Last time we saw Grant, he had just gotten the opportunity to expand the area of his booth at the expo. Now it's 15 feet long and 20 feet wide. As excited as he is about this, we know Grant. He always wants more. He wants to make this booth even bigger. The person in charge of the expo is going to continue the regulation he had before which is that if Grant's going to increase his booth size, he has to do it by the same length in each direction. Let's call that length x, just like we did earlier. However, there's one catch. The expo authorities are adding another restriction to booth size. The area of a given booth is not allowed to exceed 750 square feet. Considering this new rule that's been put in place about area, how should we express the booth area? Are we going to need an equation? Are we going to need an inequality? Or do we need something else?
We need an inequality. The fact that the booth area is not allowed to exceed 750 square feet means that we can write an inequality that's something like this. Area is less than or equal to 750.
We know that we need an inequality, but what exactly is that inequality? Please think about the new side links of Grant's booth, once he expands, and then, write an inequality relating those to this area restriction. Don't forget area is equal to length times width.
Since we know that our area needs to be less than or equal to 750 square feet. We can just replace area on the left side of this inequality with the other way we know how to express it, length times width. And then, plugin those side lengths using these quantities down here. So the new length, of Grant's booth, will be 20 feet, the original 20 feet, plus x, the amount that he's adding on. So we have 20 plus x and then that's multiplying it by the new width, which will be 15 plus x. This quantity will be the new area, which we know must be less than or equal to 750. So here's our inequality.
Now that we have an inequality it'd be really, really great if we could help Grant figure out how much he could add on to his booth and still stay under this area requirement of 750 square feet. So if we wanted to solve for x in this inequality, what are we going to end up with? Is that going to give us one solution? Two solutions or an interval of possible solutions. What kind of answer do we get from a quadratic inequality?
We look at an interval of possible solutions, a whole range of different numbers that x could equal to satisfy this inequality. However, we don't know how to handle this yet. I guess we have some learning to do.
So to help Grant out, we're going to need to be able to solve this inequality. But this is unlike other inequalities we've looked at before, because this is a quadratic inequality. You can see that if we multiplied out the terms here, we'd end up with some term that has an x squared in it. So this isn't linear anymore. Even though, when we're dealing with inequalities, we can do many of the same operations that we can And when working with equation, we do need to be a little bit more careful. Do you remember which situation might at some points cause us to have to switch the direction of inequality. Please check off any of the situations where this might happen depending on what number we are using.
We need to make sure that we're super careful when we multiply and divide inequalities.
We know that there are certain situations when we need to flip the inequality sign. But when does that happen? Dividing or multiplying by what kind of number causes us to flip the inequality sign? Please pick the best choice down here.
We have to flip the inequality sign when we multiply or divide by a negative number. This is still going to be important as we move forward working with quadratic inequalities to.
Just a little check in point to see if you remember another super important guideline for algebra. What number should we never ever try to divide by, when we're solving equations and inequalities?
We should never try to divide by zero in algebra.
So let's pick a quadratic inequality to work with. How about x squared is greater than 16. That's pretty simple, it seems like something good to start our with. What I'd like you to start off by doing, is to modify this inequality so that the only thing on the right hand side is zero. Please write the new version of the inequality in this box.
To get zero on the right hand side, we just need to subtract 16 from both sides. We end up with x grade minus 16 is greater than zero.
Now that we've go this in nice quadratic form, please factor the left hand side of this inequality. Then write the full inequality in its new version in this box.
We've seen things like x squared minus 16 several times before. This is just taking the difference between two squared numbers. And in cases like that, our two factors are almost the same, except the signs, between the two terms are switched. So we have x minus 4 times x plus 4 is greater than 0.
Let's pretend for a second that instead of this inequality of x minus four times x plus four is greater than zero, that we had an equation instead. X minus four times x plus four is equal to zero. We saw before that the way we would treat this would be to look at the two different situations that would make the left side equal to 0. This would happen if either x minus 4 is equal to 0, or if x plus 4 is equal to 0. My question for you is then whether or not we should do a similar thing in dealing with this inequality. Should our next step in moving to a solution for x be to solve the two inequalities x minus 4 is greater than 0, and x plus 4 is greater than 0?
The answer is no. This should definitely not be your next step. Why, you might ask? Well, in order to get from this inequality to either one of these, we need to effectively divide both sides by the factor that we're not using in the final inequality we're looking at. Let's say that we wanted to get from this quadratic inequality to looking at this linear one, x plus 4 is greater than zero. Well to get x plus 4 by itself on the left side you would have to divide both sides of the inequality by the quantity x minus 4, however we don't know what x is equal to here and because of that we don't know what number we are dividing both sides by, that means this variable could be a negative number and if that's the case. We don't know which direction the final sign for x plus 4 is greater than 0 should actually point. So bottom line, this is not the approach we need to take. We need to do something different.
Even though we can't split this into two separate inequalities and solve, this inequality does tell us something. It tells us that the values of x 4 and negative 4 are really important. If we did consider each of these factors by itself an inequality these two points would be the end points of the solution intervals. So let's say then that we're going to use these two points to break the entire number lineup into three sections. One over here, one in the middle and one over here. Now that we have these three different regions in our number line, I'd like you to label each of those intervals with an inequality that tells us what values of x lie in that interval. Make sure that you exclude negative four and four.
Our three inequalities describing these three intervals, are x is less than negative 4, which is over here, the region to the left of negative 4. Negative 4 is less than x which is less than 4, which is this interval in the middle. And lastly on the far right of our number line, x is greater than 4.
Now that we know that these three intervals are important in some way, I would imagine that they each represent or don't represent solutions of x in different ways. What I would like to know is whether values within each reasons satisfy or don't satisfy the original inequality we had, x squared is greater than 16. So for each interval, start out by picking a value for x that lies in that interval. And then, secondly, check to see if that value satisfies this inequality.
Values of x that are less than negative 4 do satisfy this inequality. Values of x that lie between negative 4 and 4 do not and values of x that are greater than negative 5. We plug that in to the left side at this inequality. We get negative greater than 16. Between negative 4 and 4, my favorite number would be 0. 0 squared is just equal to 0. But 0 is less than 16. So the numbers in this interval don't satisfy this inequality. Lastly, over on the right, I'll pick 5.
Now that we've talked about our regions and whether or not our inequality is satisfied in each of them. Let's make a table to figure out the value of each of the individual factors within each region and how that affects the value of the entire quantity that we're interested in, in our inequality. Remember, you don't need to actually use the inequality we're talking about in this table itself, but I'll write it here just for reference. So in each column think about the quantity at the top. Either x minus 4x plus 4, or the product of those two and then for each of these intervals or values, this on the left tell me if quantity is positive, negative, or 0. So for example If I'm thinking about x minus 4 and I let x equal something greater then 4, then x minus 4 is going to be positive. So fill in the table just like that.
Here's our filled in table. I'd like you to think for a second about how you could figure out the signs that belong in the furthest right column just based on the two columns to the left of it.
From our table here we can see that the quantity x minus four times the quantity x plus four is positive, or greater than zero, only when x is either less than negative four or greater than positive four. So, these must be the intervals of values of x that satisfy our original inequality. Since we now know that x minus or x is greater than four. Can you tell me about the solution set to sort of a related inequality. X squared minus 16 is less than or equal to 0. Of course, you can use our table down here for reference. And I'd like you to write the answer using interval notation. Remember, that involves brackets that are either square or round. Two numbers inside and then comas in between those numbers, and of course, you can combine the brackets differently depending on what you need for the interval.
And our answer is negative 4 comma 4, with square brackets on both sides. We can find this out pretty easily using the table. We want x squared minus 16, which is, of course, the distributed form of x minus 4 times x plus 4. We want that quantity to, to be less than or equal to 0. They are basically interested in all of the rows in the table, for which this right column does not contain a plus sign. So that's these 3 in the middle right here. It means our interval goes from staring at negative 4 and including that, all the way to positive 4 and including that as well. Which is what we have written here.
We haven't really used graphs to help us solve inequalities but it is possible. Thinking about how we could graphically solve equations which of these graphs do you think will help us show the inequality x squared is greater than 16. And which one do you think will help us show x squared minus 16 is greater than 0? Fill in either a or b in each box.
Graph B can help us visualize the first inequality, and graph A can help us visualize the second one. Now, we're not totally there yet in terms of the actual inequalities themselves, but at least in both cases, we now have the curves and lines graph that we need in order to show these inequalities on the graphs. Remember that when we were dealing with solving equations graphically, you could look at both sides of an equation and set each one individually equal to y, and then graph the two equations that gave us. So in this first graph over here, which we know of course belongs to x squared minus 16 is greater than zero. I've graphed the parabola, y equals x squared minus 16, and I've graphed the line, y equals zero. Just the x-axis. Over here in coordinate plane b, I've graphed the parabola y equals x squared, and the line y equals 16.
We know that the solutions to these two inequalities will each be intervals of values of x. Which if we think about these graphs is going to correspond to selecting certain portions of each of these curves. The points that lie in those regions that are selected will have x coordinates that satisfy The corresponding inequality. So for each of these inequalities, I'd like to tell you whether solutions can be found above the purple line in either graph, or below it, or maybe both or it maybe neither, you pick. You pick.
For the first inequality, we are basically looking for y-values along the pink curve that are greater than the y-value of the purple curve, since the pink corresponds with the left side of the equation and the purple corresponds with the right side of the equation, I mean, when the left to be greater than right. So, the line y equals x squared minus 16 is above the line y equals 0, well, above the purple line. That means that the x -value of any point, on the pink line that's above the purple line, should satisfy this inequality. Lets check one just to be sure. Well these end points, these points of the intersection, are the points negative 4, 0 and 4, 0. So, that means that the x-values in this region above the purple line, x is less than negative 4 in this portion of the curve, and x is greater than 4 over here. And remember, that was the solution set we found for this inequality so that's perfect. Almost exactly the same thing happens in graph b. Again, solutions are found above the purple line since we want x squared to be greater than 16 and y equals 16 is the purple line. The points of intersection, in this case, are negative 4, 16 and 4, 16. Even though the y-values of those points are different, the x-values are exactly the same. In fact, this graph is simply the same as a, just shifted upward 16. To be clear, each inequality that we're looking at only has one variable in it. We only care about x-values. What these graphs help us do is portion out the coordinate planes that we can see which x-values do satisfy these equations. So, over here in graph a, any points in either the region to the left or this region to the right, we'll have x-values that satisfy this equation. The areas that aren't shaded in are areas where we can't guarantee whether or not the x-values will satisfy this, and we have a similar thing happening in graph b. I know that connecting graphs to inequalities in this way might seem a little bit confusing at first, but hopefully you at least thought that it was interesting.
We spent quite a while on the last inequality. So let's move on to a new one, x squared plus 2x minus 35 is less than or equal to 0. Remember that the way we started out with the last inequality was to factor the expression over here. So please factor the left side and then rewrite the entire inequality in this box.
When we factor this, we get the quantity x plus 7 times the quantity x minus 5 is less than or equal to 0.
Now that we've factored, I'm interested in, how this factoring splits up our number line. So were going to work toward completing this table. And I'ld like you to start off by telling me what intervals, this splits our number line into, and then also write in the critical values of x right here. At which the splits happen. For the purposes of defining these intervals in values of x, don't include these two endpoints inside these intervals.
The way that we factor it tells us that our two critical values, the two points that are splitting our number line up are x equals negative 7 and x equals 5. So I start by writing those. Then we just need to fill in the intervals surronding those values. We have x is less than negative seven then we have the region between negative 7 and 5. So negative seven is less than x which is less than five. And then lastly, x is greater than five.
Now we got our table all set up, its time to fill it in. For each box, say whether the quantity of the column it's in, becomes positive, negative or 0 in the specified interval.
Here's a filled-in version of our table. Notice again, how we can see that the signs of the quantities for x plus 7 and x minus 5 are related to the sign of their product. If we have a negative number times a negative number, it always gives us a positive number. Anything times 0 is 0. And if we have a negative and a positive multiplied together, we get a negative number.
Looking at our original inequality, and the signs of these different quantities that we've come up with for these various intervals, I'd like you to check off which of these intervals, or values of x, does seem to satisfy our original inequality. Then I'd like you to write the solution as an inequality in this box.
We want the quantity x squared plus 2x minus 35, which is equivalent to the product of x plus 7 and x minus 5. To be either less than 0 or equal to 0, which corresponds to these 3 squares right here. The intervals that go with those rows are x equals negative 7, negative 7 is less than x is less than 5 and x equals is less than or equal to 5, awesome. That means that any values of x from negative 7 to 5 including those 2 numbers, work to make this inequality true.
Now, I think we're at a point where we can really help Grant solve this inequality. We have the quantity 15 plus x times the quantity 20 plus x, less than or equal to 750 to describe how the expanded side links of his booth area will determine a total area that needs to be less than or equal to 750 square feet. What I'd like you to do know is do a little bit of algebra. To get this in a quality in a form so that we can have 0 on the right hand side and then an expression on the left hand side, so I would like you to start out actually by simplifying the expression on the left and then getting rid of the 750 on the right side, in other words, I'd like you to simplify the left-hand side of the inequality as much as you can.
The first thing we need to do in order to simplify this and modify the equation is to multiply out the product of these two binomials. That way, when we subtract 750 from both sides we can very easily combine like terms and simplify. You'll notice I've written the expression on the left in the standard form for polynomials but the highest degree term first. Great. So we have x squared plus
Now that we have our quadratic inequality written out very nicely, I'd like you to figure out what the critical values of x that are going to divide up our number line are, and then what intervals those critical values determine. Just a hint. You should probably start by factoring the expression on the left. Also when you write your intervals out at the end, I'd like you to write them as inequalities.
Factoring the expression on the left gives us x plus 45 plus x minus 10 times x minus 10 is less than or equal to 0. So our 2 critical values for x then that are going to split up our number line are negative 45 and 10. Our interval then, that we get from these critical numbers are x is less than or equal to negative and 10 is less than or equal to x. Here, we include the critical values in each of these intervals, because we have a less than or equal to sign, not just a less than sign. We know that if x equals either of these critical values, the expression will actually equal 0 and that means it definitely would satisfy the inequality. So we want to test these intervals. We want to include those critical values in each of them.
You're so close to being done. Now we come up with our three intervals for which we need to test a value to figure out what the solution set for x is. So test values from within each of these intervals and then write what you think the solution for x is as an inequality and then also as an interval.
if we pick a value within each of these intervals for only one that satisfies this inequality this value is from the middle interval. Negative 45 is less than or equal to x is less than equal to 10. We can write this using interval notation as negative 45, 10 with square brackets.
So we just solve this inequality to come up with an interval of solutions for x. Negative 45 is less than or equal is less than or equal to 10. So to satisfy this inequality x can take on any values from negative 45 to ten, including both of those values. But remember, we're not just dealing with any numbers here. We're talking about Grant's expo booth. Remember that x is equal to the length that Grant should add to the booth dimensions in each direction. So my question is, does this solution make complete sense in the context of Grant's expo booth?
The answer is no. Not completely. If x is suppose to be a length, it probably doesn't makes sense to add negative 45 feet, to a room length. Distances can't really be negative.
Thinking about what we talked about in the last quiz, what values can x have for Grant's story, then? Write this as an inequality. Remember, that this restriction, that we're going to place here, does not in any way, change the fact that, mathematically speaking, this is the correct answer, as a solution, for this inequality. We're just trying to take real life into account now.
Know that the upper limit of length that we can add to Grant's booth is 10, and we just need to change the lower limit. It can't go below 0. He could theoretically add nothing to the length of the booth. So Grant can chose to expand each side of his booth anywhere from 0 to 10 feet. Awesome.
We've been working with quadratic inequalities and clearly linear inequalities before that, but a lot of the techniques that we've learned to use can actually be used in a number of other kinds of situations too. Now we're going to take a second to talk about rational inequalities, which are things like this, for example. The quantity 2x minus 18 divided by the quantity x minus 4 is greater than or equal to 0. The reason this is called a rational inequality is because the left side over here is an example of a rational expression. Remember when we talked about rational numbers early Julie pointed out that the word ratio is inside the word rational. So while a rational number is just the ratio of 2 integers. We will remember the denominator is not allowed to be 0, and rational expression seems to be pretty closely related. It's just the ratio of two polynomials. And like with rational numbers, the denominator is not equal to zero.
Thinking back to what I just said about rational numbers and rational expressions, especially in relation to the denominator, what number is x not allowed to equal in this particular inequality? Again, this doesn't have to do with the actual solutions. Stating this answer doesn't require actually figuring out the solution to this question. I am merely asking you, right off the bat, what can we say that x is not allowed to equal?
The answer is 4. If x were equal to 4 then the denominator would become 0, and you can't divide by 0. So great, we've already narrowed down our set of choices for the value of x.
When we talked about quadratic inequalities, it referred to things called critical numbers. These are the numbers that divided up our number line into the different intervals that we could consider to figure out which ranges of values satisfied our inequalities. For quadratic inequalities, we got these numbers from factoring and depending on what rational expressions we have that maybe the case as well. But we can generalize about it. You remember that for quadratic inequalities, we found our critical values by figuring out which values of x would cause the quadratic expression to equal 0. And we basically have the same requirement for rational inequalities except that this time, we have one polynomial divided by another polynomial. So, you have two different sources of critical numbers. In either case, whether we're talking about the numerator or the denominator, we'll come up with a critical value for every value of x that will make either one of those 0. Considering this method for finding critical numbers, what critical numbers do you find for this inequality that we were just talking about? If you happen to find more than one critical number, then please separate them by commas.
We already said that the value of X that makes the denominator equal to 0, which is the first way that we can find a critical number was 4, so that will be one of them. The other way we can look for critical numbers is to set the numerator equal to 0 and solve for X. So if we set 2 X minus 18 equal to 0 we end up with X equals 9. So that must be our other one.
We just said that the critical numbers of this rational inequality are 4 and 9. So these two values are going to split the number line into several parts, and those intervals, and these values are things that we can test to figure out what values of x satisfy this inequality. So I'd like you to start out by doing something very similar to what we did for quadratic inequalities. And filling in this column over here with intervals and values for x, starting the left side of the number line and working right. As we did before with tables, I don't want you, in this case, to include critical points in these intervals right here. I want these to be noninclusive intervals. Oh, and one other note, I'd like you to write these intervals using inequality notation.
The furthest left region on the number line starts at negative infinity and goes up until 4. So in that region, all values of x are less than 4. The next important value we have is x equals 4. Then, we have the values of x that lie between 4 and 9. Next, we have x equal to 9. And lastly, the values of x that are greater than 9.
Now finally lets fill in this table. For each quantity up here please testify you at either in the interval in the left or the value itself if its inequality and then stay in each box whether the quantity is positive, negative or zero when that x value is put in. One other note, in this far right column if any of the values over here makes the denominator equal to 0, then I want you to write the word undefined in that box over here.
Here's the filled in version of the table. You'll notice that we did end up with one spot in this far right column where we had to write undefined and that was where x equals 4. We set x equal to 4, the denominator here becomes 0. And since we know that division is not allowed, it's not something we can interpret using our algebra skills, you have to write undefined in this box. It makes the entire fraction, the entire rational expression undefined. Whereas, when we were dealing with quadratic inequalities, we were multiplying two factors together. Now, we're dividing them. But what's interesting to note is that we have the same relationship between signs of the first two columns and the sign of the third column. If we're dividing two negative numbers, we get a positive number. If we're dividing a negative by a positive or the reverse, a positive by a negative, we get a negative number. And if we're dividing two positive numbers, we get a positive. Of course, zero is the exception, 0 divided by any number, regardless of positive or negative, is 0. And any number, whether positive or negative divided by 0, is going to give us an undefined answer.
Thinking about what kind of value we want our entire rational expression to take on, which of these intervals is going to satisfy our inequality? For each row, if the inequality corresponding to it does, then check it off.
This first one, x is less than 4, and these last two, x is equal to 9 and x is greater than 9, all satisfy our rational inequality.
All we have to do now is explicitly write down the solution for the values of x that satisfy this inequality. First, I'd like you to write the solution using inequalities to show the ranges of x and then, I'd like you to write it in interval notation. Be really careful about where on a number line these intervals fall, and think back to what we learned about compound inequalities. These use either, or, in-between two inequalities or and, where things with and can be written using the change notation, or something similar to this. Remember also how compound inequalities translate into interval notation. If you need to review any of this, you can always go back to a previous lesson to look it up and refresh your memory. Or you can go to the forums, they're always there to help you.
Looking at these intervals that we said satisfy our inequality, the first one tells us that x can be less than 4. That inequality is part of our solution. However there's this other interval down here, that also has some acceptable solutions. That's from x equals 9 and onward, so we can have x is less than 4, or x is greater than or equal to 9. So you can see already, how useful and really necessary using a table like this is, to come up with the solutions for rational inequalities. It's not even always easy to tell whether or not your intervals will contain or not contain the endpoints. So we just need to be really careful and write out all of our steps. In interval notation, we can express these intervals. The first one goes from negative infinity to 4, neither endpoint being included. And as united, we write that with this capital U with the interval 9 to infinity where 9 is included, and infinity, of course is not.
Since you did so well with the last rational inequality I gave you, here's another one. We have 18 minus x over 3x minus 6 is less than 5. This is great, except for one big issue. In order to find our critical points, we're going to need zero on one side of our inequality. So, first things first. Can you tell me what we need to do to get 0 on the right hand side over here? What do we need to do to both sides of the inequality to make this happen? In this box, I'd like you to fill in the proper operator and number that we need to do to both sides. So for example, you could add a number, you could subtract a number, you can multiply a number, or you could divide by a number.
We need to subtract 5 on both sides.
When we subtract 5 from both sides, we have a new inequality, or a new version of our old inequality. We have 18 minus X over 3 X minus 6 minus 5 is less than inequality is part of a single rational expression. In order to make this all one fraction we need to start by getting this extra term, the minus five, to have the same denominator as the first term. So, what would the numerator of an equivielent fraction that's equal to five be if the denominator is 3x minus 6? Remember, I've already written the sign out here, so we're going to subtract all of whatever you write in here.
Try a fraction that's equivalent to 5, but has a denominator of 3x minus 6. We need to multiply it by 1 in the form of 3x minus 6 divided by itself. To distribute multiplication by 5 to each term in the numerator, we get 15x minus
Now take one more step and write the entire left side of our inequality as a single rational expression. So combine these two fractions into a single fraction. Be sure to simplify the numerator of new fraction as much as you can.
The first thing we need to do is just distribute this negative sign to both terms in the numerator of our second fraction. Then since these fractions have the same denominator, we can write them as a single fraction. If 18 minus x minus 15x plus 30, then we combine like terms. That gives us a rational expression of negative 16x plus 48 over 3x minus 6. And of course this is still set as less than 0, since we haven't done anything to change both sides of the equation yet, that would affect the sign in the middle.
Now that we have a single fraction on one side of our inequality, and zero on the other side, we can finally start to solve for x. First of all, what are the critical numbers here? What does x need to equal to make the numerator zero? And what does it need to equal to make the denominator zero? These will be our two critical numbers.
Negative 16x plus 48 will be equal to 0 if x equals 3. And 3x minus 6 will equal
Our critical numbers tell us how to separate the number line into different regions or different intervals for us to test, using this inequality. So check each of those three regions that are separated by these critical numbers, and also check the critical numbers themselves in the inequality. And then figure out for which intervals values of x satisfy this inequality. I'd like you to write these intervals as inequalities, however.
I always need to use a table when I'm solving rational inequalities and quadratic ones for that matter, so here is my filled out table. Now I can just read off which of these regions contains x values that make this fraction or this rational expression negative. So that's just this first reqion, which is x is less than two, and the second region x is greater than 3. [inaudible] Since on a number line these two regions do not overlap, we have to use an or since no value of x can fall inside both regions. So the solution is, x is less than 2 or x is greater than 3. Let's take a second to think about how we could visualize this inequality using graphs. So here I've plotted 2 Equations. I plotted Y equals negative 16 X plus 48, over the quantity 3 minus 6, and the purple line is just the line Y equals 0. The X axis. We want to know according to our inequality when this line, the pink one, lies underneath the purple one. We want to know when the Y values of this curve drop below this curve. So we can see that, that's every part that's going to happen for all X values except for those in the very small portion of the curve that's up here. We can see that this matches the solution we came up with. This curve lies below this one, anywhere beyond this point, where x is equal to 3. And anywhere in this part of the graph. This line right here that represents the disconnect between these 2 portions of the pink graph. Goes straight through the .20. And, in fact, this is the line, x equals 2. We can see that the entire part of the pink graph that's to the left of this, or that has x values less than 2, is underneath the purple curve. So this fits what we found. Awesome.
Since we're on a roll with rational inequalities, let's do one more. We have 1 over x plus 2 is greater than 3 over 2x minus 4. You know that in order to find critical values, we need 0 on one side of the equation. So, that means we need to pick one side of the inequality to get all of our terms on. It doesn't matter which one we pick, right and left are both totally fine, and we'll get the same answer. But I'm kind of leaning to the left today, so for the first part of the problem, I'd like you to rewrite this equation so that all of the terms are on the left-hand side, making the right-hand side equal to 0.
To get 0 on the right-hand side, we just subtract the fraction we originally had there from both sides. So we end up with 1 over x plus 2, minus 3 over 2x minus because we're just subtracting, we're not multiplying or dividing by a negative.
Now we need to combine these two fractions into a single fraction. The first step to that is figuring out the simplest common denominator we can come up with for them, then make both fractions have this denominator, and then combine them. So I'd like you to do all of this and then enter your new single rational expression on the left hand side in this box.
The least common multiple of X plus 2 and 2x minus 4 is just their product, X plus 2 times X minus 4. In order to make our first fraction have that denominator, we need to multiply it by 2x minus over 2x minus 4. And we need to multiply by the second fraction to have achieve the same end goal by X plus 2 over X plus 2. Once I've written out what each of these fractions is With the same denominator, I can think about how to combine the numerators. We'll be able to just add the numerators together since they have the same denominator. So, when we do that, the numerators going to be 2X minus 4 minus 3 times X plus 2. So, I'll simplify that and then put it in the fraction, this is just going to save me some space. This simplifies to negative X minus 10. Or if I want to pull out a factor of negative 1, negative in parentheses x plus 10. Then I can write the negative sign outside of the entire fraction and put x plus 10 as the numerator.
Awesome. Now we have a very beautiful looking rational expression on the left hand side, and that's set in this inequality, with a 0 on the right hand side. Now we're in a position where we can find the critical numbers for this rational inequality. So think back to how we find critical numbers and what they mean. Remember they're the numbers that are going to split up our number line. And please type all the critical numbers that you can find in this box.
Are critical numbers. Are found when x equals negative 10, negative 2 or 2. Remember that these are the numbers that make ither the numerator equal 0 or the denominator equal 0.
So we found 3 critical numbers and we know that we're going to have to test each of these in an equality to see whether or not it's included in this solution set. But in addition to these 3 critical numbers, how many intervals are we going to need to test? How does this critical number makes up the number line?
If we plot these 3 critical numbers on our number line, you can see that they split the number line into 4 intervals. One to the left of negative 10, the one in between negative 10 and negative 2, the one between negative 2 and 2, and the one to the right of 2. So the answer is 4.
Finally I'd like you to solve for x. Please write your answer using interval notation.
So here, I have redrawn our number line, and for each region, I've picked out a value to test to see whether it satisfies our inequality. First, let's test x equals negative 11. Now, you might think that we need to actually plug negative sign of this entire expression. We just want to know whether or not it's positive. So, what I'm going to do is just think about whether or not each of the factors here, we have 3 of them, is positive or negative. And then, of course, I also need to take into account the negative sign that's on the outside. So, I'll write that one first. We have a negative sign on the outside. If I plug in negative 11 here, negative 11 plus 10 is a negative number, so we have negative and negative in the numerator. And then in the denominator, we have a negative number for this factor and a negative number for this factor. Four negatives multiplied or divided together, gives us a positive number. So, this region works. Now, let's do the same thing for negative 5. And going through the same sorts of steps for the next 3, it turns out that only this region, the first one you found in the area between negative 2 and 2, yield positive solutions. Let's check the 3 critical values this way also. Notice that when we plug in negative 10, we end up with 0 for the numerator. That means, that this entire fraction can never be greater than 0. So, negative 10 is not part of our solution set. Negative 2 and positive 2 each make the expression on the left-hand side contain a division by 0, so these cannot be part of our solution set either. Great. So, we have these two intervals to take into account. We have the interval negative infinity to negative 10, united with the interval negative 2 to 2. Awesome. That's our solution. Great job with some really tough work on quadratic inequalities and rational inequalities. I know there are a lot of steps in all these problems, but I helped you develop a little bit of independence in figuring out how to solve these kinds of inequalities. They're pretty cool, too. I'm going to do one more quick thing and show you graphically what this inequality might look like.
As one final explanation for this lesson, let's see if there's a different inequality we can look at that can give us this same solution set. I'm just quickly going to rewrite the original inequality that we have, and see if I can modify it to simplify it a little bit. One thing I'd like you to notice, is that this negative sign outside of the entire expression, is going to effectively multiply either both terms in the numerator by negative 1 or multiply the entire denominator by negative 1. So, we can consider this a common factor of, for example, both terms in the numerator. There's one more common factor within one of the quantities here, and that's 2, in the quantity 2x minus 4. Both 2x and negative 4 are divisible by 2. Let's rewrite this. Because multiplication is commutative, I can move these factors around however I want. And I think what I'd like to do is to write this factor of negative 1 half, as a single constant outside of the rest of the terms multiplying that fraction. Now, we can very clearly see that there's this factor out here, this constant factor, that we can just divide both sides by to get rid of. However, since we will be dividing by a fraction, we can equivalently just multiply it by its reciprocal or multiply by negative 2, on both sides, of course. Now, here's the kicker. Remember, that since we're multiplying by a negative number on both sides, we need to switch the direction of the inequality. So now, we have this new inequality, x plus 10 over the quantity x plus 2 times the quantity x minus 2 is less than 0. And in fact, if you went through the entire process of solving for x here, you'd find that the solution set for this inequality is exactly the same as the solution set for this inequality up here. And we can tell why that's the case. These are equivalent. We move from this one to this one using completely legal algebraic modifications. Let's see how graphs can help us understand this inequality a little bit better.
So now as you might have guessed I've graphed the lines y equals x plus 10 over the quantity x plus 2 times the quantity x minus 2, and of course our x axis. What are we interested in finding now? Well because we have a less than 0 here we want to see where the pink curve lies below the purple curve or below the X axis. This region down here, this sort of upside down u definitely does. And as you would expect, the x values here lie between negative 2 and 2, although this curve will never actually have points with either of these two x values. We'll talk about that more in a few lessons. Get excited, it's pretty cool stuff. For the other part of our solution Where x lies between negative infinity and negative ten. We once again, have a bit of a hard time telling that the curve drops below the x axis here. But of course, you could plug in values for x that are less than negative ten, and test this out yourself. I think this is pretty cool though. You can use multiple graphs to demonstrate the same solution sets. And we can see how 2 inequalities look completely different, or at least pretty different to me, yield the exact same solutions.