INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith

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Applications of Rational Expressions

• When working on a word problem the unknown could end up in the denominator. These often lead to rational expression.
• Some common word problems that lead to rational expressions are ones that involve motion or work.
• If a job can be completed in t units of time, then the rate of work is given by 1/t.
• By multiplying the rate by the amount of time, we can figure out how much of a job has been completed at any given time.
• From working with motion we know that distance = rate x time. This leads to other equations such as rate = distance/time and time = distance/rate

Applications of Rational Expressions

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 0:07
• Applications of Rational Expressions 0:27
• Work Problems
• Example 1 2:58
• Example 2 6:45
• Example 3 13:17
• Example 4 16:37

Transcription: Applications of Rational Expressions

Welcome back to www.educator.com.0000

In this lesson we are going to take a look at applications of rational equations.0002

In some of the examples I have cooked up we will look at how some examples involves a just numbers.0009

There are ones that involve motion and some of my favorite ones that involve work.0015

This one is going to be a little bit trickier to think about, but once you get the process done0020

you will see that the work problems are not so bad.0024

Depending on the unknowns in the problem and depending on how we should go ahead and package everything up0031

there could be a few situations that actually lead to rational expressions.0037

Think about those fractions or where an unknown ends up on the denominator.0041

What we want to do is be able to solve these using many of the techniques that we have picked up for rational equations0050

and see how we can recognize these in various different situations, such as numbers, motion, and especially work.0056

The work problems are kind of unique and that you want to know how to represent the situation.0067

If you know how long it takes to complete an entire job then you know the rate of work is given by the following formula 1/T.0075

The way you can read this is by using that T for the amount it takes to do that one job.0084

Here is a quick example.0090

Let us say it takes Betty 7 hours to paint her entire room.0092

Well, that means that every single hour 1/7 of the room is going to be painted.0098

We are going to make a little chart for just keeping track of everything.0103

Maybe x will be my time and let us say how much of the room has been painted so far.0109

One hour, two hours, three hours and make sure you jump all the way to 7.0120

After she is in the room for 7 hours, she will have the entire room painted, the whole thing.0127

If you scale it back, what if she is only working for one hour then only 1/7 of the room is painted.0135

If she is in there for two hours, 2/7 of the room is painted and if she is in there for three 3/7.0144

You can see that we are just incrementing this thing by exactly 1/7 every time.0151

We can make some adjustments to our formula here and say that T would be the amount it takes to do that one for the person0156

and may be multiply it by x, that would represent how long they have been doing that particular job.0165

Watch for that to play a key component with our work problem in just a bit.0173

Let us first see our example of numbers and just see something where our variable ends up in the denominator.0180

In this one we have a certain number and we are going to add it to the numerator and subtract it from the denominator of 7/3.0186

The result equals the reciprocal of 7/3 and we are interested in finding that number.0194

Let us first write down our unknown.0202

x is the unknown number.0205

We construct a model situation here.0217

Take that unknown number and add it to the numerator, but subtract it from the denominator of 7/3.0219

Here is 7/3, so we are adding it to the numerator and subtract it from the denominator.0227

The result equals the reciprocal of 7/3, that is like 7/3 but we flipped it over.0238

This will be our rational inequality here.0245

What we have to do is work on solving it.0249

To solve many of our rational equalities we work on finding a least common denominator0253

which I can see for this one will be 3 - x and 7.0261

Let us give that missing piece to each of the fractions.0268

Here is my original (x + x) (3 – x) = 3/7 let us use some extra space in here.0272

The one on the left, it could use 7, let us put that in there.0286

The one on the right it is missing the 3 – x.0294

At that point, the denominators will be exactly the same.0302

We will just go ahead and focus on the tops of each of these.0307

7 × 7 + x = 3, 3 – x.0312

Continuing on and solving for our x we will go ahead and distribute our 7 and 3.0321

That will give us 49 7x = 9 - 3x.0328

Moving along pretty good.0336

Let us go ahead and add 3x to both sides giving us a 49 + 10x = 9.0338

We will go ahead and subtract 49 from both sides.0351

I have x = 10x is equal to -40.0359

Dividing both sides by 10, I have that x = -4.0368

Just like when we are working with equations, it has to make sense in our original.0374

Looking at the original rational expression I have a restricted value of 3 and I know that it is not 3 that will make the bottom 0.0379

I do not get any restricted values from the other faction because it is simply always 7 on the bottom.0388

Since the -4 is not 3, I'm going to keep it as a valid solution.0394

That one looks good.0402

Let us look at one that involves motion.0407

We will set these up using a table and also use that same idea to help us organize this information.0409

A boat can go 10 miles against a current in the same time it can go 30 miles with the current.0416

The current flows at 4 mph, find the speed of the boat with no current.0422

We have an interesting situation.0428

We have a boat that looks something like this and we have the flow of the river.0430

Now in one situation, it is fighting against the current, and the way you want to think of that in relation to its speed0439

is that the speed of the river is taking away some of the speed of the boat.0447

You will see a subtraction process.0452

If the boat is going in the same direction of the river, they are both helping each other out0456

and you will see an addition problem with both of their speed.0461

You will know they are both helping each other out.0464

Let us see if we organize this information so I can see how to connect it.0468

We need to think of two different situations.0492

We are either going against, or we are going with the river.0495

We will look at the rate, the time, and the distance.0502

This will help us keep track of everything.0511

And of course we are leaving unknown in here.0513

Since we are finding the speed of the boat with no current, let us set that as our unknown.0516

x is the speed of the boat with no current.0522

I think we have a good set up and we can start organizing our information.0540

In the first bit of this problem we know that it can go 10 miles where it is going against the current.0545

That is its distance.0553

It went 10 miles when it is going against the current.0554

It can do that in the same time it can go 30 miles with the current.0558

A little bit of different information, this one will be 30 when it is going that way.0564

The current of the river flows at 4 mph.0570

If we are looking at the speed of the boat and it is going against that river, probably this will be the boat - the current.0575

If we are looking at it going with the river, that will be the speed of the boat + the speed of the current.0585

They are helping each other out.0590

The only thing we do not know in here is the time, but I do know that the time was exactly the same for both of these situations.0593

Let us see what do we got here.0603

I will go ahead and create an equation for each of these.0604

x -4 × time = 10 and x + 4 × time = 30.0607

I know the times are exactly the same for each of these, let us solve them both for time.0620

This one I will go ahead and divide both sides by x - 4.0628

In this one I will divide by x + 4.0631

And I'm ready to develop that rational equation.0638

I will set each of these equal to each other since the times are equal to each other.0641

I have a rational equation then I can go ahead and try and solve.0651

10 ÷ x – 4 = 30 ÷ x + 40655

To get through solving process, we find our common denominator.0661

I'm going to give x + 4 on the left side here and I will give x - 4 to the other side.0668

We will note that we made the denominators exactly the same.0687

We just need to focus on the tops of each of these.0691

Continuing on, you will distribute 10 and we will distribute 30.0702

10x + 40 = 30x -1200710

Subtracting a 10x from both sides will give us 40 = 20x -120 and let us go ahead and add 120 to both sides.0720

160 = 200739

We can divide both sides by 20 and I have that x is equal to 8.0745

Let us make sure that it makes sense.0754

Some restricted values I have for my equations here I know that x cannot equal 4 and -40757

and fortunately both that we have found for a possible solution is neither of those.0766

We can say the speed of the boat in still water would be 8 mph and then this guy is done.0772

Do not be afraid to use those tables from earlier to organize your information.0790

Joe and Steve operated a small roofing company and Mario can roof an average house alone in 9 hours.0800

Al can roof a house alone in 8 hours.0809

We want to know how long will it take them to do their job if they work together.0812

First we need to figure out the rates of each of them individually.0818

Let us go ahead and focus on Joe.0821

Joe can roof an average house alone in 9 hours.0830

Looking at just Joe we know that every hour he will get 1/9 of that house done.0834

Steve over here can roof the house in 8 hours.0844

He is working every single hour 1/8 of that house will be done.0853

We can put in that time on it.0858

Let x be the number of hours.0861

You have 1/9 × however many hours they work and 1/8 × qualified by however many hours Steve works.0871

We want to know how long it will take them to do the job if they work together.0881

We will take each of their work that they are doing and we will add them since they are working together,0889

we want to know when they will complete one job.0895

We have all of our information here and we can go ahead and try and solve this.0899

Our LCD would be 72.0906

We will multiply that through on all parts.0911

Doing some reducing 72 and 9 = 8, 72 and 8 = 9 and now we have an equation 8x + 9x = 72.0923

Combining together what we have on the left side this would be 17x is equal to 720942

and we can divide both sides of that by 17 to get x = 72 ÷ 17.0951

It is looking pretty good.0959

That represents how long it will take them to work together.0961

If you want to represent that as a decimal, you could go ahead and take 72 ÷ 17.0964

When I did that I got about 4.24 hours, I did round it.0971

That gives me a better idea of how long it took them when working together.0980

Anytime when you are working with these types of problems, it should be less than any one of them working by themselves,0985

since they are helping each other out.0992

This one looks good.0995

In this last example we are going to look at a water tank that has two hoses connected to it.0998

Even though this is not a work problem you can see that we can set it up in much the same way.1003

The information that we have is that the first hose can fill the entire tank in 5 hours.1011

The second hose connected to this tank it can empty it in 3 hours.1018

If we start with a completely full tank and then we turned both of them open,1023

the question is how long will it take it to empty the entire tank?1029

Let us have an accrued picture of what we are dealing with here.1034

This would be our water tank and we have one hose that is going in and one hose that is going out.1039

The hose that is going in, it can fill the tank in 5 hours.1054

That means if we just leave it on every hour 1/5 of that tank will fill up.1058

I know its rate is 1/5.1063

If I look at emptying the tank it is a much smaller time to empty it, 1/3.1068

If I did have them both open I would know that the second one1076

would be able to empty the tank since it empties faster than the first one can fill it.1080

Let us see what that we can do to set this one out.1086

The first thing I want to consider was how much the tank is going to be empty every single hour.1091

Starting with the 1/3 I know that every hour that passes by the 1/3 of it will be emptied out.1097

x is the number of hours.1106

The 1/5 coming in is not emptying the tank but it is actually filling it back up.1118

We will say it is the opposite of emptying the tank by 1/5 and it will do that for every hour.1125

We want to know is when will one tank be completely empty.1132

We have a lot of similar components for this one and it often look like a lot of our work problems.1137

All we have to do is go ahead and solve it.1145

Our LCD here that would be 15.1148

Let us multiply all parts by 15 and see what that does.1153

15, 15 and 15.1159

Canceling out some extra stuff here I get 5x - 3x = 15.1163

Doing a little bit of combining on the left side I have 2x =15 or x = 15 ÷ 2 or 7 ½ hours.1173

Notice how in this case it is taking longer since both of them are open.1190

The reason for this is that they are not working with each other.1194

They are working against each other.1198

One was trying to fill the tank, and one is trying to empty the tank.1199

Use those bits of information you can find along the word problem to give you a little bit of intuition about your final solution.1205

In that way you can be assured that it does make sense in the context of the problem.1212

Thank you for watching www.educator.com.1217