### 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

### Algebra 1 Online Course

### Transcription: Applications of Systems of Equations

*Welcome back to www.educator.com.*0000

*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 is*0715

*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 house*1054

*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 bean*1178

*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 originals*1421

*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 www.educator.com.*1472

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?