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 take a look at applications of rational equations.
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In some of the examples I have cooked up we will look at how some examples involves a just numbers.
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There are ones that involve motion and some of my favorite ones that involve work.
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This one is going to be a little bit trickier to think about, but once you get the process done
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you will see that the work problems are not so bad.
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Depending on the unknowns in the problem and depending on how we should go ahead and package everything up
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there could be a few situations that actually lead to rational expressions.
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Think about those fractions or where an unknown ends up on the denominator.
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What we want to do is be able to solve these using many of the techniques that we have picked up for rational equations
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and see how we can recognize these in various different situations, such as numbers, motion, and especially work.
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The work problems are kind of unique and that you want to know how to represent the situation.
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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.
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The way you can read this is by using that T for the amount it takes to do that one job.
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Here is a quick example.
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Let us say it takes Betty 7 hours to paint her entire room.
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Well, that means that every single hour 1/7 of the room is going to be painted.
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We are going to make a little chart for just keeping track of everything.
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Maybe x will be my time and let us say how much of the room has been painted so far.
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One hour, two hours, three hours and make sure you jump all the way to 7.
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After she is in the room for 7 hours, she will have the entire room painted, the whole thing.
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If you scale it back, what if she is only working for one hour then only 1/7 of the room is painted.
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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.
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You can see that we are just incrementing this thing by exactly 1/7 every time.
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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 person
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and may be multiply it by x, that would represent how long they have been doing that particular job.
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Watch for that to play a key component with our work problem in just a bit.
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Let us first see our example of numbers and just see something where our variable ends up in the denominator.
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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.
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The result equals the reciprocal of 7/3 and we are interested in finding that number.
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Let us first write down our unknown.
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x is the unknown number.
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We construct a model situation here.
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Take that unknown number and add it to the numerator, but subtract it from the denominator of 7/3.
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Here is 7/3, so we are adding it to the numerator and subtract it from the denominator.
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The result equals the reciprocal of 7/3, that is like 7/3 but we flipped it over.
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This will be our rational inequality here.
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What we have to do is work on solving it.
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To solve many of our rational equalities we work on finding a least common denominator
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which I can see for this one will be 3 - x and 7.
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Let us give that missing piece to each of the fractions.
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Here is my original (x + x) (3 – x) = 3/7 let us use some extra space in here.
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The one on the left, it could use 7, let us put that in there.
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The one on the right it is missing the 3 – x.
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At that point, the denominators will be exactly the same.
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We will just go ahead and focus on the tops of each of these.
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7 × 7 + x = 3, 3 – x.
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Continuing on and solving for our x we will go ahead and distribute our 7 and 3.
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That will give us 49 7x = 9 - 3x.
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Moving along pretty good.
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Let us go ahead and add 3x to both sides giving us a 49 + 10x = 9.
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We will go ahead and subtract 49 from both sides.
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I have x = 10x is equal to -40.
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Dividing both sides by 10, I have that x = -4.
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Just like when we are working with equations, it has to make sense in our original.
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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.
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I do not get any restricted values from the other faction because it is simply always 7 on the bottom.
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Since the -4 is not 3, I'm going to keep it as a valid solution.
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That one looks good.
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Let us look at one that involves motion.
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We will set these up using a table and also use that same idea to help us organize this information.
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A boat can go 10 miles against a current in the same time it can go 30 miles with the current.
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The current flows at 4 mph, find the speed of the boat with no current.
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We have an interesting situation.
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We have a boat that looks something like this and we have the flow of the river.
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Now in one situation, it is fighting against the current, and the way you want to think of that in relation to its speed
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is that the speed of the river is taking away some of the speed of the boat.
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You will see a subtraction process.
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If the boat is going in the same direction of the river, they are both helping each other out
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and you will see an addition problem with both of their speed.
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You will know they are both helping each other out.
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Let us see if we organize this information so I can see how to connect it.
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We need to think of two different situations.
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We are either going against, or we are going with the river.
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We will look at the rate, the time, and the distance.
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This will help us keep track of everything.
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And of course we are leaving unknown in here.
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Since we are finding the speed of the boat with no current, let us set that as our unknown.
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x is the speed of the boat with no current.
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I think we have a good set up and we can start organizing our information.
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In the first bit of this problem we know that it can go 10 miles where it is going against the current.
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That is its distance.
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It went 10 miles when it is going against the current.
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It can do that in the same time it can go 30 miles with the current.
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A little bit of different information, this one will be 30 when it is going that way.
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The current of the river flows at 4 mph.
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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.
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If we are looking at it going with the river, that will be the speed of the boat + the speed of the current.
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They are helping each other out.
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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.
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Let us see what do we got here.
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I will go ahead and create an equation for each of these.
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x -4 × time = 10 and x + 4 × time = 30.
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I know the times are exactly the same for each of these, let us solve them both for time.
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This one I will go ahead and divide both sides by x - 4.
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In this one I will divide by x + 4.
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And I'm ready to develop that rational equation.
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I will set each of these equal to each other since the times are equal to each other.
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I have a rational equation then I can go ahead and try and solve.
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10 ÷ x – 4 = 30 ÷ x + 4
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To get through solving process, we find our common denominator.
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I'm going to give x + 4 on the left side here and I will give x - 4 to the other side.
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We will note that we made the denominators exactly the same.
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We just need to focus on the tops of each of these.
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Continuing on, you will distribute 10 and we will distribute 30.
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10x + 40 = 30x -120
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Subtracting a 10x from both sides will give us 40 = 20x -120 and let us go ahead and add 120 to both sides.
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160 = 20
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We can divide both sides by 20 and I have that x is equal to 8.
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Let us make sure that it makes sense.
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Some restricted values I have for my equations here I know that x cannot equal 4 and -4
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and fortunately both that we have found for a possible solution is neither of those.
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We can say the speed of the boat in still water would be 8 mph and then this guy is done.
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Do not be afraid to use those tables from earlier to organize your information.
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Joe and Steve operated a small roofing company and Mario can roof an average house alone in 9 hours.
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Al can roof a house alone in 8 hours.
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We want to know how long will it take them to do their job if they work together.
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First we need to figure out the rates of each of them individually.
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Let us go ahead and focus on Joe.
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Joe can roof an average house alone in 9 hours.
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Looking at just Joe we know that every hour he will get 1/9 of that house done.
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Steve over here can roof the house in 8 hours.
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He is working every single hour 1/8 of that house will be done.
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We can put in that time on it.
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Let x be the number of hours.
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You have 1/9 × however many hours they work and 1/8 × qualified by however many hours Steve works.
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We want to know how long it will take them to do the job if they work together.
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We will take each of their work that they are doing and we will add them since they are working together,
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we want to know when they will complete one job.
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We have all of our information here and we can go ahead and try and solve this.
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Our LCD would be 72.
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We will multiply that through on all parts.
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Doing some reducing 72 and 9 = 8, 72 and 8 = 9 and now we have an equation 8x + 9x = 72.
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Combining together what we have on the left side this would be 17x is equal to 72
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and we can divide both sides of that by 17 to get x = 72 ÷ 17.
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It is looking pretty good.
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That represents how long it will take them to work together.
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If you want to represent that as a decimal, you could go ahead and take 72 ÷ 17.
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When I did that I got about 4.24 hours, I did round it.
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That gives me a better idea of how long it took them when working together.
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Anytime when you are working with these types of problems, it should be less than any one of them working by themselves,
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since they are helping each other out.
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This one looks good.
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In this last example we are going to look at a water tank that has two hoses connected to it.
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Even though this is not a work problem you can see that we can set it up in much the same way.
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Let us go ahead and give it a read.
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The information that we have is that the first hose can fill the entire tank in 5 hours.
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The second hose connected to this tank it can empty it in 3 hours.
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If we start with a completely full tank and then we turned both of them open,
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the question is how long will it take it to empty the entire tank?
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Let us have an accrued picture of what we are dealing with here.
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This would be our water tank and we have one hose that is going in and one hose that is going out.
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The hose that is going in, it can fill the tank in 5 hours.
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That means if we just leave it on every hour 1/5 of that tank will fill up.
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I know its rate is 1/5.
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If I look at emptying the tank it is a much smaller time to empty it, 1/3.
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If I did have them both open I would know that the second one
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would be able to empty the tank since it empties faster than the first one can fill it.
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Let us see what that we can do to set this one out.
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The first thing I want to consider was how much the tank is going to be empty every single hour.
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Starting with the 1/3 I know that every hour that passes by the 1/3 of it will be emptied out.
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x is the number of hours.
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The 1/5 coming in is not emptying the tank but it is actually filling it back up.
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We will say it is the opposite of emptying the tank by 1/5 and it will do that for every hour.
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We want to know is when will one tank be completely empty.
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We have a lot of similar components for this one and it often look like a lot of our work problems.
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All we have to do is go ahead and solve it.
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Our LCD here that would be 15.
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Let us multiply all parts by 15 and see what that does.
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15, 15 and 15.
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Canceling out some extra stuff here I get 5x - 3x = 15.
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Doing a little bit of combining on the left side I have 2x =15 or x = 15 ÷ 2 or 7 ½ hours.
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Notice how in this case it is taking longer since both of them are open.
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The reason for this is that they are not working with each other.
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They are working against each other.
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One was trying to fill the tank, and one is trying to empty the tank.
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Use those bits of information you can find along the word problem to give you a little bit of intuition about your final solution.
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In that way you can be assured that it does make sense in the context of the problem.
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