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 go ahead and take care of rational inequalities.
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There are a few different techniques that you could use for solving rational inequalities,
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but I’m just going to focus on looking at a table and keeping track of the sign for these.
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Recall that when we are trying to solve an inequality that involves a polynomial, we want it in relation to 0.
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The reason why we are doing that is we just have to focus on whether it is positive or negative, a lot less to handle.
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This will make looking at the intervals that represent our solution a little bit easier to find.
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You will see that the process for solving these rational inequalities looks a lot like
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the process for solving our polynomials that we covered earlier.
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We will still look at factoring it down, figuring out where each of those factors are equal to 0
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and looking at tables so we can test the values around each of those 0.
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The actual process for solving a rational inequality looks like this.
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The very first thing that we are going to is we are going to set the inequality in relation to 0 on one side.
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It means get everything shifted over, so you have that 0 sitting over there.
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Then we will go ahead and solve the related rational equation.
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Solve the same thing and put an equal sign in there and see when it is equal to 0.
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We will also figure out where the denominator can be equal to 0.
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Those are some of our restricted values that we cannot use.
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The reason why it is important to solve it and find out where the bottom is 0 is at those points it could change sign.
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We will call those particular points our critical values.
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It is around those points that it could change in signs, we are interested in what happens.
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To determine what it does around those we will use a few test points
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that will help us to determine which individual interval satisfy the over all inequality.
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Until we get to finding our intervals, we are not quite done yet.
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We also have to pay close attention to the end points to see which ones should be included and which ones should not be included.
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I will give you a few tips on that, watch for that.
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Here are my tips.
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Remember that when you are working with these rational inequalities and get everything over onto one side,
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we might combine it into one large rational expression.
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If we got 3 or 4 of them, put them all down into one.
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Make sure you factor both the top and the bottom.
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You need to see where each of the factors is equal to 0.
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A lot of factoring.
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You will never include any values that make the denominator 0.
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We cannot divide by 0 even if it has an or equal in there, never include anything that makes the bottom 0.
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What that said, if we are dealing with a strict inequality like greater than or less than
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then we will not include any points on the end points.
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If we have greater than or equal to, less than or equal to, we will include where the numerator or the top will equal 0.
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Those are the only end points we have to worry about including.
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That is a lot of information, let us jump in and look at some of these examples.
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I want to solve when -1/x + 1 is greater than or equal to -2/x - 1.
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Rather than worrying about clearing up fractions or anything like that,
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let us get everything over onto one side and get it in relation to 0.
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I'm going to add 2/x - 1 to both sides.
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My goal here is to combine these fractions into one large fraction.
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Do not attempt to clear out these fractions like you would with a rational equation.
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If you clear out the fractions you will lose information about the denominators
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and we want to check around points where the bottoms could be 0.
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Do not clear those out.
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My LCD that I will need to use will be (x + 1)( x – 1)
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Let us give that to each of our fractions.
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-1 will give this one x -1.
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We will do that on the bottom and on the top then we will give the other one x +1.
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Let us combine this into a single fraction here.
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We would have to do a little bit of distributing so - x + 1 + 2x + 2 / x + 1 x - 1 greater than or equal to 0.
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Just a little bit more combining on the top that is a - x + 2 would be a single x and then 1 + 2 would be 3.
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It is quite a bit of work but we have condensed it down into a single rational expression on the left there and we also have it in relation to 0.
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That is a good thing.
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I want to figure out where would this thing be equal 0?
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It will equal 0 anywhere in the top would equal to 0.
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The top is equal to 0 at x = -3.
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We also want to check where would the denominator be 0?
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The bottom equal 0 at two spots when x =- 1, 1x =1.
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All three of these values are what I call our critical values,
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and it is around these values that we need to check the sign of our rational expression here.
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That way we can see whether it is positive or negative.
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Just like before when we are dealing with those polynomial inequalities, this is where our table come into play.
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First start out by drawing a number line and putting these values on a number line.
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I want to start with the smallest ones so -3 and then we will work our way up from there - 1 and 1.
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Along one side of this table we will go ahead and we will write down the factors that our rational inequality here.
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I have x + 3 x +1 and x – 1.
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Now comes the part where we simply grab a test values and see what it is doing around these.
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Our first test value we need to pick something that is less than -3.
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Let us choose -4, that is on the correct side and we will put it into all of our factors to see what sign they have.
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-4 + 3 = -4 + 1 still - and -4 -1=-5, negative.
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We will select a different test value between -1 and -3.
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-2 will work out just great and we will put it into all of our factors.
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Let us see what that does.
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-2 + 3 that would give us something positive, -2 + 1 that will be negative and -2 – 1, negative.
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We are doing pretty good and now we need a test value between -1 and 1.
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I think 0 is a good candidate for this.
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That will be a nice one to test out.
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0 + 3 = 3, 0 + 1 =1, 0-1=-1.
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One last test value we need something larger than 1.
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I will put 2 into all of these.
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2 + 3= 5, 2 + 1 =3, 2 -1 =1
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My chart here is keeping track of the individual sign of all the factors.
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We will look at this column by column, so we can see how they all package up for our original rational expression here.
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In this first column I have a negative and it is being divided by a couple of other negative parts.
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What I’m thinking of visually in my mind here is that this will look a lot like the following.
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I have a negative value on top and a couple of negative values in the bottom.
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When those negative values in the bottom combine they will give us another positive value.
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We have a negative divided by a positive.
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That means the overall result of that first interval is going to be negative.
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In the first interval it is a negative.
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For the next one I will have a positive ÷ negative × negative and that will end up positive.
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Then I will have a positive ÷ positive × negative = negative.
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My last I will have positive ÷ couple of other positive values, everything in there is positive.
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I know what my rational expression is doing on each of the intervals.
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I know when it is positive, I know when it is negative and things are looking pretty good.
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Looking at our rational inequality over here I can see that I'm interested in the values that are greater than or equal to 0.
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That means I want to figure out what are the positive intervals.
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In this chart that we have been keeping track of all that so I can actually see where it is positive.
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I have between -3 and - 1.
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I have from 1, all the way up to infinity.
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Both of those would be some positive values.
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Let us write down those intervals.
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I'm looking at between -3 and -1 and from 1 all the way up to infinity.
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There is one last thing that we need to be careful of what endpoints should I include, which endpoints should I not?
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Notice how we are dealing with or equals to.
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I want to include places where the top of my rational expression could have been 0.
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It is a good thing I highlighted it early on, it fact they are right here.
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I know that the top will be equals 0 at -3.
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I'm going to include that in my intervals.
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We never include spots where the bottom is 0.
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We have marked those out and I will use parentheses to show that those should not be included.
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Let us go ahead and put our little union symbol so we can connect those two intervals.
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This interval from -3 all the way up to -1 and -1 to infinity
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that would be our interval that represents the solution for the rational inequality.
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Let us try one that looks fairly small, but actually has a lot involved in it.
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This one is x -5 / x -10 is less than or equal to 3.
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Like before let us get everything over on to one side first.
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That way it is in relation to 0.
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x -5 / x -10 -3 is less than or equal to 0.
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We need to worry about getting that common denominator.
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I see I have x -10 in the denominator, let us give that x -10 to the top and bottom of our 3.
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Looking much better from now, we will go ahead and combine these rational expressions right here.
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I think we need to do a little bit of work, let us go ahead and distribute this -3, as long as we have it there.
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x -5 - 3x + 30 / x – 10 you want to know where is that less than or equal to 0.
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Just a few things that we can combine we will go ahead and do so.
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-2x from combining our x and let us see a 25 will be from combining 5 and 30.
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We want to know where is that less than or equal to 0.
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We are in good shape so far.
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Now that we have crunched it down into one rational expression, we are interested in where is it equal to 0.
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This will be equal to 0 wherever the top is equal to 0.
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We are going to do a little bit of work with this one, but we can figure out where the top is equal to 0.
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We will have to move the 25 to the other side and divide both sides by -2.
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I have 25/2.
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Let us make a little note is where the top is equal to 0.
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Okay, that looks pretty good.
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Let us figure out where the bottom equal 0, that is not so bad that will be x = 10.
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It is around those two values that we will go ahead and check to see what the sign is, around 10 and 25/2.
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Let us go to our table here.
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The smaller of these values would actually be our 10 and 25/2.
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We will put that over there.
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Let us check the sign and see how these turnouts.
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- 2x + 25 and the other one would be x – 10.
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Let us grab some sort of test value that is less than 10.
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One thing I can see here is I will plug in a 0 that is less than 10.
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If I plug it in the first one I will get 25.
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If I plug in 0 for the other one I will get -10.
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That is my sign for that one.
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On to the middle interval I need something between 10 and 25/2
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I think 11 will work out just fine.
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If I put 11 into that top one I will get -22 + 25 that will be positive.
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If I put 11 into the bottom one, 11 - 10 would be 1.
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Okay, interesting.
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I need to pick something larger than 25 / 2.
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25/2 is about 12 ½ let us go ahead and test out something like 20.
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If I put 20 into the top one I will have -2 × 20 that will be -40 + 25 that will give me a negative value.
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If I put 20 into the bottom one it will be 20 – 10 and that will be a 10.
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These two things are being divided, to look at our over all sign we will take each of these signs and go ahead and divide them.
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Positive ÷ negative = negative, positive ÷ positive = positive and negative ÷ positive = negative.
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What am I interested in for this particular one I want to know where it will be less than or equal to 0.
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Let us look for those negative values.
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I can see one interval here everything less than 10 and I have another interval here where it is greater than 25/2.
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Let us write down those intervals.
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From negative infinity up to 10, from 25/2 all the way up to the other infinity.
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One last thing, we need to figure out what end points we should include and which ones we should not include.
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This one does have or equals 2 so I will include the places where the top is equal to 0.
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We already marked those out, the top is equal to 0 at 25/2 so we will include it.
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We never include spots where the bottom is equal to 0, so I will not include the 10.
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We will finish off by putting a little union symbol in between.
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The intervals that represent the solution are from negative infinity up to 10 and from 25/2 all the way up to infinity.
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And then we can consider this inequality solved.
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It is a lengthy process, but we will use that table to help you organize your information
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and watch at what point you can and cannot include in the solution intervals.
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