Let us say we have 55 democrats, 45 republicans.
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I have one more example with these systems of equations and this involves a little bit more difficult numbers.
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We are going to end up rounding these nice whole numbers, so we do not have to deal with too much decimals.
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Be careful along the way because some of the numbers do get a little messy.
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In this problem, I have a Janet that blends coffee for a coffee house
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and what she wants to do is she wants to prepare 280 pounds of blended coffee beans and sell it for $5.32 per pound.
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The way she plans on making this mixture is she is going to blend together 2 different types of coffees.
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She is going to blend together a high-quality coffee that cost $6.25/pound.
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And she is also going to blend together with that a cheaper coffee that only costs $3/ pound.
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The question is to the nearest pound, how much high quality coffee bean and how much cheaper quality coffee bean should she mix in order to get this plan.
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Let us set down some unknowns and see if we can figure this out.
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I will call (q) the number of pounds and this will be for our high quality coffee.
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Let (c), even number of pounds for our cheap quality coffee.
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We are going to try and set up a how many pounds of each of these we need.
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One of the first bits of information that will help out with that is we know that we have a total of 280 pounds.
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The number of pounds for my high quality coffee + the number of pounds for my cheaper coffee better equal to 280 pounds.
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This entire equation just deals with pounds.
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The second part will have to deal with the cost.
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We want the final mixture to be $5.32, so we will use the cost of the high quality coffee bean and the cost of the low quality coffee bean
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and blend the 2 together and get that final cost.
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Let us see what we have here.
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Normally, high quality coffee costs $6.25 every pound of high quality coffee.
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The cheaper coffee usually costs $3/pound.
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We are hoping to make is $5.32 for that final 280 pounds of coffee.
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Let us mark this other one as just costs.
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In our system here, you can see the numbers are not quite as nice but we can definitely still work with them.
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Let us do a little bit of cleaning up then we will go ahead and try the substitution method on the system.
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I’m going to multiply these together first, so $5.32 × 280,
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I'm doing the new system 625q + $3, c =1489.6.
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To use the substitution method, I’m going to take the first equation and let us go ahead and solve it for q.
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Q = 280 – c.
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Once we have that, we can go ahead and substitute it into our second equation.
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$6.25 q is 280 – (c + $3.04c) = 1489.6.
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Our new equation only has c and we can go ahead and solve for that.
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Let us distribute through by the $6.25 and see what we will get.
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17.50 – 6.25c + 3c = 1489.6.
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Let us combine our c terms, -3.25c = 1489.6.
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We almost have c all by itself, let us get there by subtracting the 17.50 from both sides, -260.4.
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One last step, let us go ahead and divide both sides by the -3.25.
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I’m getting 80 and then something after the decimal 12307692, it just keeps going on and on.
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I can see that I’m going to have just a little bit more than 80 pounds, but we want to round this to the nearest whole pound.
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What I will say is that c is about equal to 80 pounds.
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Now, once I know how many pound the cheaper coffee I will need, we can quickly go back to one of our originals
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and figure out how much of the high quality coffee we need.
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I need the total poundage to be 280 and 80 of that will be in the cheaper coffee, then I know that the high quality coffee I will need 200 pound.
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Our rounded solution, I have that 80 pounds of cheap coffee and 200 pounds of high quality coffee.
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If you pick apart your word problems and try and set down an equation for some of the situations in there,
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you should be able to develop your system of equations just fine.
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Then use one of the methods that we have learned especially those algebraic methods to solve the system
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and do not forget to interpret your solution in the context of the problem.
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