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 inequalities with absolute values.
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In our previous lessons, we did take care of inequality separately and we did take care of absolute value separately.
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Our goal for this one is to mix the two together and see if we can solve some new types of problems.
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When you have some number A greater than 0 and you have some algebraic expression x as packaged together in the following way.
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The absolute value of x < A, or the absolute value of x > A then you can actually split these into two different problems.
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Let us look at the first one, suppose that you have the absolute value of x < A, the way that we can put this is that the algebraic expression x > -A.
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The algebraic expression of x could be less than A.
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It is interesting to note that not only do we split it into two problems
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and we will see that we have picked up our connector and it actually connected the two.
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On the other side, when we are dealing with the absolute value of x > A.
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This will still split into two problems but now we have the x < -A or x > A.
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This is a little bit different and that now we have a different connector between the problems.
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We will definitely use our tools on how to deal with these connectors and solving both problems separately
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and figure out how their solutions should be connected in the end.
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One thing that you may be looking at and be a little bit worried is keeping track of all of your signs.
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It might start with less than and then these guys one is greater than and one is less than.
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These guys over here, I have greater than and then I have less than or greater than.
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Watch for my tips on how to deal with dealing with the absolute value and the absolute value separately so we can keep them both straight.
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To understand why it splits into all of these different cases and why the signs look so funny, you have to recall two things that we covered earlier.
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One has to do with those absolute values.
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When we are dealing with absolute values this will end up being split into two different problems.
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One that handles the positive possibilities, and one that handles the negative possibilities.
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That is why we have two different things.
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Also, you have to remember the rule that when you multiply or divide by negative that you will end up flipping your inequality sign.
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That is why the direction of the inequality ends up changing in both of our cases.
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Since one is that negative possibility, the sign ends up getting flipped.
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To help you keep these both straight this is what I recommend.
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When you are looking at your connectors, if you have the absolute value being less than a number then connect using and.
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If you have the absolute value greater than a number, then use or to connect the two.
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You will see that if you use each of these rules bit by bit, rather than trying to jumble them altogether,
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it is not too bad being able to pick apart one of these absolute values and inequality problems.
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Let us start out with some very small examples and work our way up.
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That way we can keep track of everything we need to do.
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This one simply says that the absolute value of x < 3.
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Here is what I'm going to do first.
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I’m going to use my techniques for our absolute values to simply just split this into two problems.
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Remember before that we have our positive possibility that x < 3 or we have our negative possibility -x < 3.
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I have not done anything with that inequality symbol, I have not even touched it yet I simply split it into two problems.
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Because we have the absolute value less than a number, I will connect these using the word and.
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It will become especially important when we get tour final solution.
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Now I just have to solve these separately.
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x < 3 is already done, -x < 3 I have to multiply both sides by -1.
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Since I’m multiplying by a negative, we need to flip our sign.
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Here I have two things that x must be a number that is less than 3 and x must be a number greater than 3.
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Since we are dealing with and, we are looking for all numbers that satisfied both of the possibilities.
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Let us look at our number line first to see if we can hunt down everything that does that.
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All numbers less than 3 would be on this side of 3 and we will put in a big open circle.
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They must be greater than -3 and I will put in another big open circle and -3.
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We will see that it shades in everything greater than that value.
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Our overall solution is all numbers between -3 and up to 3.
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Use your techniques for splitting up into two problems.
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Be careful on how you connect them and remember that you should flip the sign when you multiply or divide by negative.
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Let us try another small one, and again watch for this process to happen.
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This one says the absolute value of z = 5.
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The first thing I'm going to do is just split this into two problems.
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I’m doing this because I'm taking care of that absolute value.
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z > or = 5, -z > or = 5 I have not touched my inequality symbol I just took care of my two cases.
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Since I'm dealing with the absolute value being greater than or equal to a number over here and I will connect these using or.
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This will become especially important when we get to our solution and we just solve each of them separately.
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Z > or = 5 is already done and for the other one, I will multiply both sides by -1.
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Now if I multiply it by a negative, this will flip its sign.
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I have z > or = 5 or z < or = 5.
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I think we are ready to start packaging things up here.
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Since I’m dealing with the connection or, as long as it satisfies one of these possibilities I will include it in my final solution.
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All the numbers greater than or equal to 5 that would be all of these numbers here on the number line.
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All the ones less than or equal to 5, that would be these ones down here.
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Now that I know more about the number line, let us go ahead and write this using our interval notation.
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Negative infinity up to 5 brackets, because we are including from 5 to infinity, our little union symbol since it could be in either one of those intervals.
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Now that we have the process down a little bit better for being able to split this up, let us look at ones that are just a little bit more complicated.
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We want to solve the inequality involving our absolute value.
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I want make sure that the absolute value is isolated on one side of my inequality sign and fortunately this one is.
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Let us go ahead and split it into our two problems.
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On this side we will have 3x + 2 < 5.
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On the other side I have -3x + 2 < 5.
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I will take note as to the direction of my inequality symbol, our inequality less than number.
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Because of its direction I will connect these using and.
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A little bit more work, let us go ahead and solve each of these.
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For this one, I will subtract 2 from both sides and then divide both sides by 3 so x < 1.
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On the other side, let us go ahead and distribute our negative sign and let us add 2 to both sides.
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We can see that on this case I only did divide by -3 so because of that we will end up flipping our sign -7/3.
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I need all numbers that are greater than -7/3 and they are also less than a 1.
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On a number line I could see where -7/3 is, there is 0 and I can see where 1 is.
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I’m just looking for all numbers in between.
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Since those would be the only ones that are greater than -7/3 and also less than 1.
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We have our interval -7/3 up to 1 and that would be our solution.
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Let us get into another one. Solve the inequality using an absolute value.
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In this one, we do a little bit of work to isolate that absolute value first.
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Let us go ahead and add 1 to both sides, 5 - 2x > or = 1.
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Now that my absolute value is isolated, we will split it into two problems.
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This will handle our positive possibility, so no changes,
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And the other one will take care of what happens when the inside part could have been negative.
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Since I'm dealing with these absolute values and inequalities, let us look at the direction of everything.
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I have the absolute value less than or equal to number.
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This is that situation where we connect those using or.
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I have two problems that are connected using or, let us continue solving.
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On the left side, I will continue by subtracting 5 from both sides, then I can divide both sides by -2.
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Since I'm dividing by negative, let us flip our sign, x < or = 2.
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Onto the other side I'm going to distribute my negative sign in there first -5 + 2x > or = 1.
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Let us get that x all alone and we are just a little bit closer being all alone by adding 5 to both sides.
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And I will finally divide both sides by 2, x > or = 3.
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I have what I’m looking for all numbers that are less or equal to 2 and all numbers that are greater than 3.
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Let us see what that looks like on a number line.
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All numbers less than or equal to 2 would include the 2 and everything less than that, so we will shade that in.
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All numbers equal to 3 or greater than would be over here and so we have a big gap between 2 and 3 but that is okay.
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Let us graph this using our interval notation.
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From negative infinity up to 2 included and from 3 up to infinity and we are looking at the union of both of those two.
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That is not too bad as long as you isolate that absolute value first before splitting it into two problems, you should be fine.
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Let us look at one last one, and this highlights why you have to be careful on how you package up your solution using those connections and and or.
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In this one, we already have our absolute value isolated to one side 3 - 12x all in absolute value.
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I'm going to split this immediately into two problems.
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3 - 12x < or = -3 and 3 - 12x < or = -3 because of the direction of our inequality symbol.
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Let us go ahead and connect these using and.
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Solving the one on the left 3 - 12x < or = 3.
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Let us subtract 3 from both sides then we will divide both sides by -12.
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I’m flipping my signs because I'm dividing by negative.
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We can reduce this fraction as normal, just ½ so x > or = ½.
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Working with the other one, I will distribute through with my negative sign and I will add 3 to both sides.
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It looks like one last step and I can go ahead and divide both sides by 12, 0 ÷ 12 is simply 0.
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Let us see exactly what this would include.
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On a number line, I want to shade in all numbers that are greater than 1/2 and they must be less than 0.
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Of course, that poses a huge problem.
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There are lots of numbers greater than ½ and there is lots of numbers less than 0 but unfortunately I can not find numbers that satisfied both of them.
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There are no numbers that satisfied both conditions.
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In fact this one, I can say has no solution.
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Now, you might have caught that earlier and if you did that is okay, you can definitely save yourself a lot of work.
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If we look all the way back here at the original problem, we have a bit of an issue.
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Remember that the absolute value makes everything positive and sitting on the other side of our inequality is a negative number.
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We simply can not have something positive being less than something negative, that is not going to work.
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That is another good reason why this one has no solution.
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Be careful when going through these problems and make sure you split them up because of your absolute value.
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Watch the direction of that absolute value so you know how to connect them using, say, either, and or or .
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Remember to solve each of them separately and connect very carefully when you get to your solutions.
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