 Welcome back to our lecture series math 1050 college algebra for students at Southern Utah University. As usual I'll be a professor today Dr. Andrew Missildine. In lecture 11 we're gonna start the first of actually like four lectures honestly about the topic of systems of linear equations. Lectures 11 and 12 will focus on the systems of linear equations. Lecture 11 will mostly focus on two by two linear systems while lecture 12 will focus on three by three linear systems. And then the next two lectures after that we're gonna focus on using matrices to help us with the systems of linear equations. So I keep on saying this word systems of linear equations what are we talking about here? So a system of equations without the necessarily adjective of linear here. A system of equations is a set of equations with common variables. So you can see example of such a thing right here like this right here is a two by two system of linear of system of equations it's a linear equation linear system we'll talk about in just a second. It has two equations we got this one right here 2x minus y equals 5. We have a second one here x plus 4 y equals 7. We have two equations and of course we have these two variables x and y. Alright and so you often describe a system in this case two by two to suggest there's two equations and two variables. You always mention the number of equations first and then the number of variables second so if one was describing an m by n system this would suggest that you're gonna have m many equations and you'd have n many variables in that system. Now we're not gonna do anything there that's that's huge but in theory that that's that's what this terminology is meaning here. So a system of equations is a set of equations with common variables here. A solution to a system of equations is any assignment it's an assignment to each of the variables such that each of the equations is satisfied here. So if we were to take like a single equation like an equation with one variable take something like 2x plus 1 equals 0. The solution to this equation is negative one half because notice if you plug in negative one half for x you're gonna get two times negative one half plus one as working with the left hand side simplifying that you're gonna get there a negative one plus one that does turn out to be zero which is the right hand side. The assignment that x equals negative one half that gives us a solution to the to the equation this is that's what it means there the assignment makes the left hand side equal to the right hand side. Now in a system of equations a solution would be an assignment of each of the variables because there might be more than one variable here using this example right here we would need an assignment for x and for y and that assignment needs to make the first equation true it needs to make the second equation true if there was a third equation it needs to satisfy that one and the fourth one and the fifth one and all of them. So a solution to a system of equations is going to be an assignment of all the variables which satisfies all of the equations in the situation. Now we're gonna be focused on systems of linear equations so we said so it's a system of linear equations of every equation in the system is linear that is with regard to the variable the exponent is always one so you don't see anything because we don't write exponents of one here but the variables always one and you never have products of these things together so this should be linear equations with multiple multiple variables mind you but it's gonna be a system of linear equations sometimes we call it a linear system for short. Conversely the system is nonlinear if there does exist a nonlinear equation inside the system. We'll explore nonlinear systems much later in this semester I should say in this lecture series but for now we're gonna talk about linear systems they're a lot more tame compared to what we can see for nonlinear systems. So let's discuss what is a solution to a linear system look like so we keep on looking at this example right here what I'm first going to do is take an equate or take a take an assignment that's not a solution like if I were to test the value say 1 comma 0 notice that if you put 1 comma 0 into the first equation and we're gonna follow the usual convention that if we call our variables x and y then if I write a point an ordered pair the first coordinate is the x-coordinate the second coordinate is the y-coordinate so if I plug in x equals 1 and y equals 0 that's what this means here x equals 1 and y equals 0 if you plug it into the equation the left-hand side you're gonna get 2 times 1 minus 0 is equal to 2 which is not equal to 5 that does not satisfy the right-hand side like so well sorry the right-hand side is equal to 5 I should I should clarify that the right-hand side which is 5 is not equal to 2 it didn't satisfy the first equation in fact if you check the second equation as well you're gonna get 1 plus 4 times 0 which is going to equal just 1 that's not 7 so again that is violated as well this is not a solution to the system because it didn't satisfy the equations in fact it didn't satisfy any of the equations now on the other hand if we did something like the following let's test the point this time I'm gonna take 0 comma negative 5 notice what happens here that if I plug in x is 0 and y is negative 5 if I plug that into the first equation I'm gonna get that the left-hand side becomes 2 times 0 minus negative 5 well that would simplify just to be 5 that is the right-hand side so that actually certifies the first equation but if you look at the second equation there if you take 0 plus 4 times negative 5 that's gonna give you negative 20 negative 20 is not equal to 7 so it turns out it didn't work on the second equation this is what's important about a solution to a system of equations that each equation must be satisfied by that assignment to count as a solution so now let's get to the point that was suggested here on the screen 3 comma 1 what's so special about that well it turns out this is in fact a solution to this system of equations if we were to plug in x equals 3 and y equals 1 into the first equation we get the following the left-hand side would become 2 times 3 minus 1 2 times 3 is 6 minus 1 is in fact 5 that is the right-hand side so the first equation is satisfied how about the second equation if I take the left-hand side there there's an x plus 4 y I replace the x with a 3 I replace the y with a 1 so we're gonna get 3 plus 4 times 1 4 times 1 of course is 4 3 plus 4 is of course equal to 7 that is the right-hand side and so Bob's your uncle 3 comma 1 satisfies the first equation and it satisfies the second equation those only two equations involved in this linear system and therefore it is in fact a solution so we have now demonstrated that this linear system has a solution and in fact it's the only solution to this linear system how do I know that well there is a very nice geometric argument you can make about this because after all our linear system which I know is no longer on the screen so let me rewrite it we had we had the equations 2 x minus y equals 5 that was the first one and we had the second equation x plus 4 y is equal to 7 these this is a linear equation both of them with two variables here if I solve for y into the first equation here you can move the 2x to the other side so you get negative y equals 5 minus 2x times both sides by negative 1 you're gonna end up with this equation right here you're gonna get y equals 2x minus 5 this is now the slope intercept form of that equation for which this is a as a little as a line it has a y intercept of negative 5 so while the scale on my x and y axis aren't included here we can assume negative 5 would be down here somewhere I mean as a slope of positive 2 so we're gonna go up to every time we go up one and so while this is drawn to scale I didn't label the scale you get a line that looks something like that that's why equals 2x minus 5 now if you take the second equation right here x plus 4 y equals 7 the same thing move the x to the other side you get 4 y equals 7 minus x divide both sides by 4 we end up with this equation right here y equals negative 1 4th x plus 7 4th 7 4th is a teeny bit smaller than 8 4th which of course is 2 so you'd expect a wiener set somewhere around here our slope this time is negative 1 4th so every time we go down one we're gonna go over by 4 and so you get a graph of a line that looks something like the following so I'm approaching this problem geometrically I can graph the solution set of this equation that's its line that's the graph is the line I can graph this equation the graph is the solution set to the equation so a solution to a system of equations is the simultaneous solutions of all of the equations so if the first graph is just an illustration of its solution set and the second graph is just an illustration of the solution set of the second equation this actually suggests that the intersection of the graphs will then be the simultaneous solutions to these two equations that's the solution to the linear system and when we graph these two lines we can see very clearly that there is only one point of intersection these lines only hit at one point and that one point of intersection has got to be 3 comma 1 because we've already determined 3 1 is a solution and geometrically there can only be one solution so that's got to be the unique solution this happens actually a lot when you work with linear systems we get this unique solution now if we want to think about this geometrically it turns out there are other options as well one possibility is that when you graph your lines you could you get a line like this one but you might also get a line that's parallel to it and so much that they actually never intersect whatsoever because parallel lines by nature are going to be equidistant they never touch they never intersect so it's possible that there is no solution to a linear system at least for this when you have two equations two unknowns because maybe there's no solution because they're parallel right so we can get zero solutions we can get one solution third and final possibility is you actually can get infinitely many solutions it turns out that when you graph the first line you get something like this but then when you graph the second line you get the exact same thing again it could be that the two lines completely overlay each other they overlap as is illustrated with this dashed line you see the yellow line and the blue line are overlapping and so you actually get every solution to the first line is a solution to the second line and as each line contains infinitely many points you can get infinitely many possibilities now it turns out that for a two by two linear system these are the only three options you either have no solutions a unique solution or infinitely many solutions as we explore linear systems in the future we start looking at higher dimensions like what if we start looking at three by three linear systems we'll do that in the next lecture we'll see that the same pattern holds that a linear system has only no solution one solution or infinitely many solutions