Systems of Linear Equations

Systems of Linear Equations

Definitions.

A "system" of equations is a set or collection of equations that you deal with all together at once. Linear equations (ones that graph as straight lines) are simpler than non-linear equations, and the simplest linear system is one with two equations and two variables.Think back to linear equations. For instance, consider the linear equation y = 3x 5. A "solution" to this equation was any x, y-point that "worked" in the equation. So (2, 1) was a solution because, plugging in 2for x:

3x 5 = 3(2) 5 = 6 5 = 1 = yOn the other hand, (1, 2) was not a solution, because, plugging in 1for x:3x 5 = 3(1) 5 = 3 5 = 2

which did not equal y (which was 2, for this point). Of course, in practical terms, you did not find solutions to an equation by picking random points, plugging them in, and checking to see if they "work" in the equation. Instead, you picked x-values and then calculated the corresponding y-values. And you used this same procedure to graph the equation. This points out an important fact: Every point on the graph was a solution to the equation, and any solution to the equation was a point on the graph.Now consider the following two-variable system of linear equations:y = 3x 2y = x 6

Since the two equations above are in a system, we deal with them together at the same time. In particular, we can graph them together on the same axis system, like this:

A solution for a single equation is any point that lies on the line for that equation. A solution for a system of equations is any point that lies on each line in the system. For example, the red point at right is not a solution to the system, because it is not on either line:

The blue point at right is not a solution to the system, because it lies on only one of the lines, not on both of them:

The purple point at right is a solution to the system, because it lies on both of the lines:

In particular, this point marks the intersection of the two lines. It solves both equations, so it solves the system. And this is always true: For systems of equations, "solutions" are "intersections". You can confirm the solution by plugging it into the system of equations, and confirming that the solution works in each equation. Determine whether eitherofthe points (1, 5) and (0, 2) is a solution to the given system of equations.

y = 3x 2y = x 6To check the given possible solutions, I just plug the x- and y-coordinates into the equations, and check to see if they work. Copyright Elizabeth Stapel 2006-2008 All Rights Reservedchecking (1, 5):

(5) ?=? 3(1) 25 ?=? 3 25 = 5 (solution checks)(5) ?=? (1) 65 ?=? 1 65 = 5 (solution checks)Since the given point works in each equation, it is a solution to the system. Now I'll check the other point (which we already know, from looking at the graph, is not a solution):checking (0, 2):

(2) ?=? 3(0) 22?=? 0 22 = 2 (solution checks)So the solution works in one of the equations. But to solve the system, it has to work in both equations. Continuing the check:(2) ?=? (0) 62 ?=? 0 62 ?=? 6

But 2does not equal 6, so this "solution" does not check. Then the answer is:the solution is the point (1, 5)

Systems of Linear Equations:

A) Solving by Substitution

This method works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this into the other equation, "substituting" for the chosen variable and solving for the other. Then you back-solve for the first variable. Here is how it works. Solve the following system by substitution.

2x 3y = 24x + y = 24The idea here is to solve one of the equations for one of the variables, and plug this into the other equation. It does not matter which equation or which variable you pick. There is no right or wrong choice; the answer will be the same, regardless. But some choices may be better than others.For instance, in this case, can you see that it would probably be simplest to solve the second equation for "y =", since there is already a y floating around loose in the middle there? I could solve the first equation for either variable, but I'd get fractions, and solving the second equation for xwould also give me fractions. It wouldn't be "wrong" to make a different choice, but it would probably be more difficult. Being lazy, I'll solve the second equation for y:4x + y = 24y = 4x + 24

Now I'll plug this in ("substitute it") for "y" in the first equation, and solve for x:2x 3(4x + 24) = 22x + 12x 72 = 214x = 70x = 5

Now I can plug this back into either equation, and solve for y. But since I already have an expression for "y =", it will be simplest to just plug into this:y = 4(5) + 24 = 20 + 24 = 4

Then the solution is (x, y) = (5, 4).If I had substituted into the same equation as I'd used to solve for "y =", I would have gotten a true, but useless, statement: Copyright Elizabeth -2008 All Rights Reserved4x + (4x + 24) = 244x 4x + 24 = 2424 = 24

Twenty-four does equal twenty-four, but who cares? So when using substitution, make sure you substitute into the other equation, or you'll just be wasting your time. Solve the following system by substitution.

y = 36 9x3x + y/3 = 12We already know (from the previous lesson) that these are actually both the same line; that is, this is a dependent system. We know what this looks like graphically: we get two identical line equations, and a graph with just one line displayed. But what does this look like algebraically?The first equation is already solved for y, so I'll substitute that into the second equation:3x + (36 9x)/3 = 123x + 12 3x = 1212 = 12

Well, um... yes, but...I did substitute the first equation into the second equation, so this unhelpful result is not because of some screw-up on my part. It's just that this is what a dependent system looks like when you try to find a solution. Remember, when you're trying to solve a system, you're trying to use the second equation to narrow down the choices of points on the first equation. You're trying to find the one single point that works. But in a dependent system, the "second" equation is really just another copy of the first equation, and all the points on the line work.In other words, I got an unhelpful result because the second line equation didn't tell me anything new.solution: y = 36 9xThis is always true, by the way. When you try to solve a system and you get a statement like "12 = 12" or "0 = 0" something that's true, but unhelpful (I mean, duh!, of course twelve equals twelve!) then you have a dependent system. We already knew, from the previous lesson, that this system was dependent, but now you know what the algebra looks like.(Note that your text may format the answer to look somethinglike "(t, 36 9t)", or something similar. This form means the same thing as "y = 36 9x".) Solve the following system by substitution.

7x + 2y = 1621x 6y = 24Neither of these equations is particularly easier than the other for solving. I'll get fractions, no matter which equation and which variable I choose. So, um... I guess I'll take the first equation, and I'll solve it for, um, y, because at least the 2 (from the "2y") will divide evenly into the 16.7x + 2y = 162y = 7x + 16y = (7/2 )x + 8

Now I'll plug this into the other equation:21x 6((7/2 )x + 8) = 2421x + 21x 48 = 2448 = 24

Um... I don't think so....In this case, I got a nonsense result. All my math was right, but I got an obviously wrong answer. So what happened?Remember that, when solving, you're trying to find where the lines intersect. What if they don't intersect? Then you're going to get some kind of wrong answer when you assume that there is a solution(as I did when I tried to find that solution). We knew, from the previous lesson, that this system represents two parallel lines. But I tried, by substitution, to find the intersection point anyway. Since there wasn't any intersection point, my attempt led to nonsense.solution: no solution (inconsistent system)This is always true, by the way. When you get a nonsense result, this is the algebraic indication that the system of equations is inconsistent.Note that this is quite different from the previous example. A useless result (like "12 = 12") is quite different from a nonsense result (like "48 = 24"), just as two identical lines are quite different from two parallel lines. Don't confuse the two. A useless result means a dependent system which has a solution (the whole line); a nonsense result means an inconsistent system which has no solution of any kindSystems of Linear Equations:

B) Solving by Addition / Elimination

The addition method is also called the method of elimination. This method is similar to the method you probably learned for solving simple equations. If you had the equation "x + 6 = 11", you would write"6" under either side of the equation, and add down to get "x= 5" as the solution.x + 6 = 116 6x = 5

You'll do something similar with the addition method. Solve the following system using addition.

2x + y = 93x y = 16Note that, if I add down, the y's will cancel out. So I'll draw an "equals" bar under the system, and add down:2x + y = 93x y = 165x = 25

Now I can solve for x = 5, and then backsolve, using either of the original equations, to find the value of y. The first equation has smaller numbers, so I'll back-solve in that one:2(5) + y = 910 + y = 9y = 1

Then the solution is (x, y) = (5, 1).It doesn't matter which equation you use for the backsolving; you'll get the same answer either way. If I'd used the second equation, I'd have gotten:3(5) y = 1615 y = 16y = 1y = 1

which is the same result as before. Solve the following system using addition.

x 2y = 9x + 3y = 16Note that the x-terms would cancel out if only they'd had opposite signs. I can create this cancellation by multiplying either one of the equations by 1, and then adding down as usual. It doesn't matter which equation I choose, as long as I am careful to multiply the 1through the entire equation. I'll multiply the second equation.

The "1R2" notation over the arrowindicates that I multiplied row 2by 1. Now I can solve the equation"5y = 25" to get y = 5. Backsolving, I get:x 2(5) = 9x 10 = 9x = 1

Then the solution is (x, y) = (1, 5).A very common temptation is to write the solution in the form "(first number I found, second number I found)". Sometimes, though, as in this case, you find the y-value first and then the x-value second, and of course in points the x-value comes first. So just be careful towrite the coordinates for your solutions correctly. Copyright Elizabeth Stapel 2006-2008 All Rights Reserved Solve the following system using addition.

2x y = 93x + 4y = 14Nothing cancels here, but I can multiply to create a cancellation. I can multiply the first equation by 4, and this will set up the y-terms to cancel.

Solving this, I get that x = 2. I'll use the first equation for backsolving, because the coefficients are smaller.2(2) y = 94 y = 9y = 5y = 5

The solution is (x, y) = (2, 5). Solve the following system using addition.

4x 3y = 253x + 8y = 10Hmm... nothing cancels. But I can multiply to create a cancellation. In this case, neither variable is the obvious choice for cancellation. I can multiply to convert the x-terms to 12x's or the y-terms to 24y's Since I'm lazy and 12is smaller than 24, I'll multiply to cancel the x-terms. (I would get the same answer in the end if I set up the y-terms to cancel. It's not that how I'm doing it is "the right way"; it was just my choice. You could make a different choice, and that would be just as correct.)I will multiply the first row by 3and the second row by 4; then I'll add down and solve.

Solving, I get that y = 5. Neither equation looks particularly better than the other for backsolving, so I'll flip a coin and use the first equation.4x 3(5) = 254x 15 = 254x = 40x = 10

Remembering to put the x-coordinate first in the solution, I get:(x, y) = (10, 5)

Usually when you are solving by addition, you will need to create the cancellation. The most common mistake is to forget to multiply all the way through the equation, on both sides of the "equals" sign. Be careful of this. Solve the following using addition.

12x 13y = 26x + 6.5y = 2I think I'll multiply the second equation by 2; at least this will get rid of the decimal place.

Oops! This isn't true! So this is an inconsistent system (two parallel lines) with no solution (with no intersection point).no solution Solve the following using addition.

12x 3y = 64x y = 2I think it'll be simplest to cancel off the y-terms, so I'll multiply the second row by 3.

Well, yes, but who cares? I already knew that zero equals zero. So this is a dependent system, and, solving for "y =",the solution is:y = 4x 2(Your text may format the answer as "(s, 4s 2)", or something like that.)

Remember the difference: a nonsense answer (like "0 = 2" in the previous problem) means an inconsistent system with no solution; a useless answer (like "0 = 0" above) means a dependent system where the whole line is the solution.Some books use only "x" and "y" for their variables, but many use additional variables. When you write the solution for an x,y-point, you know that the x-coordinate goes first and the y-coordinate goes second. When you are dealing with other variables, assume that they are written in alphabetical order. For instance, if the variables in a given system are aand b, the solution point would be (a, b); it would not be (b, a). Unless otherwise specified, the variables are written in alphabetical order.Systems of Linear Equations:

C) Solving by Gaussian Elimination

Solving three-variable, three-equation linear systems is more difficult, at least initially, than solving the two-variable systems because the computations involved are more messy. You will need to be very neat in your working, and you should plan to use lots of scratch paper. The method for solving these systems is an extension of the two-variable solving-by-addition method, so make sure you know this method well and can use it consistently correctly.However, though the method of solution is addition/elimination, just trying to do addition would get very messy, so there is a systematized method for solving the three-or-more-variables systems. This method is called Gaussian elimination. Note: While Gaussian elimination reduces the messiness of solving, you will still need plenty of scratch paper for doing the steps. Let's start simple, and work our way up. Solve the following system of equations.

5x + 4y z = 010y 3z = 11 z = 3It's fairly easy to see how to proceed in this case. I'll just back-substitute the z-value from the third equation into the second equation, solve for y, and then plug z and yinto the first equation and solve for x.10y 3(3) = 1110y 9 = 1110y = 20y = 2

5x + 4(2) (3) = 05x + 8 3 = 05x + 5 = 05x = 5x = 1

Then the solution is (x, y, z) = (1, 2, 3).The reason this system was easy to solve is that the system was "triangular"; this refers to the equations having the form of a triangle, because of the lower equations containing only the later variables.

The point is that, in this format, the system is simple to solve. And Gaussian elimination is the method we'll use to convert systems to this upper triangular form, using the row operations we learned when we did the addition method. Solve the following system of equations using Gaussian elimination.

3x + 2y 6z = 65x + 7y 5z = 6 x + 4y 2z = 8No equation is solved for a variable, so I'll have to do the multiplication-and-addition thing to simplify this system. In order to keep track of my work, I'll write down each step as I go. But I'll do my computations on scratch paper. Here is how I did it:The first thing to do is to get rid of the leading x-terms in two of the rows. For now, I'll just look at which rows will be easy to clear out; I can switch rows later to get the system into upper triangular form. There is no rule that says I have to use the x-term from the first row, and, in this case, I think it will be simpler to use the x-term from the third row. So I'll multiply the third row by 3, and add it to the first row. I do the computations on scratch paper:

...and then I write down the results:

(When wewere solving two-variable systems, wecould multiply a row, rewriting the system off to the side, and then add down. There is no space for this in a three-variable system, which is why weneed the scratch paper.)Since I didn't actually do anything to the third row, I copied it down, unchanged, into the new matrix. I used the third row, but I didn't actually change it. Don't let this confuse you.To get smaller numbers, I'll multiply the first row by one-half:

Now I'll multiply the third row by 5 and add this to the second row. I do my work on scratch paper:

...and then I write down the results: Copyright Elizabeth Stapel 2006-2008 All Rights Reserved

I didn't do anything with the first row, so I copied it down unchanged. I worked with the third row, but I only worked on the second row, so the second row is updated and the third row is copied over unchanged.Okay, now the x-column is cleared out except for the leading term inthe third row. So now I have to work on the y-column. If I add twice the first row to the second this will give me a leading 1 in the second row. I won't have gotten rid of the leading y-term in the second row, but I will have converted it (without getting involved in fractions) to a form that is simpler to deal with. (You should keep an eye out for this sort of simplification.) First I do the scratch work:

...and then I write down the results:

Now I can use the second row to clear out the y-term in the first row. I'll multiply the second row by 7 and add. First I do the scratch work:

...and then I write down the results:

I can tell what zis now, but, just to be thorough, I'll divide the first row by 43. Then I'll rearrange the rows to put them in upper-triangular form:

Now I can start the process of back-solving:y 7(1) = 4y 7 = 4y = 3

x + 4(3) 2(1) = 8x + 12 2 = 8x + 10 = 8x = 2

Then the solution is (x, y, z) = (2, 3, 1).Note: There is nothing sacred about the steps I used in solving the above system; there was nothing special about how I solved this system. You could work in a different order or simplify different rows, and still come up with the correct answer. These systems are complicated enough that there is unlikely to be one right way of computing the answer. So don't stress over "how did she know to do that next?", because there is no rule. I just did whatever struck my fancy; I did whatever seemed simplest or whatever came to mind first. Don't worry if you would have used completely different steps. As long as each step along the way is correct, you'll come up with the same answer.Note also: I could have been more thorough-going in my row operations, and cleared out all the y-terms other than that in the second row and all the z-terms other than that in the first row. This is what the process would then have looked like:

This way, I can just read off the values of x, y, and z, and I don't have to bother with the back-substitution. This more-complete method of solving is called Gauss-Jordan elimination. Many texts only go as far as Gaussian elimination, but I've always found it easier to continue on and do Gauss-Jordan.Note that I did two row operations at once in that last step before switching the rows. As long as I'm not working with and working on the same row, this is okay. In this case, I was working with the first row and working on the second and third rows.Systems of Linear Equations:

D) Solving by Graphing

Thinking graphically, when you are solving systems, you are finding intersections. For two-variable systems, there are three possible types of solutions:Case 1Case 2Case 3

The first graph shows two distinct non-parallel lines that cross at exactly one point. This is called an "independent" system of equations, and the solution is always some x,y-point.Independent system:one solution pointCase 2Case 3

The second graph shows two distinct lines that are parallel. Since parallel lines never cross, then there can be no intersection; that is, for parallel lines, there can be no solution. This is called an "inconsistent" system of equations, and it has no solution.Independent system:one solution andone intersection pointInconsistent system:no solution andno intersection pointCase 3

The third graph appears to show only one line. Actually, it's the same line drawn twice. These "two" lines, really being the same line, then "intersect" at every point along their length. This is called a "dependent" system, and the "solution" is the whole line.Independent system:one solution andone intersection pointInconsistent system:no solution andno intersection pointDependent system:the solution is thewhole line

This shows that a system of equations may have one solution (a specific x,y-point), no solution at all, or an infinite solution (being all the solutions to the equation). You will never have a system with two or three solutions; it will always be one, none, or zillions.

Probably the first method you'll see for solving systems of equations will be "solving by graphing". As I said before, you have to take these problems with a grain of salt. The only way you can find the solution from the graph is IF you draw a very neat axis system, IF you draw very neat lines, IF the solution happens to be a point with nice neat whole-number coordinates, and IF the lines are not close to being parallel.Copyright Elizabeth Stapel 2006-2008 All Rights ReservedFor instance, if the lines cross at a shallow angle it can be just about impossible to tell where the lines cross.

And if the intersection point isn't a neat pair of whole numbers, all bets are off.(Can you tell by looking that the displayed solutionhas coordinates of (4.3, 0.95)? No? Then you see my point.)

On the plus side, since they will be forced to give you nice neat solutions for "solving by graphing" problems, you will be able to get all the right answers as long as you graph very neatly. For instance: Solve the following system by graphing.

2x 3y = 24x + y = 24I know I need a neat graph, so I'll grab my ruler and get started. First, I'll solve each equation for "y=", so I can graph easily:2x 3y = 22x + 2 = 3y(2/3)x + (2/3) = y4x + y = 24y = 4x + 24

The second line will be easy to graph using just the slope and intercept, but I'll need a T-chart for the first line.xy = (2/3)x + (2/3)y = 4x + 24

48/3 + 2/3 = 6/3 = 216 + 24 = 40

12/3 + 2/3 = 04 + 24 = 28

24/3 + 2/3 = 6/3 = 28 + 24 = 16

510/3 + 2/3 = 12/3 = 420 + 24 = 4

816/3 + 2/3 = 18/3 = 632 + 24 = 8

(Sometimes you'll notice the intersection right on the T-chart. Do you see the point that is in both equationsabove? Check the grayed-out row.)Then I grab my ruler and graph neatly, and look for the intersection:Even if I hadn't noticed the intersection point in the T-chart, I can certainly see it from the picture.

solution: (5, 4)Systems of Linear Equations:

E) Warnings

Most "solving by graphing" problems work nicely, but sometimes they'll give you an inconsistent system (that is, two parallel lines) or a dependent system (that is, two forms of the same line equation). This is what these cases will look like: Solve the following system by graphing.

y = 36 9x3x + y/3 = 12As usual, I'll first solve the equations for "y =". The first one is already solved, so now I'll solve the second one: Copyright Elizabeth Stapel 2006-2008 All Rights Reserved3x + y/3 = 129x + y = 36y = 36 9xThese are really both the same line! So the algebra tells me thatthis is a dependent system, and the solution is the whole line. Of course, this is a "solving by graphing" problem, so I still have to do the graph, but I already know the answer.

solution: y = 36 9xThe solution to this system is the whole line, so, in my classes, you could give the answer as being"y = 36 9x". However, most books do something like this: You are looking for an x,y-solution, and, in this case, x = x and y = 36 9x, so the solution "point" is of the form(x, 36 9x). But then the book does this weird thing with "a" (or "t" or "s" or some other variable). Instead of using x, which is a perfectly good variable, they pull out thisnew variable from behind their left ear and give the solution as being"(a, 36 9a)". I have no idea why they do this, but if they do, then this is the format that your teacher will want on the test, so memorze the variable that your particular book uses (which was "a" in this example). Solve the following system by graphing.

7x + 2y = 1621x 6y = 24As usual, I'll first solve each equation for "y =":7x + 2y = 162y = 7x + 16y = ( 7/2 )x + 8

21x 6y = 2421x 24 = 6y( 21/6 )x 4 = y( 7/2 )x 4 = yThese lines have the same slope (m = 7/2 )but different y-intercepts, so they are parallel. Since parallel lines never cross, the algebra tells me that this is an inconsistent system; that is, there is no solution. But this is a "solving by graphing" problem, so I still have to draw the picture.

solution: no solution (inconsistent system)Note: When the algebra tells you that you have two parallel lines, for heaven's sake, draw the lines on your graph so they look parallel!

Note: The solution to a dependent system, being all the points along the line, contains infinitely-many points. But don't make the mistake of thinking that "infinitely-many" means "all". Any point off the line is not a solution; only the infinity of point on the line will solve the dependent system.Also note: The pictures on the firstpage are very useful for explaining "what's going on" with linear systems, but pictures are not terribly useful for finding actual solutions to systems. For instance, in the picture at right, is the solution point at (3, 2) or at (3.15, 1.97)?You can't tell!

Or, in the picture at right, are the lines really parallel, so there's no solution?Or are you just looking at an unuseful portion of the graph?

You can't tell!

In this case, zooming out shows that the lines in the previous picture do indeed cross, at the point (450, 449.5). But this was not at all apparent in the "standard" viewing window shown above.

So you can see thatthe pictures can be useful, especially for the concepts, but you should take "solving by graphing" with a grain of salt, and should keep in mind that the algebraic techniques (not the pictures)are the tools you need for sure answers.

The above discussion was specific to the two-equation, two-variable case, because you can draw pictures of the two-variable case to illustrate what is going on. But the terminology and basic concepts are the same, no matter how many variables you have. You could have four equations in four variables or twelve equations in twelve variables, and you would still be looking for where the "lines""intersect" you just couldn't draw a picture of it.Formatting note: For reasons which will become apparent when you start working with matrices, equations in systems of equation are generally written with the variables on the left-hand side of the "equals" sign and the numbers on the right-hand side. Sometimes you'll find a question formatted differently, but variables-on-the-left will be the norm.Systems of Linear Equations:

F) Examples

Solve the following system of equations using Gaussian elimination:

2x + y + 3z = 12x + 6y + 8z = 36x + 8y + 18z = 5I think I'll use the first row to clear out the x-terms from the second and third rows:

Technically, I should now divide the first row by 2to get a leading 1, but that will give me fractions, and I'd like to avoid that for as long as possible. Instead, I'll move on to using the second row to clear out the y-term from the third row:

I can divide the third row by 4:

To be technically correct, I'll now divide the second row by 5and the first row by 2:

(You might want to check with your instructor regarding how particular he's going to be about proper form. Sometimes it's nice to be able to avoid fractions!)Back-solving, I get:y + (0) = 2/5y = 2/5x + ( 1/2 )( 2/5 ) + ( 3/2 )(0) = 1/2x + 1/5= 1/2x = 3/10Then the solution is (x, y, z) = ( 3/10, 2/5, 0).I didn't show my scratch work on this last problem, but I did do it. Please use scratch paper and write things out; don't try to do this stuff in your head; there are just way too many opportunities for errors. (Ya wanna know how many mistakes I made while writing this lesson? Don't even get me started!)Here's another example: Solve the following system:

3x + y 6z = 102x + y 5z = 86x 3y + 3z = 0I think I'll use the second row to work on the x-terms in the first and third rows. I'll be able to clear out the third row, and I'll be able to produce a 1x as the leading term in the first row. This will let me finish the job of clearing out the x-column, and I'll be able to do it without having to deal with fractions:

(Many instructors would teach you always to divide through on one of the rows to get a leading coefficient of 1, but I would rather take an extra step or two and use addition to get a leading 1. That's just a personal preference, but I'm sure you can see the advantage of avoiding fractions for as long as possible. It can really cut down on computational errors.)Now I'll use that nice leading xin the first row to clear out the leading term in the second row. While I'm at it, I'll also divide the third row by 6, to get smaller numbers:

Hm... The second and third rows are the same. I can use the second row to clear out the third row entirely: Copyright Elizabeth Stapel 2000-2005 All Rights Reserved

Thinking back to the two-variable case, getting a line like "0 = 0" (which is true, but unhelpful) means that this is a dependent system, and the solution is going to have variables in it. If you get into linear algebra much, you will learn that the answer abovemeans that the solution is a line in three-dimensional space rather than a single point. For now, all you need to know is how to write the solution. To find the solution, I have to solve the two remaining equations for xand yin terms of z:x z = 2x = z 2

y 3z = 4y = 3z 4

(x, y, z) = (t 2, 3t 4, t)Remember that your book may use some variable other than "t", and that tis just standing in for z. This form of the solution just says that zis whatever value you chose, and then xis two less than that and yis four less than three times as much as z. Solve the following system of equations:

x + z = 1x + y + z = 2x y + z = 1You should be getting the hang of things by now, so I'll just show the steps that I used:

As soon as I get a nonsense row (like "0 = 1"), I know that this is an inconsistent system, and I can quit.inconsistent system: no solutionRemember the difference between the two special cases: A trivial row (such as "0 = 0") means a dependent system with a solution that contains variables; a nonsensical row (such as "0 = 1") means an inconsistent system with no solution whatsoever. Don't confuse these; they are common trick questions on tests.Depending on the course, you might now move on to using matrices for solving systems of equations. If you do, the techniques you'll be learning for matrices will likely be very similar to what you have seen in this lesson.