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Lecture Comments (5)

0 answers

Post by Khanh Nguyen on October 2, 2015

I cannot see the practice questions.

0 answers

Post by Ardeshir Badr on January 4, 2015

Do we not have any Practice Questions for "Applications of Systems of Equation" ?

2 answers

Last reply by: Khanh Nguyen
Fri Oct 2, 2015 10:57 PM

Post by Ardeshir Badr on January 4, 2015

Can someone make the corrections to the mistakes on the Practice exercises?

Applications of Systems of Equations

  • Some word problems lead to the development of a system of equations.
  • Identify both unknowns used so you can quickly reference them later.
  • Separate situations are a clue that more than one equation can be used to connect the information.
  • Once a system is developed, you may use the substitution or elimination method to solve it.
  • Make sure you check to see if the solution makes sense in the context of the problem.

Applications of Systems of Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:12
  • Applications of Systems of Equations 0:30
    • Word Problems
  • Example 1 2:17
  • Example 2 7:55
  • Example 3 13:07
  • Example 4 17:15

Transcription: Applications of Systems of Equations

Welcome back to

In this lesson we are going to look at applications of systems of equations. 0003

Think about word problems involving the systems of equations that we saw earlier. 0007

Our main goal will be looking at these word problems and figuring out how we can interpret them and build the system from there.0015

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.0022

A good way to think of how to start picking these apart is to take every little bit of information in there.0033

Say 2 different situations and create an equation for each of those situations.0039

You will notice some of these problems look similar to ones we did in the earlier applications of linear equations.0045

In that lesson we tried to write everything in terms of one variable. 0051

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.0055

It will actually make things a little bit neater.0062

Once we have built our system of equations, we are getting down to being able to solve that system.0067

We will use our techniques for solving a system such as elimination method or the substitution method.0072

We can also use tables like we did earlier to help you organize this information 0079

but I will mainly focus on just the methods of substitution and elimination to tackle the systems of equations.0083

One very important part to remember is that all of these are going to be word problems.0094

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.0099

Always interpret these in the context of the problem. 0108

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.0112

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.0122

That way you can easily say here is how it fits in the context of the problem.0130

Let us jump into our examples and see how we can actually start creating a system of equations.0140

In this first one, we have a restaurant that needs to have 2 seat tables and 3 seat tables.0146

According to local fire code, it looks like the restaurants maximum occupancy is 46 costumers.0151

If the owners have hardly enough servers to handle 17 tables of customers, how many of each kind of table should they purchase?0158

Notice that we are under a few different constraints.0165

One has to do with simply the number of people that you can fit in this restaurant.0168

We want to make sure that we have no more than 46 customers.0173

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.0177

From the owner's perspective, we do not want to go less than these numbers. 0187

After all, if we have less than 46 customers, we are not making as much money.0190

If you have less than 17 tables then we have servers that are not doing a lot of work.0195

We are going to try and aim to get these exact when we set up our equations.0200

Let us hunt down some unknowns and first write those down.0205

I'm going to say let x that will be the number of 2 seat tables.0210

Let y be the number of 3 seat tables.0228

We can use these unknowns to start connecting our information gathered.0243

In the first situation, we are going to focus on the number of people that we want to fit in this restaurant.0248

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.0254

There will be 2 people for every 2 seat table so we can represent that by saying 2 × X.0264

There will be 3 people at every 3 seat table or 3y.0272

That expression right there just represents a number of people and we want it equal to 46 customers.0277

That is dealing with our people, maybe I will even label that people.0285

The other restriction is what we can do with our servers in handling all of these tables.0291

We want the total number of tables and that is our 2 seat tables and our 3 seat tables to equals 17.0298

You can say we have encapsulated all the information to one equation or the other.0311

We have actually set up the system and it is just a matter of going through and solving it.0316

You could use the elimination method or you could use the substitution method, both of them should get you to the same answer.0321

I’m going to go through the elimination method.0327

I will do this by taking the second equation here and we will multiply that one by -2.0331

Let us see what the result will be.0340

The first equation will remain unchanged and everything in the second one will be multiplied by -2.0341

We have done that, we can go ahead and combine both of those equations together.0365

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.0371

3y + - 2y will give us 1y, and now we have 46 - 834 and that will give us 12.0379

One of the great thing is about writing down our variables at the very beginning is not only do I know that y = 12, 0397

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.0402

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.0410

It is not so bad, just always remember that the total number of tables must be equal to 17.0415

I will borrow one of our original equations down here and just substitute in the 12 for y.0421

To solve that one we will subtract 12 from both sides and get that x = 5.0433

Now we can fully answer this problem.0441

The number of 3 seat tables is 12 and the number of 2 seat tables that will be 5.0446

We will keep that as our final answer.0462

Set down your unknowns, set up your system, solve it and interpret it in the context of the problem.0468

Let us try another problem, it has a lot of the same feel to it.0478

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 $13.25.0483

We are interested in finding the total cost of just one hamburger.0495

First, we need to go ahead and label our unknown. 0499

We have the number of hotdogs and we have our hamburgers.0504

Let us work on that. 0509

I will say that we will use and see d for hotdogs and let us use h for our hamburgers.0512

Both of these situations here are involving costs.0529

In the first one, we are looking at so many hotdogs, 5 of them and 2 hamburgers all equal to $9.50.0534

In the next situation, we have a different combination of hotdogs and hamburgers to equal the $13.25.0543

We will take each of these and package them into their own equation. 0549

5 hotdogs 2 hamburgers equals to $9.50.0554

2 hotdogs 5 hamburgers equals to $13.25.0569

There is our system of equations.0581

When we are done solving we should be able to get the price for just one hotdog and a price for just one hamburger. 0583

In this system, you could solve it using elimination or substitution. 0591

I think I'm going to attack this using our elimination method.0595

I will have to multiply both equations by something to get something to cancel out.0599

Let us multiply the first one here or multiply everything in there by -2.0604

We will take our second equation and I will multiply everything there by 5.0614

Let us see what the result is, -10 - 4h -19.0621

Then we will take our second equation 10 d + 25h (5 × 13.25).0639

Adding these 2 equations together, let us see what our result is.0661

The 10 d and -10 d both will cancel each other out and we will have 25h - 4h = 21h.0665

We can add together the 66 + 25 -19.0679

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 21.0689

This will give me that h = $2.25.0702

By looking at what we have identified earlier, I know that this is the cost of one hamburger $2.25.0708

Even though the problem is not asking for it, let us go ahead and figure out how much a hotdog is0715

by taking this amount and substituting it back into one of our original equations.0719

You can see that I have put this into the first equation, now I need to multiply 2 by $2.25 that will $4.50.0737

Subtract $4.50 from both sides and divide both sides by 5.0750

Now we have our entire solution. 0768

I know for sure that the hamburgers will cost $2.25.0769

I also know that one hotdog will cost exactly $1. 0774

Try and pick out each situation there and interpret it into its own equation.0780

Let us try this one out.0789

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?0791

One assumption that we are going to use here is that all representatives are either democrats or republicans.0799

We are not even going to consider any third parties here.0804

Let us see if I can break this down but first let us identify some variables.0808

Let us say r is the number of republicans and we will let d be the number of democrats.0813

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 100.0846

We can interpret that as the number of republicans + the number of democrats that should be equal to 100.0856

We need to interpret that there are exactly 10 more democrats than republicans.0870

To look at that, let us compare the 2.0879

If I was to take republicans and democrats and put them on each side of the equal sign,0881

they would not exactly be equal because I have 10 more democrats than republicans.0887

The way you want to interpret is that right now this side got a lot more, it has 10 more than the republican side.0893

If I want to balance out this equation, I need to take something away from the heavier side.0901

Since it is exactly 10 more, I will take away 10 and that should balance it just fine.0906

My second equation is r = d -10.0914

I will get that relationship that there are exactly 10 more democrats than republicans.0917

When we are done setting up the system, notice how it set up a good for substitution.0924

The reason is if you notice here our r is already solved.0929

Let us use that substitution method to help us out.0940

We will take the d – 10 and substitute it into the first equation.0943

In this new equation, we only have these to worry about.0958

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.0961

One last step, let us go ahead and divide both sides by 2.0979

This will give me that the number of democrats should be 55.0987

Since I also know that there are exactly 10 more democrats than republicans, 0992

I can take this number then subtract 10 and that will give me my republicans.0997

Let us say we have 55 democrats, 45 republicans.1011

I have one more example with these systems of equations and this involves a little bit more difficult numbers.1038

We are going to end up rounding these nice whole numbers, so we do not have to deal with too much decimals.1045

Be careful along the way because some of the numbers do get a little messy.1050

In this problem, I have a Janet that blends coffee for a coffee house1054

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.1058

The way she plans on making this mixture is she is going to blend together 2 different types of coffees.1066

She is going to blend together a high-quality coffee that cost $6.25/pound.1072

And she is also going to blend together with that a cheaper coffee that only costs $3/ pound.1080

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.1088

Let us set down some unknowns and see if we can figure this out.1098

I will call (q) the number of pounds and this will be for our high quality coffee.1105

Let (c), even number of pounds for our cheap quality coffee.1126

We are going to try and set up a how many pounds of each of these we need.1149

One of the first bits of information that will help out with that is we know that we have a total of 280 pounds.1153

The number of pounds for my high quality coffee + the number of pounds for my cheaper coffee better equal to 280 pounds. 1160

This entire equation just deals with pounds.1170

The second part will have to deal with the cost.1176

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 bean1178

and blend the 2 together and get that final cost.1186

Let us see what we have here.1191

Normally, high quality coffee costs $6.25 every pound of high quality coffee.1191

The cheaper coffee usually costs $3/pound.1200

We are hoping to make is $5.32 for that final 280 pounds of coffee.1206

Let us mark this other one as just costs.1215

In our system here, you can see the numbers are not quite as nice but we can definitely still work with them.1221

Let us do a little bit of cleaning up then we will go ahead and try the substitution method on the system.1227

I’m going to multiply these together first, so $5.32 × 280,1234

I'm doing the new system 625q + $3, c =1489.6.1248

To use the substitution method, I’m going to take the first equation and let us go ahead and solve it for q.1265

Q = 280 – c.1274

Once we have that, we can go ahead and substitute it into our second equation.1280

$6.25 q is 280 – (c + $3.04c) = 1489.6.1289

Our new equation only has c and we can go ahead and solve for that.1307

Let us distribute through by the $6.25 and see what we will get.1312

17.50 – 6.25c + 3c = 1489.6.1323

Let us combine our c terms, -3.25c = 1489.6.1338

We almost have c all by itself, let us get there by subtracting the 17.50 from both sides, -260.4.1354

One last step, let us go ahead and divide both sides by the -3.25.1378

I’m getting 80 and then something after the decimal 12307692, it just keeps going on and on.1396

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.1407

What I will say is that c is about equal to 80 pounds.1415

Now, once I know how many pound the cheaper coffee I will need, we can quickly go back to one of our originals1421

and figure out how much of the high quality coffee we need.1427

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.1433

Our rounded solution, I have that 80 pounds of cheap coffee and 200 pounds of high quality coffee.1446

If you pick apart your word problems and try and set down an equation for some of the situations in there,1454

you should be able to develop your system of equations just fine.1459

Then use one of the methods that we have learned especially those algebraic methods to solve the system 1463

and do not forget to interpret your solution in the context of the problem.1467

Thanks for watching