 Hi, I'm Zor. Welcome to Unisor Education. Today we will talk about linear equations again but not in theoretical but more practical standpoint. I have a couple of problems which I would like to share with you and obviously you are encouraged to do them just by yourself first. They are on the web and this is basically an explanation of what these problems are all about. Alright, so I have four different problems and I'll just do one by one. Let's start with the first one. We have an equation of 5.5 and plus 11 equal to zero. Okay, why did I decide to present this particular equation? Well, here is a small condition which I would like to impose in the beginning. I would like to solve this equation not in just any real number but in integer numbers. Again, specification of domain is always very important. If I specify a domain for instance of positive numbers obviously this particular equation doesn't have any solutions because if you will apply the regular transformation minus 11 first so you will get 5.5x equals minus 11 and then divide by 5.5 which will give you at 11 divided by 5.5 which is equals to 2. So we do have an equation which has a solution x equals 2. So if my condition in the very beginning is I would like to solve this equation in the domain of integer numbers. Well, I'm sorry, that's minus 11. And this particular solution is a real solution because minus 2 is an integer number. If however my condition in the very beginning is I would like to solve this equation in the domain of positive numbers or positive integer numbers or something like this this is not an element of that domain which means equation would be just not solvable. Equation would not have any solutions. So that was basically the purpose of this particular exercise to emphasize that there is always a condition what's the domain where you are looking for the solution of this particular equation and if domain does not contain your solution which you have obtained well it means it's the reason of solution basically, that's what it is. Okay, so that's as much as we have for the problem number one. Sometimes it's a sub-problem number one. Now, the next one is another illustration of the importance of the domain. I have decided that I will solve equation among the complex numbers. So, here is my example. Plus 4i x plus 4 minus 3i equals to zero. Okay, so if you remember complex numbers are always comprised from a real part and imaginary part and the imaginary part contains this number i which is a square root of minus one. Introduced purely theoretically there is no real number which corresponds to square root of minus one but it's very convenient mathematical theory and the complex numbers are very very interesting set. We will definitely be involved more with complex numbers when we will go to equations of higher order. In any case complex numbers as you know can be added, subtracted, multiplied, divided by each other and I'm in my in my rights asking to solve this particular equation in the domain of complex numbers. Now how to do it? Well exactly the same thing as before. The first we will apply invariant transformation by subtracting a complex number 4 minus 3i from both sides. That gives me 3 plus 4i x equals minus 4 plus 3i. I reverse the signs here because it goes to the right. And obviously next is I divide both sides of the equation by a non-zero which makes it invariant transformation by non-zero number 3 plus 4i. And I can say x is equal to minus 4 plus 3i divided by 3 plus 4i. Is this a solution? Well in a way it is. However it's not a common way of representing complex numbers. Complex numbers are usually represented by a combination of real part, so this is supposed to be a real number and imaginary part which is a product of real times i and plus or minus in between. So all these numbers are represented in their traditional form. This is not a traditional form. Now how can we convert this into a more traditional form when I have a real number plus or minus and another real number times i? Well first of all this is a good opportunity to press the pause button and try to do it yourself. That's my pause. Okay and now I will basically tell how it can be done very easily. You know that if you have a ratio of two numbers you can always multiply numerator and denominator by the same number and you will get the same result obviously. So if I will multiply let's say by two both numerator and denominator it's exactly the same fraction. So now I will have to multiply it by something which will get rid of the imaginary number in the denominator and the standard approach is the following. If you have a complex number a plus bi and you multiply it by a minus bi let's see what happens in this case. It's very simple. A times a is a square. Bi times a is adi. Now a times minus bi is minus adi and bi times minus bi is b square and minus and i square. Now i square is minus one so it will be a square minus and minus one which means it's just b square which is equal to obviously reducing these two things a square plus b square. Okay so that's the standard way of converting this particular thing into more traditional format. So I will multiply both numerator and denominator by three minus four i. So what will I have in the denominator? a is three b is four. So I will have three square plus four square which is nine plus sixteen which is twenty-five. So this is equal to and I will have twenty-five in the denominator. Now here I will just multiply minus four times three is minus twelve plus three i times three is nine i. Minus four and minus four i is plus sixteen i and three i minus four i is three times four minus twelve and i square is minus six plus twelve which is equal to you see how easy it is. So it's twenty-five i times twenty-five divided by twenty-five which is i. So that relatively complex expression actually weighs to i. Well considering we did a certain number of relatively complex transformations even if all the transformations were invariant I think it's a good practice to do the checking. So let's try to put i into the beginning in the original equation and see if that actually is the solution. Well checking is always a good thing you know. Sometimes it's mandatory you know that from the lecture sometimes it's just a good practice to do. Well it's always a good practice to do but sometimes it is mandatory. Alright so x is equal to i. What do we have? We have three i, four i and i. i square is minus one so it's minus four plus four minus three i and this reduced and this reduced it is indeed equal to zero. So we were right with our solution x is equal to i. Alright so that's second self problem mini problem. The third one, five minus ten x equals to zero and I'm asking to solve it graphically. Now as you remember we have to graph the function five minus ten x which is okay let's start from the very beginning. y is equal to x. Step number one is this. y is equal to ten x. This is ten times steeper so it will be like this but also crossing the zero. And now y is equal to minus ten x. Well this is a symmetrical so it goes this way. Instead of steepness that way it will be steepness down if you wish and plus five. That means the whole graph will go up five units so instead of crossing in five in zero I will cross in five. So let me just get rid of all this and the result graph will be x, y, zero, five so it will be here. Now as I said the steepness is characterized by this number ten which means this segment is ten times greater than this segment so if this is five this is one half. This is one. So the scale is not actually the same just to make sure that you have a good grasp on the picture collages rather than particular dimensions so here five units will be something like this and here so the scale is different this is one unit on the x and this is one unit on the y slightly different but that's okay. In any case we have the crossing of this graph with y equals to zero which is the x-axis at x is equal to one half. Well that's the graphical solution let's check it out. Put one half instead of x which is definitely ten times one half is five, five minus five is zero everything is fine so that's the graphical solution. Well most important here is what? To draw a line and exactly know what's the ratios between these two segments where it crosses the y-axis and where it crosses the x-axis and that ratio is characterized by this number ten in this particular case and then the crossing of the y-axis is characterized by this three member, five in this particular case and everything else basically follows from this ratio. Alright so much for this mini problem and we have the last one which I planned for this particular set of problems for linear equations I was thinking that linear equations are really so simple what can be really challenging as a nice problem well I didn't really come up with anything real challenging but anyway the last one seems to be a little bit unorthodox I have two x plus four times two x minus six equals to zero well you obviously can tell me hey this is not a linear equation so why are we talking about this? You're right it's not a linear equation however the fact that this is a multiplication of two expressions each of them are linear actually makes it two linear equations at the same time because think about it this way if something which is a product of two different expressions is equal to zero when the product of two numbers is equal to zero well obviously either one is equal to zero or another is equal to zero and both cases are equally right if none of them is equal to zero then the product is also not equal to zero so if we are looking for solutions what are the values of a and b we should really look among a equals to zero or b equals to zero and both have the same rights in this particular case what makes this expression which is a product of two linear expressions equal to zero either this particular component or this particular component is equal to zero which means we have really two linear equations one linear equation is two x plus four equals to zero and the solution is x is equal to minus two and another is two x minus six is equal to zero and solution is equal to three so what I can say is this actual quadratic equation can really be broken down if you wish into two linear equations and each of them has a corresponding solution which means that both of these numbers minus two and three represent solution to the original equation it's just the way how we get to these two solutions is by solving two linear equations and this is actually a very good point because in theory I'm jumping forward really significantly in this particular case any equation of the power of n equals to zero now this is a general equation with different coefficients different multipliers and x to the nth degree and to the x to the m minus first etc so all the different powers of x are represented with different multipliers and the total sum together is equal to zero this is a general equation of the nth degree or nth power so there is a very important mathematical theory that any expression of this type can be really transformed into expression of this type when you have a bunch of linear expressions and not just a bunch, exactly n linear expressions and the result of the multiplication is the same as this polynomial expression and therefore solving any equation of the nth degree is basically equivalent to transforming it into this form when it's actually a multiplication of different linear expressions and then solving each linear expression which is really very simple the only little detail about this is that it's true among the complex numbers and this is one of the very very interesting examples of how something quite artificial like complex numbers come handy because this particular transformation of any polynomial into a bunch of into product of linear expressions it's true only within the domain of complex numbers so we will talk about this a little later but right now I just wanted to make a very quick glimpse into something which if you remember in the beginning we had a couple of lectures about beauty, harmony, etc this is a perfect example of a harmonious approach to equations when the equation of any power can be expressed as the corresponding number of linear equations multiplied by each other linear expressions rather okay so that basically completes these many problems for linear equations very simple indeed but nevertheless important thank you very much