College Algebra - Lecture15 - Equations in One Variable

welcome to unit 3 now we're going to look at as you see on the screen the equations and inequalities we'll take all the information we learned from functions and their graphs and everything else that's come previously in this course and we'll put it all into the service of equations and inequalities of one variable we'll stick to one variable because that's all we need to do in college algebra so I guess we'll go ahead and get started as we begin this unit and I you see that I have a carefully marked equations in one variable the first thing we want to look at is what your graphing device can do for you and what makes it theoretically possible for that to be true so the first topic will look at is solving equations approximately with the graphing device and the intermediate value theorem now that would be the theoretical underpinning for what the graphing devices allows us to do so on that note let's go ahead and look at solving equations approximately the first thing I want to do is make a few points before we get into complete detail I want to make some general points again the equations I said this before but let me say it again it's worth mentioning the equations are in one variable only and as we saw even in the last unit whenever you end up with an equation with more than one variable you need to eliminate one of the variables in order to proceed secondly if you can solve an equation exactly if you can solve and the equation exactly well do so by all means try and solve it exactly however if you cannot or if it turns out that solving exactly would be tremendously difficult and we'll see examples of that later you want to try and use the graphing device to get to an approximate kind of solution so if not a graphing device any graphing device whatever can lead to an approximate solution and frankly in some cases the only solution you can get is an approximate solution whether you use a graphing device or some other numerical technique so there are two types of solutions you can get approximate solutions and exact solutions we try for exactness if we can't get exactness we go for approximation and you want to be clear on how each one works now since we'll be doing approximations in this course we ought to have an agreement on how approximate we want to be so for the sake of this course here is our agreement we'll call this agreement on approximations now sometimes we'll alter this but we'll say that specifically at that time this agreement is in all other cases where it isn't mentioned so when we approximate answers to problems when we approximate it will be unless of course as I said we make a specific change when we approximate it will be correct and what I'm going to mean by that is truncated I'll explain that in a moment correct to two decimal places now two decimal places is fine for us it gives us enough accuracy for everything that we're going to do if you're in other areas you may want more decimal places but if you can do two you can do as many as you want so this is a good round number for us to work with now what does this mean and in particular what does this truncated mean if you don't know well here's what I mean by this meaning okay if say the exact answer to whatever it is you're working on the exact answer is say two point six five eight one nine then what do I mean by correct to two decimal places well correct or as I said above truncated and you'll see now why that is the correct word truncated to two decimal places is two point six five what I have done is I have truncated truncate means cut off I have cut this decimal off at the second place I have not done any rounding all I've done is cut it off that is our agreement all right that's what we will do in this course we will simply chop it off at the second decimal place while just to give you a an example of what happens when you don't truncate if you had if we had rounded up say to two decimal places that is 2.66 now why is that well two point six five the next number is eight so five eight is closer to six Oh than it is to five oh so we replace this with a six this would be rounded upward we are not going to do that in this course so we won't be rounding when we talk about finding something correct to two decimal places will simply mean truncation okay that's just a matter of convention for this course and now how would you actually get that degree of accuracy on your graphing device well here's a nice way to see how you can sometimes it takes a little bit of work in fact it's usually better to try to find the exact answer and then approximate that with the calculator but if you're trying to approximate from a graph which may be the only way you can go here's a way to think about it let's see I've made a little table here for accuracy to these decimal places : I'm gonna go all the way across here what we want to do is then once you know what decimal place you want you want to choose the appropriate X scale usually abbreviated that way on the calculators all right if you want it approximately to zero decimal places just to the nearest whole number approximation you choose X scale equal 1 now Y where does the one come from well let me give you the background this is 1 over 10 to the 0 remember anything to the zero power is one name I think this is a funny way to write one but it'll be the pattern that follows down below so you choose X scale equal one what does that mean that means when you look at your graph the scale will mark off and have tick marks only at the whole numbers in both directions okay going off to infinity there if you want accuracy to one decimal place that is to say you want to choose x scale to be 0.1 the reason for that is that's 1 over 10 to the 1 notice that the 0 here in the power 0 here match the 1 here and the power 1 here match that's where these come from so you choose the X scale to be 1/10 so when you look at your calculator the tick marks whether they're marked or not will stand for 0 1/10 2/10 etc and also in the other direction so what's happened is the 0 to 1 or for example well part of this has have been expanded 0 to 1 now has expanded out that way you're now going to a tenth of the size you were before gives you more accuracy once more for accuracy to two decimal places two decimal places remember is what we're going to do in this course you would have an X scale of 0 one point zero one and of course that's the same thing as 1 over 10 squared 1/100 and when you look at your scale on your calculator it'll go from 0 to 0.01 to 0.02 etc so it's like the tenths here have been expanded and now you're down to the point where if you know that your answer is in here say you know that it is point zero one something and if we're truncating then point zero one is the correct truncation so now this sometimes requires a triple process where you look at the graph with scale one then you zoom in and look at scale one tenth and then one one hundred to finally get to the point where you can estimate from the graph the answer to two decimal places that have been truncated now that can be a little tedious but if there is no other way to do it it's Creole II quite fast okay now that we've said a few words about approximation let's get back to the whole idea of solving equations this is something that we actually mentioned back when we were doing basics but I'm going to say it in a slightly different form now since I have functional notation fact all equations I'm abbreviating again as I always do all equations in one variable say just for the record here X can be written as follows can be written f of X equals zero in other words to make completely clear what's on the Left all nonzero terms everything all nonzero terms on the left and what's on the right zero on the right now you can always do this you can push everything to the left and leave zero on the right now with that in mind I can write down certain facts about equations assuming that they are in this standard form so the first fact I'd like to talk about is how do you solve this in other words how do you find what X values make this zero well let's talk about that solving f of X equals zero means well in this course by now you've caught on that there are two ways to look at almost everything one is the algebraic way and what is the graphical way algebraically what would solve this mean well it's what I said a moment ago it is finding X values well what's the next value anyway it's a number finding X values which make the statement f of X equals zero true so if you put in these X values for X you get something that's true you get zero equals zero is what you get graphically what is the picture that goes along with this that we want to get used to the picture is solving f of X equals zero means pictorially finding x-intercepts now remember what those are those are points of the form X comma zero because they intersect the x axis their y-value is nothing it's zero finding x-intercepts on the graph of the function y equals f of X so what we've done in this case is if we reinterpreted f of x equals zero to mean take the function y equals f of x set it equal to zero find out where the y-value is zero those are also called x-intercept points so graphically finding the x-values which make the function makes make the equation true is the same thing as finding the x-intercepts on the graph of y equals f of X and this is the key to linking these two the algebraic and the graphical ways of seeing things so to fix that in your mind let's do a couple of examples we'll take them very simply because we really haven't talked about solving such equations yet but I think these are things that you've seen before they're simple enough that we can figure out what's going on without too much extra technique solve 2x equals 6 very easy looking problem let me go ahead and break the page up into two parts let me do the algebraic part first and I have a couple of points to make here as we go to algebraically well your first instinct is probably the following write down 2x equals 6 divide both sides by 2x equals 3 end of story and that's perfectly fine however let me show you another way to think about this and I'm gonna say it's better now it's not going to be better on this problem it's not going to be simpler than what you've just done there isn't anything simpler than this however in problems that get more complicated this better way will save you from losing solutions and making other sorts of errors the better way to think about this is to try and get back into that f of X equals zero form we talked about a moment ago so if I started with 2x equals 6 and I rewrote it as 2x minus 6 equals zero pushing everything over to the left then I have at this stage that f of X equals zero form now how can I solve this well I notice on the left that 2 is common to 2x + 6 so i factor that out 2 times X minus 3 is 0 then when I look at this I say the only way this product can be 0 since these are numbers is if one or both of the numbers is 0 2 can't be 0 X minus 3 must therefore be 0 that you may recall we called something like the zero product principle or the zero product fact and if X minus 3 equals 0 of course x equals 3 we end up with the same answer of course but we've followed a procedure putting everything to the left and then using that zero product principle which will hold us in good stead later okay algebraically we found that the only solution here is 3 so let's see what happens when we look at this graphically graphically or again I'll write down two x equals six and again I will push the six to the left so I have 2x minus 6 equals zero except now I'm going to label the left hand side I'm going to ahead and go ahead and call it Y so this becomes an equation but it's also a function we can think of it as y equals 2x minus 6 this is an f of X if you like but we've seen these before this is a line it's the equation of a line which we talked about earlier it is also in that f of X equals zero form here and what we want to know this is a line and this equaling zero we want to know where this line crosses the x axis so if we look at a picture and I've gone ahead and worked this out so I won't pull out the calculator for this 1 2 3 here and then 1 2 3 4 5 6 so minus 6 is where it crosses here and 3 is where it crosses here and there is the line y equals 2x minus 6 the X and the y axis this point is the point 3 comma 0 that means that x equals 3 is the solution that is the x intercept point of course and that of course agrees back here with the other one we expected that but this is a graphical way of looking at the problem this says solve the equation and don't really care what the picture is this one says look at 2x minus 6 remember it's a line and graph it where it crosses the x-axis will give you an x point that an X comma zero point and the x value will be the solution to the equation so this is a very nice graphical way of seeing the solution to this algebraic equation let us solve X cubed is equal to 25 X so that's a cube there alright this is more complicated than doing a line and you might say well I don't know how to solve cubics well this cubic is easy enough that we can solve it even with what little we know at this point once again let me split the page up into two parts on the left I'll do algebraically and on the right I'll do graphically now algebraically I'll do the same procedure as I did before but now I have experience behind me X cubed equal 25 X I will put everything on the Left X cubed minus 25 X is zero I immediately see that X can be factored out so I'll pull that out x squared minus 25 and now this is where your experience comes in x squared minus 25 is the difference of two squares you will recognize that from the work we did in basics the difference of two squares factors very nicely so this is x times X plus 5 times X minus 5 equals 0 now we have three things multiplied together all of them standing four numbers that means that one or all of them or any two of them say must be 0 so if x is 0 we get one solution X can be 0 when X plus 5 is 0 we get another solution minus 5 when X minus 5 is 0 we get another solution x equals 5 and so this cubic of degree three you notice has three real solutions now that was the algebraic technique we pushed everything to the left to get f of X equals zero and then we use some other facts from before difference of squares can be factored and that zero product principle came up again all right that was the algebraic way of doing this once again let's look at the graphical solution now doing these both ways in these first two examples may seem like a waste of effort because the algebraic solution here on the left is really not very difficult but you have to remember there will be times where there is no easy algebraic solution or no algebraic solution at all so the graphical solution on the right becomes the only way to go so you want to practice this on these simple ones X cubed equals 25 X well just as before X cubed minus 25 x equals 0 pushing everything to the left and now I label this Y so this becomes a function of X that I will graph and where that function equals zero in other words its x-intercepts will be I hope these three points now if we graph this and I went ahead and did this already as I said before and we run to the calculator on this one we get a picture that looks roughly like this passes through these points these points as it turned out turns out our -5 0 and 5 the x axis and the y axis by the way if you want to check this on your calculator the window I'm using here is minus 6 2 6 that takes me just a little beyond the minus 5 and 5 and the Y I had to do a little work with is minus 50 - 50 now that's not essential to this problem but I thought if you want to look at the picture you can look at that yourself these points correspond to -5 0 the origin point which is 0 0 and the point here which is 5 0 and of course you see that then X is equal to minus 5 from there 0 from here or 5 from here and of course these are the same as what we got algebraically so we can see that graphically where this curve X cubed minus 25 X crosses the axis gives us the same answers the same 3 X values as we had gotten algebraically so you want to always look at an equation like this as an algebraic object and is a graphical object the two different insights are very valuable now why did these two examples work there really was some theoretical mathematics here that I left out well I want to be fair and tell you what those pieces are we'll call them to facts and these are to facts we used in the examples there are two facts that we really can't prove or demonstrate at this level of a course but we can show you the idea certainly because the ideas is elementary first of all we used what I'm going to call a working definition don't even put that in quotes a working definition of continuous and continuous it's what what is being defined here I'll tell you why it has to be a working definition and not a strict precise definition here's the definition we used without really giving much thought a function y equals f of X is what we will call continuous if it's graph contains no what you might call holes gaps jumps etc whatever word makes you happy the working definition of continuous that we will use is it is a function whose graph has no holes in it another way to say that colloquially is to say it's a graph you can draw without lifting your pen or pencil from the page now to do this precisely to precisely define this calculus is required or the least a theory that is introduced in calculus so calculus is required to make this precise because there are some mathematical subtleties here but for our purposes in this course this is a perfectly good definition it is a continuous function if it's graph has no holes in it now that's one of the two facts that we used in the examples the other one is this one it's this one's called a theorem this one's even called the intermediate value theorem we will not prove this theorem but I think it seems like a theorem that you will look at and say isn't that common sense well it's common sense if you have the hypothesis of the theorem and you have a proof for us we'll just simply take it as a statement of fact intermediate value theorem now this is an example of what's called an existence theorem it tells you that something is there and it's an if-then statement so it's a one directional arrow if F is continuous now we can use continuous in our working definition sense but this works with the actual precise definition of continuity if F is continuous if also a is less than B so a and B or two numbers and when you evaluate F at a and F of B we're assuming a and B are in the domain of F these two numbers have opposite plus or minus signs if all of that happens then again this is a one directional theorem if you have the hypothesis then the conclusion follows then the graph of F well say the graph of y equals f of X has at least one now this is the existence part I mentioned existence if the graph of y equals f of X has at least one what one x-intercept between a and B now I'll bring this statement back in a moment once we have the picture let me read it one more time if you have a continuous function and a is less than B and F of a and F of B have opposite plus or minus signs so one of them's a positive number one of them is a negative number then the graph of the function y equals f of X has at least one x-intercept between those two numbers now you'll see from the picture that this is one of those theorems that's visually obvious it turns out to be subtle to prove but the picture I think is clear the proof idea in a picture and that's all we'll do in this course the proof idea in a picture is the following let us suppose say remember the hypothesis said that F of a and F of B had different plus or minus signs let's just suppose that f of a is the one that's less than zero and F of B is the one that's greater than zero doesn't matter which one but we'll just say this so I can draw a picture all right here's my picture here say is a here say is B let me mark that below then let's say that this is where F of a is so that's the height F of a so this direction is F of a and you can see from the picture that F of a is less than zero it goes down and let me put F of B up here so this will be a high def of B up here and so that would be F of B and it is greater than zero those are the conditions and the function f which goes from here to here is continuous so if I draw a continuous function without lifting up my pen I get a picture like that it's visually obvious that it must cross the x-axis at least one such x-intercept exists it's clear you can't go from here to there without crossing the x-axis now that turns out to actually use a subtle property of the real numbers but we won't go into that here we'll just say it's visually obvious so it seems like the theorem is true and then we can use that fact but remember too that there may be more than one this may be a function that crosses many times before getting to the top so in that case there be many places where it crosses the theorem simply says there's at least one this this theorem justifies our use of graphing devices our use of graphing devices as we have used them above that is to say we look and see where the graph seems to cross the x-axis we even use one of the buttons on the calculator and a menu item to find the x-intercept and the number that it gives us we can assume is a good approximation because we know that there is such a number there and the theorem here says there is such so this is just theoretical background this says it is okay to do what we're going to do so I won't need to talk about it again and I said I'd bring the theorem back for a moment you see if F is continuous a less than B and F of a and F of B of opposite signs it's clear that the graph has at least one intercept so now you can read this theorem and remember the picture okay let me show you one example of how this might easily be used just to illustrate it in this simple case here's the problem so that the graph of f of x equals x to the fifth minus x cubed minus 1 a very high degree polynomial fifth degree we have no clue on how to solve this at this point show that that graph has has an x intercept between 1 & 2 you need not approximate it so all I want to do is show that this x-intercept exists I don't have to find it I just need to show that it exists well for my solution I since I don't know how to solve this algebraically I'm going to go right ahead to my graphing device and to do that as you know I have the functional form here that I can enter into the calculator I also know I want to look between 1 & 2 so that means my window should be 1/2 for the X values but I haven't got a clue about the Y values now I can hunt I can use experimentation I can fiddle as I said before I can also use auto scale or auto scale or zooming in an automatic fashion or whatever else your calculator allows you to do let's go ahead and look at the calculator now and see what I can do in this case so let me turn this on and let me type in the function so I'll have to go back here and delete the previous one clear that off and I'll type in X to the fifth this calculator types it in very nicely minus X cubed and then what did I have minus 1 and I can go to the window I know that I need to be between 1 & 2 so 4 X min I'll enter 1 X max I'll enter 2 and as I said I don't know what I need to do for my WiMAX and y-min well I have to confess I've actually worked this out before but before I use the numbers that I came up with let's go ahead and see what Auto scale would do now on my calculator I have to go to the zoom key and then see there's a command it says auto I'm going to hit enter it's auto scaling and remember what that does i've given it an x value so from 1 to 2 has been given and then it will find the height the y value that works best to show off every y value in the curve between 1 and 2 and this is what auto scale is given to me now that's perfectly correct and you can see sort of toward the left there that there is a place where it crosses the x axis it's a little bit hard to see because it flattens out there isn't a nice crossing point so with this in mind i'm going to change the scale i am going to change the scale based on i if you trace it you can see where you need to go just beyond that point turns out you only need to go to 1 so Y min minus 1 it looks good to me my max UC is 23 I'm gonna change that to one that looks like a height that's good enough for me and I make that change and then i graph it again now I like this picture better it is exactly the same function but notice that the crossing is less flat it's not parallel to the x-axis it really looks like it's closer to being vertical and so it's easier to see now you can use trace to move over there and find the x intercept or if you have a command like I do take you to that menu you come over here to intersect no what I want is x intercept and I hit enter says it's calculating and then it will give me the coordinates at that point now I don't need the coordinates at that point for this problem do I you notice that it is X X is about one point two three remember the problem only asked to show that there exists a solution a number an x-intercept in the interval 0 1 to 2 the fact that I know it's about 1 point 2 3 did not matter in this problem but I thought that since we here I'd show you two how to go ahead and do that once again so let's go back to the pad and on the pad I go ahead and draw the picture because every time you do one of these for homework or for any kind of problem set you have to produce the picture that is on the screen now the picture I had at the end there where I went from let's see one to two in the X direction and then minus one to one in the Y direction and that point there I discovered was approximately one point two three even though I didn't need that for the problem in this problem all I needed to know is that such a point exists and so it does from the picture okay well that was a simple example and I didn't have to get an approximation let's do one where the point is to get an approximation and show you how you might go through the procedure on the graphing calculator that is to move from zero to one to two decimal place accuracy since this is a little bit new I'm going to walk through one example like this in as much detail as I need so here's the example I want to look at and let me just cover this bottom part up for a moment what I want to do is approximate the smaller of the two solutions of x squared minus 6x plus 7 equals 0 now there must be two solutions and so we want to look at the smaller one the one that will be furthest left in the picture and I want to approximate that correct to two decimal places now all I've done here at the bottom is I've written out that the solutions of this equation are the x-intercepts of f of x equals the function x squared minus 6x plus 7 so that's something we've already discussed the equation and the x-intercepts the solutions to the equation in the x-intercepts coincide now with that in mind let's see what you could do with this we'll pause now and then when we come back I'll show you how I attacked it now onto the solution what I will do here is use my graphic device of course and I will enter the function and the function is x squared minus 6x plus 7 and what I will do here as I go on I will write out what I'm going to do and then I'll go to the calculator and show you how it works but I want to give you the outline first so I'll enter the function I will start with my X scale being equal to 1 as recommended earlier I will experiment now I've already done this experiment to find a window and the window that I will find is minus 1 7 and minus 2 8 so that's my X&Y now later on we'll have more of a technique especially for equations or functions like this we can actually do a much better job of picking a window without guessing all right we will then have a graph on the screen where the X scale is 1 I will then use the zoom box commands everyone has a zoom key on their calculator and I will zoom box around the smaller solution because that's what I was asked to do find the approximate smaller solution solution then once I've done that and I will have a new picture then I will change to a better X scale a finer X scale X scale equal point 1 then I will repeat that process repeat and end with X scale equal point 0 1 so that is the finest and that will give me very easily the solution I'm looking for to two decimal places which is what I wanted to get to so let me keep that there and bring it back in a moment but now I'm gonna go to the calculator here's the calculator and you can see that I've already entered the x squared minus 6x plus 7 up there at the top and let's go hit the window here and you can see that the minimum is already set to minus 1 the maximum to 7 the scale set to 1 the minimum y-value is sent to minus 2 the maximum Y value is set to 8 and I don't care about the Y scale so I'll just leave it at 1 now let's see what kind of graph I get here here's the graph of the function coming down and it is a quadratic it's a parabola and it's striking twice as we expected from our results and the question in this problem was to approximate the smaller of the two roots now that of course is the one to the left so to do that I'm going to do follow the procedure I recommended so you can practice this I'm gonna do what's called zoom box so hit my zoom key you'll have a zoom key somewhere and our mind box is the second thing so I'll go to zoom box and hit enter now you see that nothing's really happened on the screen except that there's a cursor suddenly appeared in the center and it the reflection of what that cursor what position that cursor is at is on the bottom you can see the x and the y values are indicated down here but that doesn't matter to me what I'm going to do is move my cursor over till I get close to the lower root the smaller root now that looks pretty good to me now what am I trying to do here I'm trying to create a box around this smaller root that I will then zoom in on so now that I'm in the zoom box mode if I hit enter of course you don't see anything but what's happened is that there has been a point placed on the screen right behind the cursor if I move the cursor downward now the cursor you see is drawing a line behind it and if I go over to the left it's creating a box and I can tighten that box up by going back up a little bit and now I'm happy it is a box that encloses where the second root is where the curve has an x intercept on the x axis so now if I hit enter the screen will zoom in on to what is inside that box so here I go hitting Enter now this is what was inside the box it is now filling the screen having done this once I go back to window and immediately change my scale now to point 1 which is finer than the one scale and then go back to the graph so there's the graph and you can see that it's striking between a couple of tick marks but that's not going to help me because the scale is point 1 and I want to get down to a resolution of 0.01 so I'm going to priests repeat the same process I did before here zoom go over to box hit enter and then come on up here to somewhere near the root it doesn't matter where exactly say there I hit enter to pin the point behind the cursor to the screen then I go over to the right and you see it's drawing a line and then I go down and I have encompassed the root button go a little further to the right if I want and the root is entirely within the box so now I hit enter and the screen zooms in on that root and then the first thing I will do is change the scale now to 0.01 which is the finest scale that I want point-o 1 enter and then go back to the graph and now the tick marks are 0.01 apart so now I know that whatever the coordinates of the curve are and let me there's my cursor all I have to do is pass the tick mark the first tick mark there that the line is going between and you see that X says 1.58 the next tick mark will make it 5 9 5 8 is where I stop if I want to get truncate this at 2 and so I'm finished so from here we'll go back to the pad alright this is the these are the directions I gave you before and now here is the last window this is the last window that was on the screen go ahead and draw that here this is the kind of thing you'll be doing when you do calculations the last window there were a bunch of tick marks but I noticed that the curve looks something like that and that this tick mark stood for 1.58 now that is going to be the x-intercept of course x-intercept and that is approximately one point five eight there we go and that is truncated that is cut as we said we would do after the second decimal place I might also note here that the curve when I zoomed in this far looks almost like a straight line and that in fact is a very important property of continuous functions if you zoom in far enough they will appear to be straight lines with that in mind one can develop calculus so that is an observation worth remembering but I wanted to show you that's just to give you an idea of how this works let me give me one final note while we're here talking about the calculator so a final note that if the x-intercept after you've done all of that the x-intercept appears to lie on a tick mark number the tick mark number then I would recommend that you try that number in your equation f of X equals zero because if it works you've found an exact solution and an exact solution is always nice to have so that's just a little tip while we're there okay well let's go ahead and pause now and we'll come back to another segment now we're going to continue we've looked at solving equations approximately just to get our approximation juices moving and now we're going to go ahead and start looking at equations in particular types so let's first look at solving linear equations and under this category I've also written something called the linear formula which I'll explain or by graphing so let's see what I mean by all of this so solving linear equations and remember we're doing this in one variable all right let's recall first of all a linear function now this was a function that came in when we talked about the library of functions back under functions this is a linear function is the name that was given to it it looks like this it's f of X equals a X plus B so that was the form and if you remember the a stood for slope actually we used M in those cases but now we're going to use a and of course this is a function we want to get to the point where we talk about equations don't worry we will get there this can also be called and these are words I'm trying to prepare you for the use of later this is a polynomial function of degree one because that's the highest power of the X in this particular function and all right so with that in mind let's go ahead and define what a linear equation would be definition a linear equation now one thing you should remember is an equation is a mathematical sentence a function for example is a mathematical noun or pronoun so when we're talking about an equation we're talking about a sentence with an equal sign in it a linear equation in one variable and one can call this if you like a degree one equation in standard form they remember standard form in mathematics is very important in order to read things off and to have certain formulas work in standard form is going to be a X plus B equals 0 now you remember that's in that familiar f of X equals 0 form that I said was going to be helpful that's what we have here an f of X on the Left equals 0 now in order for this to be meaningful just for the record we assume that a is not 0 because if a is 0 we just have a constant here and there's only one constant that equals 0 that's 0 so they better not be 0 and otherwise the a and the B are real numbers so we are not talking about the situation where they might be complex numbers for us they are simply real numbers ok well that's the definition of a linear equation well let's go ahead and do a little derivations an algebraic derivation of course I'm doing this with algebra and let me show you where it leads and then what we'll call it suppose you start with ax plus B equals 0 then we could rewrite that as ax equals minus B moving the B over to the right we can then divide by a because remember we took a naught to be 0 so X would be equal to minus B over a this if you like you could call the linear formula now depending on your experience you may have heard of something called the quadratic formula well this is the analog of that for linear equations this is the linear formula now you may not have heard of the linear formula before and I'm going to tell you why that's true in a moment now the linear formula of course gives you one solution there is only one solution to a linear equation now why haven't you heard of this before well because it is not nearly as useful as the quadratic formula is not nearly as useful as other formulas we'll see now why would that be well because frankly linear equations are so easy to solve that having a formula is not much of an improvement and remember to use this formula we would have to be able to start out with the linear equation in this standard form and frankly most of them don't start out that way and they're not worth getting into that form so although this is technically correct it is the linear formula it is not all that helpful let me go ahead and point out what the graphical solution would look like here in a sort of generic way the graphical solution remember what you're given you're given a ax plus B equals zero that equation now what you want to do is reinterpret this as f of X equals ax plus B and ask when will that be zero well that's asking when this y value f of X is zero in other words we're asking for x-intercepts find x-intercepts that is to say where this crosses the x axis so if you think of this as a generic example if this is the X and the y axis then the equation of a linear our linear function has an equation that is a line and whatever at this point is would be the X solution well let's see that would be the solution and that would correspond to x-intercept so generically that's what you'd have if this function this linear function is graphed again since there is only one solution and everything is so simple to do algebraically people do not usually solve this graphically they use the graphical picture for confirmation only well let's go ahead and get into one now let's go ahead and solve for P ah new letter instead of X we'll be solving for P here is the equation we're looking at one half times P plus five minus three equals there is the equation part it's got to have an equal in it one-third times two P minus one there is a linear equation it's linear because it has a variable P that is taken only to the power one and I'll jump right in with a solution here well there's some things here that can occur often enough that it's worthwhile mentioning first of all first we want to simplify always a good thing to do because the ideal simplification will give you the answer of course usually you can't simplify all in one blow like that but that's the basic idea first we want to simplify my advice in this case an advice that occurs a piece of advice that's used quite a bit is to do what's called clear out the fractions of course these are numerical fractions fractions with numbers in them clear out the fractions by multiplying by two times three equals six because you see the fractions have denominators two and three and multiplying them together gives me a number that I can multiply the whole equation by to eliminate those fractions and that's exactly what I'm going to do now so I'm going to take that multiplier six and I'll multiply it by the left-hand side being careful to rewrite the whole thing just as it stands so that's 1/2 P plus 5 minus 3 that will be equal to 6 times the same thing on the right 1/3 times 2 P minus 1 all right now we multiply it through doing the operation that we designed it to do 6 times 1/2 will leave me with 3 so 3 times P plus 5 minus the 6 of course must also multiply by the 3 so that's 18 the 6 times the 3 here gives me 2 by design times 2 P minus 1 well I can simplify further by multiplying all these parentheses out so I have 3 P plus 15 minus 18 is equal to 4 P minus 2 now in solving for P so I want to get ideally P equals a number at the end of the problem so I want to pull all of my P's together so I'm going to move my 3p over to the right in other words subtract 3p from both sides and I'll go ahead since I'm pulling the piece to the right I'll go ahead and pull my constant minus 2 over to the left so what will I end up with well I'll have the 15 minus 18 which I had before the minus 2 has been moved over that's now plus 2 then for P minus 3 P of course is just P and then if I reverse these because I prefer to have the P on my left just by convention and see what I have 15 minus 18 is minus 3 plus 2 is equal to minus 1 lo and behold I have a solution let me point out two notes about this first note this solution is P equal minus 1 a very nice integer please don't expect that all your answers are to be integers that doesn't have to be the case secondly I never did if you look through all of the various versions of the equations here all these equivalent equations at no point that I actually have a X plus B equals zero that form as I warned you earlier it usually will not turn up in problems because linear equations are so easy to solve all right now there's one other thing one needs to do in these kinds of equations and it's a practice that you will sometimes drop out but very often need to do especially in more difficult equations coming up you need to check your answer and how do you check if a solution is really a solution you check it in the original equation that means you take your number you substitute it in the original equation let me rewrite it was one-half P plus 5 minus 3 equals 1/3 times 2 P minus 1 if your solution P equals minus 1 is really correct it should fit into that equation and give you equality let's see if it does 1/2 minus 1 plus 5 minus 3 equals 1/3 2 times a minus 1 minus 1 and let's go ahead and simplify this the -1 plus 5 in here is 4 1/2 plus 5 is 2 so we have 2 minus 3 there inside here we have a minus 2 and a minus 1 that's minus 3 times 1/3 is minus 1 so we get minus 1 equals minus 1 and yes that is true so that means this number P equals minus 1 actually does work in the original equation it solves the original equation it makes the original equation into a true statement all right let's do another example and let's pick one with harder non integer coefficients here's one solve for X going back to our favorite variable two point seven eight X plus two over seventeen point nine three one equals fifty four point zero six and I want to solve it correct to two decimal places as we saw in the last segment that is truncation to two decimal places all right there's the problem let's look at it for a moment two point seven eight not an integer times X plus two over seventeen point nine three one an awful looking fraction equals fifty four point oh six another non integer and then the answer we get by saying correct to two decimal places it means find an approximate solution now we're gonna need to know how to deal with that let me give you a tip before I start solving this problem that you will find useful I hope later on also tip in these sorts of problems where you're trying to solve us something approximately solve for X and another way to say that is to say isolate X that's another phrase that's often used before definitely before approximating anything in other words do all the algebra you can before introducing the calculator to approximate anything this will avoid cumulative errors every time you approximate you enter an error into the problem if you approximate several times you've entered several errors those errors can accumulate and give you a wrong answer so the moral is solve for X entirely algebraically and then then and only then once in every given problem then substitute it in your calculator and try to find an approximation so let me show how that would work in this case solution here's the original equation let me write it out one more time for you two over seventeen point nine three one equals fifty four point oh six then two point seven eight X what I'm going to do is I'm going to take this number the two over seventeen point nine three one and move it to the right by subtraction of course so two point seven eight X is going to be fifty four point oh six minus two over seventeen point nine three one I will not attempt to simplify this at all then I will solve for x by dividing by two point seven eight so X will be one over two point seven eight times fifty four point zero six minus two over seventeen point nine three one you might say that doesn't look very pretty and you're absolutely right it doesn't however I have introduced no errors at this point I've simply algebraically reorganized the equation so that I have X alone and everything over here now what I want to do is approximate this calculated approximately to correct to two decimal places how will I do that well I will enter it in my calculator I'm going to have X then approximately equal to some number but since I wanted to show you how this works let's go to the calculator so now I'm going to move the pad and we're going to go to the calculator and once on the calculator we will go ahead and look at an actual calculation here so I'll turn this on and I go to my calculation screen and on my calculation screen I will see if I can enter these numbers and this calculator allows me to do some nice things so I'll go ahead and show it to you one over two point seven eight and you see this calculator allows me to write it the way I would enter it on paper times parentheses five 4.0 6-2 over seventeen point nine three one and then I'll go to the right and put another parenthesis so that's exactly what I had on the screen you can see over to the left and two point seven eighths a little bit has disappeared but it's really still there now I will hit enter and I got out 9.40 something I'm going to cut it off after the second decimal place I'm sorry nineteen point four oh and so that would be the number that I use so now I'll use I'll go back to the pad and I'll go ahead and show you the number we just got the number was nineteen point four zero and that was truncated that was cut off after the second decimal place and that's the approximation we were looking for so there we are so when you do this on the calculator you it's really quite nice just remember to do it at most once so you don't introduce errors that you really don't want to do okay now we're not done with this problem yet very important thing we need to do in this and I need to show you why it works out the way it does we need to check our answer so check the first thing I will do is give you a warning about checking in this kind of an approximation problem warning checking and approximate solution will only yield of course an approximate equation so when you check your approximate number in the equation you will only get an approximate equality because you're using a number that is not in fact correct it's an approximation so here's the check the actual check down here 2.78 and remember we that's where X was so it'll be nineteen point four zero that I substitute plus two over seventeen point nine three one equals question mark five four oh six point oh six now I'd like this on the left to be exactly equal to five four oh six I'm going to substitute this now in my calculator and I'm not going to show you I'll just go ahead and write down what I got when I did this my calculator truncating again after the second decimal gave me five four point oh four now that's only approximately five four point oh six it's not equal but then if we wanted an approximate answer we should expect that that's what we'll get here is an approximate equation in approximate equality and that's the best you can do if it is reasonably close you should be happy with the approximation you gave of course if these are very far apart then you probably have made an error earlier but be aware that you will get approximate solutions here so with these examples behind us let me write up a summary for you and then you'll have a chance to try some problems on your own here's a summary of some of the aspects to solving linear equations in one variable and the variable of course we're using here is X but it can be any letter so first thing if needed clear fractions that happens often a noun so it's worthwhile putting down secondly simplify simplify each side now what a simplify mean well it means things like multiply out parentheses I can't list every possible thing but I'll give you some ideas multiply out parentheses combine like terms etc the sorts of standard algebraic simplifications that you do then once you've got everything as simple as you can on each side you want to start working side to side so that you can isolate the variable and I'll say I call it X here to get an expression that looks like this x equals here's a technical term stuff okay that's the idea x equals stuff stuff without exit okay numbers the next thing you can do if you desire it you can't approximate if desired if you don't like the way the answer looks and you want an approximation and then finally you want to check check your solution and you always check your solution in the original equation I stress this because there are many equivalent equations along the line when you solve the original equation you have a whole series of equivalent equations your check must be in the original and as we'll see later you will find out that B by doing algebra and solving a given equation you can end up with candidate solutions that do not solve the original equation so we have to be very careful to check these things okay well why don't we go ahead and pause and you can try this summary of material out yourself on solving linear equations in one variable now we've looked at solving linear equations sometimes there are nonlinear equations that can be reduced to linear equations after a step or two and so they're worth mentioning here so let's go ahead and look at those for a moment solving nonlinear equations that lead to linear equations so what I will do is do this by example so we'll just start out with an example here all right here's the example 2y plus 1 times y minus 1 equals y plus 5 times say 2y minus 5 now if you look at this at first glance it is definitely nonlinear why because if you think about it on the left you have going to have a 2y x y in here which gives you 2y squared and that's degree 2 that also occurs on the left and on the right so on both sides you have to what you have a y squared occurring and so this is a degree two equation it's not linear however after is step or two it reduces to linear so really it's nonlinear in form but linear in practice so here's what we can do with the solution we will multiply both sides out how do we do that well you need to remember what we did in basics on how to multiply binomials I'm going to take the 2y here and multiply it by the Y in the second binomial and by the minus one there so I will end up with 2y squared and minus 2y then I'll take the 1 and multiply it by the Y and the 1 and multiply it by minus 1 of course 1 doesn't change anything so this will be plus y minus 1 I do the same thing over here the Y times 2y and the y times minus 5 so I have 2y squared minus 5 y and then the 5 times 2y which gives me 10 Y and then the 5 times minus 5 which gives me minus 25 so there we have everything multiplied out and you see what happens with these squared terms notice you have two Y squared here two Y squared here if I subtract those from both sides I have made them disappear and there is no longer a problem that is nonlinear this is now a legitimate linear equation all right well let's start simplifying some things here the minus 2y and the Y give me minus y minus 1 equals the minus 5y and the 10 y give me 5y minus 25 and I'm going to need to move things around here so let me move to another page you bring that last part up here so you can see that last bit what I will do here is I want to combine the Y's on one side so what I'll do is bring the minus y over to here of course I'm subtracting minus y in other words adding Y to both sides so that will leave me on this side with a 6y and on the other side I'll bring the 25 over to the other side so that'll be 25 over here minus 1 is 24 so I have 24 equals 6y of course that means Y is equal to 4 and that is what I think is the correct solution I will check to make sure if I put it in the original two times four plus one times four minus one equals question mark always put a question mark at first because you don't know if this is going to work out equals four plus five times 2 times four minus five so I've gone through the original I replaced the Y everywhere by the four if I then simplify this 2 times 4 is 8 plus 1 is 9 times 4 minus 1 which is 3 question mark 4 plus 5 is 9 and then here I have 2 times 4 that's 8 minus 5 is 3 is 9 times 3 equals 9 times 3 sure it is so that means that that number I found y equals 4 is indeed a solution so that original equation that appeared to be nonlinear was in fact linear let me show you another one like that 3 over X minus 2 equals 1 over X minus 1 plus 7 over X minus 1 times X minus 2 not only is this nonlinear it's not even a polynomial equation it's a rational equation so how could this possibly turn into something that's linear well let's see first of all as always in these rational expressions the first thing you need to do is find out where these denominators would be 0 because those are numbers that are not allowed as solutions so first I will write down that the domain of x equals the set of all the X's in the are in the real numbers and it's that set builder notation I've used before such that X is not equal to 2 because that would make these X minus 2's go to 0 and it's not equal to 1 because that would make the X minus 1s go to 0 and either one of those going to 0 would cause division by 0 which we won't allow alright then having noted the domain clear of in this case their variable fractions but the principle is the same clear of fractions by multiplying by let's see X minus 1 times X minus 2 those are the those are the only factors that occur in the denominator so if we multiply through by this we'll get everything to cancel out so let's go ahead and see what that works out to be so we're multiplying X minus 1 times X minus 2 times the original left-hand side was 3 over X minus 2 now that is equal to X minus 1 times X minus 2 times 1 over X minus 1 plus the rest of that sides over X minus 1 times X minus 2 so let's work this out in the first expression here on the left hand side we sue there's an X minus 2 on top and an X minus 2 on bottom of course that makes that 1 which is another way of saying they cancel so what we're left with is 3 times X minus 1 I reverse the order of the 3 in the X minus 1 but it's still the same thing over here X minus 1 times X minus 2 times the first term the X minus 1 over X minus 1 cancels but do not get in the habit you've noticed that I have I'm trying to avoid writing slashes when I cancel things I didn't write anything here I'm trying to avoid that because in cases like this it is mistaken to do so the X minus one indeed cancels with this but this entire term here needs to multiply by the second one and you don't this X minus 1 being gone here doesn't mean it's gone for the second one so multiplying times the first one leaves us with simply X minus 2 plus here the X minus 1 X minus 2 cancels entirely with the denominator leaving us simply with 7 and let me note here that we have finally reached the stage where this is now linear everything up to this case was nonlinear it's suddenly reduced to linear so if we multiply this out we have 3x minus 3 equals x plus 5 minus 2 and 7 is 5 we move the x over to here say to get 2x the minus 3 over there to get 8 x equals 4 is the apparent solution this is not one of the numbers that we omitted in our domain remember we said the domain consisted of X not equal to 2 or 1/4 certainly not 2 or 1 so that is needed legitimate solution and now I'm reaching the point where I'm going to leave the check to you so I'll say left to you abbreviating as L to Y so I'm leaving the check left to you in this problem all right well having seen that and having seen the rest of the work we've done let me pose one to you actually I already have it written out let me bring this up to you here and pose it to you here's the problem I'm posing solve for X 3x over X minus 2 equals 1 plus 6 over X minus 2 now you take some time to work on this and when we come back I'll show you what I would have done okay let's see what I'm going to do solution well of course you know the first thing I need to indicate is the domain of that variable X so it'll be all the X's in the real numbers such that what well there's only one denominator here X minus two so I don't want X to be 2 so X not equal to 2 all right with that in mind I will multiply by X minus 2 that's to clear the fractions so let me show you the work there X minus 2 times 3x over X minus 2 equals X minus 2 times 1 plus 6 over X minus 2 okay well this is simple enough I should be able to slip this in down here at the bottom X minus 2 X minus 2 leaves me with 3x on that side the X minus 2 multiplied by the 1 gives me X minus 2 plus X minus 2 again will cancel with the X minus 2 here and give me 6 so let me move to another page and I now have let me repeat that so I can work with it 3x equals X minus 2 plus 6 so that's 3 X is equal to X plus 4 and of course I can then move the x over here and get 2x equals 4 and so x equals 2 now I'm done right I have found the solution x equals 2 but wait a minute let me go back here let's see the domain of X X can't be equal to 2 but I found that the only solution possible is to what do I conclude from this do I conclude that I made a mistake well usually that's a good idea to check but in this case I'll assure you no I didn't make a mistake I will say since 2 is not in the domain of X the original equation has no solutions in other words this is not a solution now that should not be a surprise to anyone there is no reason on earth that every equation ought to have a solution just because we can write it down this equation that we had originally here the 3x over X minus 2 equal 1 plus 6 over X minus 2 simply states something that is impossible in the real numbers that's all so when you solve it you end up confirming that by finding a number that is impossible in the domain so it's time for you to try a few of these nonlinear type problems and reduce them to linear problems

Channel: UMKC

Views: 31,986

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Keywords: Solving equations, Solving algebraically, Solving graphically, Linear Equations, Linear function, Linear Formula, Non-Linear Equations.

Id: 43an6fuCIU4

Length: 83min 5sec (4985 seconds)

Published: Tue May 05 2009