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 work on solving rational equations.
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Two things that we will look at is I can solve a equation that contains a rational expression.
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I will throw in an example or we deal with a formula that also involves a rational expression.
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Recall that when you are working with equations that have a lot of fractions in them we will often use the least common denominator
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to go ahead and clear out all of those fractions.
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These particular types of equations you definitely want to make sure that you check your solutions.
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This will be especially important for these types of equations this is because we are dealing with expressions that involve fractions,
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we will have some restricted values, values that we cannot have.
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Some of these values may try and show up as our solution.
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We simply have to get rid of them.
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They are not valid and they will make our denominator 0.
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When working with these expressions that have a lot of fractions we like to keep the least common denominator throughout the entire process
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or that way we can go back in and make sure that makes sense in the original.
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Let us go ahead.
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When you see my example for formulas here again and know that the same process plays out.
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We like to use that least common denominator to go ahead and try and clear things out
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Of course, our goal is to isolate which ever variable we are trying to solve for.
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When we are all done with that formula we still have an equation of these words
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and for more variables still floating around in there.
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That is okay as long as we get that variable we are looking for completely isolated.
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That is how we will know it is solved.
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Let us take a look at one of these.
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I have 1-2 / x + 1 = 2x / x +1 and we can see that this is a rational equation because looking at all these rational expressions I have,
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think of those fractions.
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To help out I’m going to work on getting a common denominator.
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Already these 2 parts of it have an x + 1 in the bottom.
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We are just going to focus on trying to give an x + 1 to the one over there on the left.
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I can do that by taking 1 / 1 and then multiplying the top and bottom of that by x + 1.
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(x + 1)(x +1) - 2 / (x + 1) = 2x /(x +1)
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On the left side that allows us to finally combine things now that we have a common denominator.
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In doing so, we just put the top together.
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x + 1 - 2 / x +1
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Doing a little bit of simplifying let us go ahead and subtract 2 from 1, x - 1 now x + 1 = 2x / x +1.
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At this stage notice how we have two fractions that are set equal to one another and the bottoms are exactly the same.
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Since the bottoms are the same then I know that the tops of these must also be the same.
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Let us just focus on the top part for a little bit and see if we can find a solution out of that.
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x - 1 = 2x this one is not so bad to solve.
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Let us just go ahead and subtract an x from both sides and I think we will be able to get a possible solution.
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It looks like a possible solution is that -1 = x.
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Be careful this can sometimes happen with these rational equations.
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It looks like we have done all of our steps correctly and it looks like we have found a solution, but this guy does not work.
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To see why it does not work, take this possible solution all the way back to one of our original denominators
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and you can see that if you try and plug it in and it makes the bottom 0.
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We have to get rid of that possible solution.
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Since we do not have any other possibilities left, we can say that this particular equation has no solution.
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With these types of equations, even when you find something that looks like it is a solution you have to go back
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and check to make sure that it make sense in the original.
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Oftentimes you may end up getting rid of things that will simply make that denominator 0.
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Let us try a different one.
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This one is 2 /(p² - 2p) = 3 /(p² – p)
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Before working with this, I need to see what is in the denominator.
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To get a better idea we are going to go ahead and factor.
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In this side I can see that they both have a p in common, we will go ahead and take that out.
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I get p × p -2.
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On the right side here, it looks like it has a p in common as well, p × p -1.
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They almost have the same denominator, but I need to essentially give each one its missing factor.
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Let us give the left side p -1 and we will give the right side p -2.
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Just copy this down and see where the extra pieces are coming from.
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On the left I will put in p -1 on the bottom and on the top.
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The one on the right we will give p - 2 in the bottom and p -2 on top.
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Distributing this will give us (2p -2) / and now we have this nice common denominator equals (3p – 6) / (p) (p -1) (p -2)
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This is right back to that situation we had before.
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The bottoms of each of these fractions are exactly the same I know that the tops of the fractions will also end up being the same.
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Let us extract out just the tops and focus on those.
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If we have to solve this what can we do?
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I will go ahead and add 2 to both sides 2p = 3p - 4 and let us go ahead and subtract 3p from both sides, -p = -4.
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One last step, let us go ahead and multiply both sides by -1.
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My possible solution is that p = 4.
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We cannot necessarily just assume that that is the solution, not until we go back to the original
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and make sure that it is not going to give us 0 in the bottom.
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Let us check to see what factors are in the bottom.
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Some of our restricted values that we simply cannot have is 1p = 0, we cannot have that one.
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We also cannot have p = 2 and we cannot have p = 1.
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Those three values are restricted.
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Fortunately when we look down at what we got, it is not any of these restricted values.
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We will go ahead and keep it as our solution.
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The solution to this one is p = 4.
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Let us look at a little bit large one, one that has a few more things in the denominator.
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This one is 1 / x -2 + 1/5 =2/5 × (x² – 4)
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Quite a bit going on the bottom and it looks like we definitely need to factor one of our pieces
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just so we can see what total factors we have in here.
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1/x – 2 + 1/5 = 2/ I have 5 and then the x² – 4 that happens to be the difference of squares.
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(x + 2)(x – 2)
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It looks like this will be our LCD and we are going to give it to these other fractions here with missing pieces.
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The 1/x – 2 and see what we can put into that one.
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It needs to have a 5, we will put that down below and up top and it also needs to have an x + 2 and below and up top.
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Onto the next fraction here, this is 1/5 and it looks like it is missing an x - 2 piece so we will throw that in there.
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We are missing x + 2 missing both of those pieces.
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On the other side of our equal sign it already has the LCD, so we just leave it just as it is.
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We spread out quite a bit but now we are going to focus on the tops of everything since all the bottoms are now exactly the same.
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We will also use our distribution property a little bit to help us out by multiplying 5 × 2
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and we will take the two binomials over here and we will foil them so we can put those together.
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5x + 10 would be that first piece on the top of the first fraction.
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We have our first terms x², outside terms 2x, inside terms – 2x and last terms -4.
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We want that all equal to 2.
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You can notice this is just the tops of all of our fractions.
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Let us go ahead and combine some things.
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My + 2x and my - 2x will go ahead and cancel each other out and we will have x², I will take care of that one.
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5x took care of that one and combining the 10 and -4, +6 = 2.
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We are almost there. Let us just go ahead and subtract 2 from both sides and see that we need to solve a quadratic.
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We are going to solve x² + 5x + 4 is all equal to 0.
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This quadratic is not too bad I can end up factoring it using reverse foil without too much problem.
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Our first terms x and x and two terms that need to multiply to give us 4 but add to give us 5.
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I am thinking of 1 and 4, those will do it, both positive.
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This will give us two possible solutions.
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When I take x + 1 = 0 and x + 4 = 0, x could equal -1 or x could equal -4.
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There are two possible things that I'm worried about.
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Before we put our seal of approval on them,
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let us go back to some of those original fractions and see what some of our restricted values are.
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Looking at this first one, the bottom would be 0 when x is equal to 2 so I know one of my restricted values x can not equal 2.
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The bottom of this one is a number 5, so that one is not is not going to be 0.
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Over here I can see that when x =-2 I will run into a 0 on the bottom, x cannot equal -2.
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This factor here is exactly the same as the first one.
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We already have that list as one of our restricted values.
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As long as I know my possible solutions are 2 and -2, it looks like we are okay.
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Since I'm looking at -1 and -4, both of these are going to be valid solutions, none of them are restricted.
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The process of solving these rational equations you just have to work on combining them and focus on the tops for a bit.
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Let us look at that formula as I promised.
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To see how you could solve the formula involving some these rational expressions.
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To make things interesting, let us go ahead and solve for y.
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In order to get rid of a lot of these things in the bottom we will identify our lowest common denominator.
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One has an x another one has y and the other one has z.
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It looks like we will need all three of those parts, x, y and z in our new denominator.
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We will go through and give each of them their missing pieces.
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Here we have our originals, let us throw in the missing.
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The first one it already has an x in the bottom, we will give it y and z on the top and the bottom.
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The next one already has a y, so give it the x and z.
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For the last one we have z so xy.
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This will ensure that all of the bottoms are exactly the same.
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We will go ahead and turn your attention to just the tops of all of these and we will continue moving forward.
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2yz = 1 × x × z that will be just xz + xy.
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We are trying to get these y isolated all by himself and looks like we have two of them, one on the left side and one on the right side.
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Let us get them on the same side by subtracting an xy from both sides.
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2yz – xy
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We can factor out that common y that way we will have 1y to deal with.
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Let us go ahead and factor out the common y out front.
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y and then left over we will have 2z – x.
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We are almost free with this one.
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To get the y completely isolated, let us finally divide both sides by that 2z – x.
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y = xz / 2z - x and this one is solved because we have completely isolated the variable we are looking for.
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In a sense we have created a new type of formula.
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Be careful in solving many of these different types of rational equations.
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Make sure you always check your solutions and use that least common denominator to help yourself out.
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