7. Systems of Equations and Inequalities

Systems of Linear Equations in Two Variables

# Solve a Linear System û by the Substitution Method

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Hi, my name is Rebecca Muller. During this session, we're going to look at the topic of linear systems. Now, a system of equations is a set of equations for which you want to find a common solution or solutions. If the system is a system of linear equations, that means that all of the equations that we're going to be looking at are linear, which means that we have a variable raised to the first power. We don't have variables in a denominator or underneath the radical sign. And we don't end up having a term where we have variables multiplied together. Here are our tops, or our subtopics, I should say. First we'll look at solving linear equations in two variables. We're going to look at doing this by graphing, by the substitution method, and by the elimination method. And then we're going to move on to solving linear systems in three variables, and we'll look at our last two methods by the substitution method and by the elimination method in order to accomplish this. So we're going to begin by reviewing a little bit about linear equations. So you may recall that if we're looking at vertical lines, we have an equation that has the format x equals a number a. If we have horizontal lines, then our equation format would be y equals a number b. For other types of lines, we can always write it in what's called slope intercept form, that is, y equals mx plus b. Now, of course, a horizontal line can also be written like this, where the slope is equal to 0. So when we're thinking about what can occur graphically with linear systems, let's just think about the two variable situation where we have an xy plane. And I'm going to just draw two lines on this xy plane. So what can occur? Well, we could have the two lines-- let's say one comes in like this, and then we have another line that comes in and intersects it in a single point. This is going to give us a system of two equations that have a common solution. We can find out all the points that lie along the first line, all the points that lie along the second line, but the single point that is going to lie on both lines is going to give us the solution to the system. This is going to be called a consistent system when this occurs. What else can occur if we have an equation of two lines? Well, we could end up having the first line looking something like this. And then maybe there's another line that comes in and is parallel to it. If we're looking for common solutions, that would be something that's going to make the first equation true and the second equation true at the same time, notice there's not going to be a single point that's going to work there. In this case, what we end up having is what's called an inconsistent system. And an inconsistent system is not going to have any solutions. And then finally, our third case when we have two lines drawn on an xy plane, is I could be giving you two equations, but it turns out that when I draw the first equation in, that's what I end up getting. And when I draw the second equation in, it comes right on top. It superimposes. So in other words, what I have here is the same line. And maybe that's obvious by the two equations and maybe it's not right off the bat. But when I have the same line, this is called a dependent system. And so graphically, if we're looking at what can occur, the typical things that you're going to see is that if I don't put them all in slope intercept form, you're looking for what would happen if it's inconsistent, what would happen if it's dependent. And then most of the time, your solutions are going to end up being a consistent system, where you're trying to find that single point. For our first example we're going to look at solving a linear system in two variables using graphing. So we have two equations given to us. We have x plus 2y equals 4, and the second equation is x minus 2y equals 0. So we're going to review graphing straight lines as we're doing this process. I'm going to start with the x plus 2y equals 4, and I'm going to use the intercept method in order to graph this straight line. So again, for x plus 2y equals 4, recall that we can set up a little chart, and substitute in a 0 for x, and solve for y, and then vice versa. So if I take this equation and substitute in for x, we're going to have 0 plus 2y equals 4. So 2y equals 4, and y is equal to 2. So when x is 0, y equals 2. When I allow the y value to equal 0, then I'm looking at the equation where I substitute in 0 for y. So x plus 2 times 0 equals 4. 2 times 0 is 0, so I end up with x plus 0 equals 4, or x equals 4. And so when y is equal to 0, x is equal to 4. Now, these two points I can place onto my xy plane and draw in the graph, as we can know that any two points will determine a line. So 0, 2 gives us this point, and 4, 0 gives us this point. And we're going to draw in a straight line. I'm going to have to kind of stand in front of it in order to try to get a straight line going on here. Eh, it looks pretty good. I would suggest you use a ruler at home on your paper. For the second line, we have x minus 2y equals 0. And what I can do is I can solve for y in this equation, and I'm going to show you how to graph it using the slope and the intercept. So we're going to have negative 2y equals negative x. And then dividing both sides of this equation by negative 2, we'll end up with y equals. And when I divide by negative 2, it is true it equals negative x over negative 2. But I'd like to write that in slope intercept form. And notice that I can use the coefficients. In front of the x, I have a negative 1. I now divide that by negative 2. Negative 1 divided by negative 2 is 1/2, and then that's multiplied times x. So in this case, recall that my slope is equal to 1/2, and that means that's my rise over run. My y-intercept, it doesn't look like I have one, but of course I do. The y-intercept is equal to 0. So I'm going to plot the y-intercept at 0, 0, the origin. And then I'm going to use the fact that I can rise one and then move to the right two units to end up with a second point on that line. Now I'm going to connect the dots. And once more, let me stand in front of it to try to get it straight. And now I have the graph of the second line. Well, remember that when we're graphing a system of equations and we end up noticing that we have a single point of intersection, then we have what's called a consistent system. Now, in graphing it, it looks like both lines are going to intersect at the point that is located two units to the right and one unit up, or in other words, the point whose coordinates are 2, 1. Now, right now, that's really kind of a guess. Like, did I do a very good job of graphing it or not? We want to be able to check our answer, and it's very easy to check our answer, because what we can do is substitute those values for x and y in our original equations to see whether or not they give us true statements. If they do, then this is going to be the solution to the system that we're working on. So let's start by taking the value of 2 and substituting for x in the first equation. So that's going to give me a 2, plus. Substitute the value of 1 for y in the first equation. So that's 2 times 2, 2 plus, that is, 2 times 1. That's going to be 2 times 2, 2 plus 2, which is going to equal the value of 4. In the second equation, we're going to substitute in the value 2, 1, but I'm going to substitute the value of 2 for x here. That's going to be 2, minus, and then we're going to have 2 times 1. That's 2 minus 2, which is equal to 0. So in both cases, I see that it does satisfy the two equations and I was looking at. And therefore, the point 2, 1 is the solution to this system. The graphical method of solving a linear system is effective if we can graph the equations easily and if the solution turns out to be integer values. If not, it can be fairly difficult to determine what the solution is. So we need other alternatives. I'm next going to show you one of those alternatives, and it's called substitution. Here's the system I'm going to use to illustrate this. x plus 6y equals negative 7 is one equation, and 3x minus 4y equals negative 32 is the second equation. So the idea of substitution is exactly what it sounds like. We're going to solve for one of the variables in one of the equations, and then we're going to substitute that into the other equation that we didn't solve for the variable. When I look at this system, I have an x that has a coefficient of 1, and then a y with a coefficient of 6 in my first equation. In my second equation, the x has a coefficient of 3, and the y has a coefficient of negative 4. In deciding what to solve for here, I want to choose the method that's going to be the easiest in solving for the variable and also in the substitution process. The easiest variable to solve for in these two equations is going to be the x from the first equation, since its coefficient is equal to 1. Were I to solve for any of the other three variables that I see here, I would have to do a division by that coefficient, which would give me a fractional setup. And if you're like most people, avoiding the fractions is a good idea. Even though I don't feel intimidated by fractions, I still avoid them because it's easy to make careless mistakes if you do so. So I'm going to begin by taking my first equation and solving for the variable x. I can accomplish this by subtracting 6y from both sides of the equation. So here's what I end up getting. I end up with x equals. Subtracting the 6y gives me negative 6y, and then I have the minus 7 that was there originally on the right-hand side of the equation. Now that I've solve for x in the first equation, I'm going to take what x equals, the negative 6y minus 7, and substitute for x in the second equation. In the second equation, I have three multiplied times x. So now I'm going to write down 3 multiplied times negative 6y minus 7. Then I'll write down the rest of the equation. Minus 4y equals negative 32. Now, before I go on and solve this, let's think about what we've accomplished. We started off with two equations and two variables. I now have been able to combine those into a single equation and a single variable. And that means I'm able to solve for y in the resultant equation. To use distributive property first gives us a negative 18y minus 21, and then minus 4y equals negative 32. Let's combine like terms. We have negative 18y. I'll minus the 4y. I'll just rewrite it like this. Minus 21 equals negative 32. That's going to give us a negative 22. I'm going to go ahead and add 21 to both sides of the equation. So that's going to end up giving us a-- if I add 21 here, I'm going to end up with a negative 11. And that's going to give us y equals negative 11 divided by negative 22, which is going to be 1/2. Now that I have a value for y, I can substitute back into my formula where I had solved for x. We had x equals negative 6y minus 7. So again, using substitution, we can have x equals negative 6 times the value of y, which I determined was 1/2, minus 7, and we'll evaluate. Negative 6 times 1/2 is going to be negative 3, minus 7, and that gives us x equals negative 10. So at this point, I can say that it looks like the negative 10 for x and the 1/2 for y is going to be the solution to this system. It's always a good idea to check your work, and it's very easy to do so when we're working with systems of equations. We can do that simply by substituting into each equation to see whether or not we end up with the result that we have in the original format. So again, go back to the original format. We had our x plus 6y was equal to negative 7. When we substitute, we end up with negative 10 plus 6 times 1/2. And we're questioning whether or not that equals negative 7. Well, 6 times 1/2 is going to give us a positive 3. So negative 10 plus 3 does indeed equal negative 7. So that checks. In our second equation, we had 3x minus 4y equal to negative 32. Let's substitute a negative 10 for x. That's going to be 3 times negative 10, minus the 1/2 for y. So that's minus 4 times 1/2 equals negative 32. That's what we're trying to decide. 3 times negative 10 is negative 30. We have minus 4 times 1/2. That's minus 2. And does that equal negative 32? Sure it does. So that means that the solution to this system is the ordered pair negative 10, 1/2. It's time for a quick quiz. Where do the lines 3x plus y equals 4 and 3x plus y equals 5 intersect? A. At the point 0 comma negative 1. B. At the point 3 halfs comma negative one halfs. C. They do not intersect. Choose A, B, or C. You're correct, they do not intersect. Sorry the answer is C, they do not intersect. By inspection, we see that the left hand side of these equations are exactly the same 3x plus y so you can't have 3x plus y equal to 4 and 3x plus y equal to 5 exactly the same time. Hence these 2 lines cannot intersect each other. Now, there's another method we want to use in order to solve systems of equations, and it can be called the elimination method, or sometimes it's called the addition method. Either term is acceptable, because really in combination, that's what you're trying to do. You're trying to add the two equations together, but in the process, you want to eliminate one of the variables. So when I look at the system that's given here, which is 5x minus 2y equals 4, and 4x plus 3y equals 17, notice that if I decide to add the two equations together at this juncture, we would end up with 5x plus 4x, which is 9x, and we'd have minus 2y plus 3y, which would give us a y, equals the value 21. So even though I can add the two equations together at this point, they don't give me a single equation with a single variable. So always remembering what is your purpose in doing the addition is important. Now, if I can't do it in the current situation, then what can I do in order to help myself? Well, I need to get a coefficient on either the x's or the y's that is exactly opposite each other, or in other words, the negative of each other. I can notice that for my y values, one of them is already a negative value-- that's negative 2y-- and the other one is a positive 3y. That doesn't make it particularly easier, but I can use this in order to decide that what I'm going to try to do is to eliminate the y's when I add the two equations together. That requires me to have the same coefficient, only opposite signs. So I can accomplish that by looking for the least common multiple of my coefficients. If I look at the negative 2 and the 3, and just disregarding the sign for right now, we can say the least common multiple of 2 and 3 is the value 6. How can I end up with a negative 6y in the first equation and a positive 6y in the second equation? Well, we need to remember a property of equality. That is, I can multiply both sides of any equation by a non-zero value, and end up with an equivalent equation. So having said that, if I'd like to get the first equation to have a negative 6y, I can accomplish that by multiplying both sides of the equation by positive 3. So let's just write that down. We're going to have 3 multiplied times the 5x minus 2y on the left-hand side, equals 3 multiplied times the 4 on the right-hand side. Let's go ahead and simplify that. We'd have 15x minus 6y equals the value of 12. So I've ended up with my minus 6y, which was my purpose. In the second equation, I'd like to get a positive 6y. And I can accomplish that by multiplying both sides of the equation by positive 2. So again, writing this down we're, going to have 2 multiplied times the left-hand side. That's 2 multiplied times 4x plus 3y equals 2 multiplied times the right-hand side-- that's 2 times 17. Simplifying by using distributive property on the left, we end up with 8x plus 6y equals 34. So now if I take the two equations that were the result of my multiplication, we end up rewriting this system as 15x minus 6y equals 12. And the second equation can be written as 8x plus 6y equals 34. Now I'm just going to draw a line and simply add the two equations together. When I add 15x plus 8x, I end up with 23x. When I add negative 6y and positive 6y, I accomplish the purpose of my multiplication here. That is, I end up eliminating the variable y, because negative 6 plus 6 is 0, and 0y is 0. On the right-hand side, adding the two numbers together, 12 plus 34 is going to give me 46. So now I've been able to combine the two equations into a single equation with a single variable. That allows me to solve for the variable x here. We can divide both sides of the equation by 23, and that gives us x equals 2. Now, I could go through the process once more. And this time, instead of eliminating the value of y, I can eliminate the value of x. But once we have one of the solutions, the simplest thing to do is to substitute back into one of the equations. You can choose either equation you'd like to, and substitute in the value for x So if I just go with the first equation, we had 5x minus 2y equals 4, and I substitute in the value of 2 for x. We end up with 5 times 2 minus 2y equals 4. That's going to give us 10 minus 2y equals 4. Subtracting 10 from both sides of the equation, negative 2y is going to equal negative 6, and dividing both sides of the equation by negative 2 gives us y equals 3. So I come up with what I think is the correct solution that is going to have the ordered pair where x is equal to 2 and my y value is equal to 3. But again, it is so easy to check this work, and you definitely want to do it every time you solve a system. So let's check by substituting into the original equations, and make sure that we end up with the correct solution to this system. So we're going to have 5 times 2 minus 2 times 3. That would give us 10 minus 6, which does equal 4. So that checks. In the second equation, we would have 4 times 2 plus 3 times 3. And that's going to give us 8 plus 9, which does equal 17. And so this checks, which means that the ordered pair 2, 3, is going to be the single point where these two lines intersect. It's time for another quick quiz. Where do the lines x minus y equals 3 and 2x minus 2y equals 6 intersect? A. At the point 3,6. B. At all points on x minus y equals 3. C. At all points in the plane. Choose A, B, or C. You're correct, the answer is B, At all points on x minus y equals 3. Sorry the answer is B, At all points on x minus y equals 3. If we take line x minus y equals 3 and we multiply both sides of that equation by the value of 2 we end up with 2x minus 2y equals 6 which is the second equation that were given. So these 2 lines are actually the same line. Now in part A, the possible answer was 3 comma 6 but in order for part A to be correct I would have to substitute the 3 for x and the 6 for y and end up with a true statement 3 minus 6 is equal to negative 3 and negative 3 does not equal 3 so A is not the correct answer. And in answer C, these 2 lines are not going to intersect at all points in the plane because they don't cover all points in the plane they only cover points that are all on the line itself hence B is the correct answer. So our next thing that we're going to look at will be linear systems of three variables with three equations. Now, I have an equation here. We have x plus y plus z equals 4. You may or may not be familiar with graphing a system where you end up having-- or an equation, I should say, where you end up having three variables. So let's first of all think about the fact that when you're in two variables and you're graphing the x and the y on the xy plane, you're ending up getting a straight line. So it's a linear equation because when we look at the graph, it's going to be a straight line. When we talk about three variables, we're looking at a 3D space. And just very quickly, I'm going to think about what this might look like graphically. So to grab something in three dimensions, we use an optical illusion. So I'm still going to do an axis, which is going to be an x-axis. But I'm going to draw in a line here. And that's actually going to be our y-axis. And then going straight up, that's going to be our z access. Now, notice I'm not including the what would be considered the negative portions of these in my picture. This is, as I said, an optical illusion. I'm drawing on a flat surface, but I'm trying to make it look like it's three-dimensional. If I'm trying to graph this particular equation, then I can think about, well, can I mimic something that I did previously? Yes, we can try to do intercepts. And we can do that this time by thinking about instead of ordered pairs, we have ordered triples. We have to know what the x value is, the y value is, and z value. Now, if I look at the x value of 4, what would the y and z value have to be to make this statement true along the x-axis? Well, that means that I'm not moving off of the axis. That is, I'm going to move to the right four units. I'm not going to move up. I'm not going to move out. So my y and z values are both going to be 0. This is an ordered triple. This point right here corresponds to the point 4, 0, 0. And notice that that would satisfy the equation. So I'm going to plot that point. What would the y-intercept look like? Well, I would be looking at going along the y-axis, but I would not be moving off of the axis. So I wouldn't be moving right or left. I wouldn't be moving up or down. So that means my x and my z would have to equal 0. And if x and z are both 0, then y has to equal 4. So going out in this direction, 1, 2, 3, 4, I have the point which corresponds to an ordered triple 0, 4, 0. And finally, what would the z-intercept look like? Well, in the same manner, if I stay on the z-axis-- that means I don't move right or left, I don't come out this way toward the y-- and therefore, my x and y are both 0, but my z value then has to equal the value 4 in order to make the equation true. So right here, I plot this point. I end up with the coordinates for that ordered triple, which would be 0, 0, 4. Now, what in the world does that look like? Well, I'm going to pretty much just connect the dots in order for you to see what kind of shape that could possibly be. Now, when I connect the dots, of course, it looks like a triangle. But if you extend this, this is going to be all three points lying in the same plane. So when we graph a linear equation in three variables, we do not end up with a straight line. Instead, we end up with a plane of values, all of which would solve the given equation. And again, what we're looking at are ordered triples. Now, having given you all this information, you can see that it's tough enough just getting the intercepts for a plane, and I really don't think I want to start trying to graph a system of three equations and three unknowns in order to try to find the solutions. But we can look to what the solution to three equations and three unknowns might look like. So if our graphical process is not the one that we're going to try to use, we still have two other methods that we can employ. The first one is substitution, and I'm going to illustrate that using the following example. We have three equations and three unknowns. The first equation reads 3x minus y plus 4z equals 11. The second equation, x minus 2y minus z equals 3. And the third equation is 2x plus 3y plus z equals 2. Now, if I want to employ substitution to solve this system, I'd like to substitute by solving first for one of the variables in one of the equations. Well, you have a number of choices. But as what happened when we had the two equation, two variable systems, I'm going to look for an easy variable to solve for. In my first equation, I notice that all of my coefficients are not equal to 1. In my second equation, I notice that the x value has a coefficient of 1, which would make it easy to solve for. In my third equation, the z value has a coefficient of 1, which would make it easy to solve for. I'm just going to pick one of those. I'm going to solve for x in the second equation. So in the second equation, I can write it as now x equals, and I will add 2y to both sides of the equation. That's going to give me x equals 2y, plus we're going to add z to both sides of the equation. So that's going be plus z and then plus 3. So again, taking the second equation, I can rewrite it as x equals 2y plus z plus 3. Now, I have two other equations. What will happen when we end up with three equations and three unknowns is I'll solve for one of the variables, and then substitute that into both of the other two equations I haven't worked with yet. So let's look at it now. We'll have our first equation, which will read, in this case, 3 times. We'll use the value of x that I solve for. That's going to be 2y plus z plus 3. And then write the rest of the equation down. Minus y plus 4z equals 11. And now I want to simplify this. So it's going to give us 6y plus 3z plus 9 minus y plus 4z equals 11. Let's combine like terms. We notice we have a 6y minus y. That's going to give us 5y. We have a 3z plus 4z. That's going to give us 7z. And I'm going to go ahead and subtract 9 from both sides of the equation. And so I'll have 5y plus 7z equals 2. Now, what's important is that you don't lose track of what you're doing when you're working with three equations and three unknowns. Remember, I solved for x from the second equation. I've substituted into the first equation, and come up with an equation that doesn't have an x in it anymore. I now need to do the same thing with that third equation that I have in my system. That is, I want to substitute for x in the third equation. So we'll end up having our 2 times the value of x, which was 2y plus z plus 3. And then write the rest of the equation. Plus 3y plus z equals 2. So again, this is the third equation I'm working with. Simplifying, we end up with 4y plus 2z plus 6 plus 3y plus z equals 2. Combine like terms. 4y plus 3y is 7y. 2z plus z is going to equal 3z. And I will subtract 6 from both sides of the equation to end up with 2 minus 6, which is negative 4. So now let's look at what we've accomplished. We now have an equation that has only a y and a z in it, and we have another equation that has only a y and z in it. We can now see that we have a system of two equations and two unknowns. So just rewriting, we have 5y plus 7z equals 2, and then we have 7y plus 3z equals negative 4. We're now going to concentrate on solving this system. Again, since I'm using substitution, let's just continue with this idea. I'm going to choose in my first equation to solve for y. Notice that none of my coefficients are equal to 1 here, so it's pick anything, and solve for it, and then substitute. So now in the first equation, if I solve for y, I'll end up with 5y equals 2 minus 7z. And then I can divide both sides of the equation by 5 to give us y equals 2 minus 7z divided by 5. What do we do at this point? We have solved for y in the first equation. We substitute for y in the second equation. So in my second equation, we're now going to have 7 times the value of y, which is going to be the 2 minus 7z over 5. And then we have the plus 3z equals our negative 4. Let's draw a line here to separate. Now, you have some options here. You can go ahead and multiply through. But I'm going to decide just to go ahead and rid the equation of fractions in this step by multiplying everything by 5. When I multiply the first term by 5, notice that I have two things multiplied together, and then that's been multiplied times 5. So I'm going to end up with the 7 times the 2 minus 7z. That's been multiplied by 5, so I lose my denominator. Plus, multiplying by 5 gives me 15z. And multiplying by 5 gives me negative 20. Distributive property yields 14 minus 49z plus 15z equals negative 20. Let's go ahead and combine like terms. We have a negative 49z plus 15z. That's going to give us a negative 34z. If I subtract 14 from both sides of the equation, I end up with negative 20 minus 14, which is negative 34. And now dividing both sides of the equation by negative 34 gives me z equals the value of 1. Well, the minute I find a value for one of my variables, I can now start what we could call back substitution. I can solve for y in the equation where I have y equals the value 2 minus 7z divided by 5. So my y value is now going to equal 2 minus 7 times 1, which is going to be 7, and then divide by 5. That's going to be negative 5 divided by 5, which is negative 1. Once I solve for my y value, I can now go back and solve for my x value using the original equation where I solve for x. My x is going to equal 2 times the y value-- that's going to be 2 times negative 1-- plus the z value, which I determined was positive 1, plus 3. So x is going to equal negative 2 plus 1 plus 3. That's going to be 4 minus 2, which equals 2. Now, let's put it all together. What have we determined here? We determined that if our x value is 2, our y value is negative 1, and our z value is 1, then I'm going to end up having a solution. So I have an ordered triple, which I'm going to go ahead-- and let's just write it up here next to the system. The ordered triple is going to have coordinates 2, negative 1, positive 1. And that ordered triple is tentatively our solution. Now, I'm going to leave the check to you, but what you should do next is to substitute those three values in for x, y, and z, and ensure that it is true for all equations that we end up equaling the constants that are to the right. Now, it's also possible to use the method of elimination, or addition, in order to solve a three equation, three variable system. And so I'm going to illustrate this on this next example. We have three equations. The first reads x minus 4y plus z equals negative 7. The second is 2x minus 6y minus z equals negative 7. The third, 4x plus 2y minus 3z equals negative 5. Now, in order to work with elimination, we need to look at the equations and we're going to be putting them together pairwise, trying to eliminate the exact same variable. The key to being successful using this method is to have some structure to what you're doing. So when I look at, for instance, the first two equations, I notice that I have z and I have a negative z. Those equations can be easily added together in order to eliminate the value of z. And then I notice that the first and the third could also be worked with if I do a little bit of manipulation. So I make a determination of what variable I want to eliminate, and then go with it. And so I'm going to start with my first two equations and eliminate the value of z. So this is going to be equation 1 plus equation 2. We're going to have x minus 4y plus z equals negative 7. Our second equation, 2x minus 6y minus z equals negative 7. Adding together, we end up with 3x, that's going to be minus 10y. The z's will subtract out, and that's going to equal negative 14. And so I end up with the first equation, which now does not have a z in it. I also need to eliminate z, but I need to do a pairwise type of a grouping. And I notice that if I take my first equation and my third equation, I can add those together and eliminate z if I multiply the first equation by 3. So I'm going to write that down. It's going to be 3 multiplied times equation 1, and then add that to equation 3. So again, I'm just trying to be structured here and keep track of what I'm trying to do. So in my first equation, I'm going to rewrite it as 3x minus. 4 times 3 is going to be 12y. And then multiply by 3, plus 3z is going to equal the negative 7 times 3, which is negative 21. My third equation I'm going to leave as is. That's going to give us 4x plus 2y minus 3z equals our negative 5. And let's add those together. 3x plus 4x is going to be 7x. When I add negative 12y and 2y, I end up with negative 10y. When I add 3z and minus 3z, I end up with 0, which eliminates the z. That's great. And on the right-hand side, negative 21 minus 5 is going to give us negative 26. So let's just look at what we've accomplished here. We've ended up taking our equations pairwise, and we have now two equations that only have an x and a y in them. I'm just going to rewrite that, but in doing so, notice that if I think a step ahead, I can save myself a little bit of work. In that, if I look at the fact that the coefficient in front of y is a negative 10 in both of them, if I multiply one of those by a negative 1, I'm going to end up with a perfect solution for adding them together in the next step. So I'm going to take my first equation that I ended up with here, the 3x minus 10y equals negative 14. I'm just going to multiply through by negative 1. Let's rewrite it. We end up with negative 3x. We're going to multiply by negative 1, so it's going to be plus 10y. And then we're going to multiply by negative 1. That's going to equal positive 14. So again, just notice that I'm going to rewrite this in another format that's equivalent. My second equation now, if I write it right underneath here, we have 7x, we have minus 10y equals negative 26. Now, these worked out pretty nicely. It was pretty easy to come to this point where now when I add the two equations together, I'm going to end up with an elimination of the variable y. If it wasn't that easy, we would do some manipulation as we did when we had a system of two equations and two unknowns in order to come up with an elimination process. Adding negative 3x plus 7x is going to give us 4x. 10y minus 10y is going to be 0. On the right-hand side, 14 minus 26 is going to be a negative 12. And now dividing both sides of the equation by the value of 4 gives us that x equals negative 3. Now, once I have a value for x, I can substitute back into one of the equations that only has an x and a y in it in order to solve for y. So let's do that now. I'm just going to take my second equation here. It's 7 times my x value, which is negative 3, minus 10y equals negative 26. We're going to have negative 21 minus 10y equals negative 26. We're going to add 21 to both sides of the equation. That's going to give us negative 10y equals. That's going to be a negative 26 plus 21, which is going to be a negative 5. And dividing by negative 10 gives us y equals negative 5 divided by negative 10, which is equivalent to 1/2. So we have an x value equal to 3, a y value equal to 1/2, and now I can solve for z. I can take those two values for x and y, and substitute back into one of the equations that had all three variables in them. So let's just take the first one. We're going to have the equation x minus 4y plus z equals negative 7. I'm going to substitute in the value of x, which is negative 3. So that's going to be negative 3. Minus 4 times the value of y that I found, which is 1/2, plus z equals negative 7. And now notice I have a single equation and a single unknown. Let's simplify. Negative 3. 4 times 1/2 is going to be 2, so I end up with negative 3 minus 2, plus z equals negative 7. Negative 5 plus z equals negative 7. And if I add 5 to both sides of the equation, we end up with z equals negative 7 plus 5, which is negative 2. So putting these together, we end up with an ordered triple where my x value is negative 3, my y value is 1/2, and my z value is negative 2. And as in all other cases when we're working with our linear systems, we want to make sure we perform a check. I've run out of space. I'm going to leave that to you. So you should go back to all three equations, substitute in, and ensure that this indeed is a solution. It's time now for you to try some more problems on your own. Good luck.

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