Solving Systems of Equations in Three Variables
 Use substitution and elimination to solve a system in three variables.
 A system in 3 variables can have a unique solution, infinitely many solutions, or no solution.
Solving Systems of Equations in Three Variables
(a) − 2x + 4y + 4z = − 4
(b) − 2x − 5y − 3z = − 12
(c) 2x − 2y + z = − 8
 The strategy to solve a system of equations in three variables is to divide and conquer.
 Break down 3x3 system into 2x2 systems, solve for one variable, then work backwards to solve for all three variables. This is done in several steps.
 Choose the easiest variable to eliminate.
 The Easiest variable to eliminate is x because we can add (a) + (c) to eliminate x, and we can add (b) and (c) to eliminate x
 (a) + (c):
(− 2x + 4y + 4z = − 4) + (2x − 2y + z = − 8) = (2y + 5z = − 12)  (b) + (c):
(− 2x − 5y − 3z = − 12) + (2x − 2y + z = − 8) = (− 7y − 2z = − 20)  Combine and solve for one of the variables. The easiest variable to eliminate is y.
 (2y + 5z = − 12) + (− 7y − 2z = − 20)
7*(2y + 5z = − 12) + 2*(− 7y − 2z = − 20)
(14y + 35z = − 84) + (− 14y − 4z = − 40)
31z = −124
z = −4  Use z = − 4 to solve for y
 2y + 5z = − 12
2y + 5( − 4) = − 12
2y − 20 = − 12
2y = 8
y = 4  Using y = 4 and z = − 4, solve for x using (a), (b), or (c)
(c) 2x − 2y + z = − 8
2x − 2(4) − 4 = − 8
2x − 12 = − 8
2x = 4
x = 2
(a)2x + 5y + 3z = − 9
(b)6x + 2y − 3z = − 18
(c) − x − 12y − 3z = − 18
 The strategy to solve a system of equations in three variables is to divide and conquer.
 Break down 3x3 system into 2x2 systems, solve for one variable, then work backwards to solve for all three variables. This is done in several steps.
 Choose the easiest variable to eliminate.
 The Easiest variable to eliminate is z because we can add (a) + (b) to eliminate z, and we can add (a) and (c) to eliminate z
 (a) + (b):
(2x + 5y + 3z = − 9) + (6x + 2y − 3z = − 18) = (8x + 7y = − 27)  (a) + (c):
(2x + 5y + 3z = − 9) + (− x − 12y − 3z = − 18) = (x − 7y = − 27)  Combine and solve for one of the variables. The easiest variable to eliminate is y.
 (8x + 7y = − 27) + (x − 7y = − 27)
9x = − 54
x = − 6  Use x = − 6 to solve for y
 x − 7y = − 27
− 6 − 7y = − 27
− 7y = − 21
y = 3  Using y = 3 and x = − 6, solve for z using (a), (b), or (c)
 (a) 2x + 5y + 3z = − 9
2( − 6) + 5(3) + 3z = − 9
− 12 + 15 + 3z = − 9
3 + 3z = − 9
3z = − 12
z = − 4
(a)− 4x + 5y − 5z = − 18
(b)− 6x − 5y − 4z = − 11
(c)− 6x + 5y + 13z = − 4
 The strategy to solve a system of equations in three variables is to divide and conquer.
 Break down 3x3 system into 2x2 systems, solve for one variable, then work backwards to solve for all three variables. This is done in several steps.
 Choose the easiest variable to eliminate.
 The Easiest variable to eliminate is y because we can add (a) + (b) to eliminate y, and we can add (b) and (c) to eliminate y
 (a) + (b):
(− 4x + 5y − 5z = − 18 )+(− 6x − 5y − 4z = − 11)=( 10x − 9z = − 29)  (b) + (c):
( − 6x − 5y − 4z = − 11 ) + (− 6x + 5y + 13z = − 4 ) = (− 12x + 9z = − 15)  Combine and solve for one of the variables. The easiest variable to eliminate is z.
 (− 10x − 9z = − 29) + (− 12x + 9z = − 15)
− 22x = − 44
x = 2  Use x = 2 to solve for y
 − 12x + 9z = − 15
− 12(2) + 9z = − 15
− 24 + 9y = − 15
y = − 1  Using y = − 1 and x = 2, solve for z using (a), (b), or (c)
 (a) − 4x + 5y − 5z = − 18
− 4(2) + 5( − 1) − 5z = − 18
z = 1
(a)3x − y + 4z = 18
(b) − 4x − y − 2z = 10
(c)2x − 2y + z = 10
 The strategy to solve a system of equations in three variables is to divide and conquer.
 Break down 3x3 system into 2x2 systems, solve for one variable, then work backwards to solve for all three variables. This is done in several steps.
 Choose the easiest variable to eliminate.
 The Easiest variable to eliminate is y because we can add (a) + (b) to eliminate y, and we can add (b) and (c) to eliminate y, however, (a) and (b) need to be multiplied by 2 and − 2 in order to eliminate y.

(a)2*(3x − y + 4z = 18)
(b)2*(− 4x − y − 2z = 10)
(c)2x − 2y + z = 10 
(a)6x−2y+8z=36
(b)8x+2y+4z=−20
(c)2x − 2y + z = 10  (a) + (b):
(6x − 2y + 8z = 36) + (8x + 2y + 4z = − 20 ) = (14x + 12z = 16)  (b) + (c):
(8x + 2y + 4z = − 20 ) + (2x − 2y + z = 10) = (10x + 5z = − 10)  Combine and solve for one of the variables. The easiest variable to eliminate is z.
 (14x + 12z = 16) + (10x + 5z = − 10) =
5*(14x + 12z = 16) + −12*(10x + 5z = − 10) =
(70x + 60z = 80) + (− 120x − 60z = 120) =
− 50x = 200
x = − 4  Use x = − 4 to solve for z
 10x + 5z = − 10
10( − 4) + 5z = − 10(
− 40 + 5z = − 10
z = 6  Using x = − 4 and z = 6, solve for y using (a), (b), or (c)
 3x − y + 4z = 18
3( − 4) − y + 4(6) = 18
y = − 6
(a) − 3x + 2y + 3z = − 2
(b)3x + y − 3z = − 1
(c)x + 2y − 3z = − 6
 The strategy to solve a system of equations in three variables is to divide and conquer.
 Break down 3x3 system into 2x2 systems, solve for one variable, then work backwards to solve for all three variables. This is done in several steps.
 Choose the easiest variable to eliminate.
 The Easiest variable to eliminate is x because we can add (a) + (b) to eliminate x, and we can add (a) and (c) to eliminate x, however, (c) need to be multiplied by 3 in order to eliminate x.
 (a) − 3x + 2y + 3z = − 2
(b)3x + y − 3z = − 1
(c)3*(x + 2y − 3z = − 6)  (a) − 3x + 2y + 3z = − 2
(b)3x + y − 3z = − 1
(c)3x + 6y − 9z = − 18  (a) + (b):
( − 3x + 2y + 3z = − 2 ) + (3x + y − 3z = − 1) = (3y = − 3)  (a) + (c):
(− 3x + 2y + 3z = − 2) + (3x + 6y − 9z = − 18 ) = (8y − 6z = − 20)  Solve for y using (3y = − 3)
 y = − 1
 Use y = − 1 to solve for z using (8y − 6z = − 20)
 8y − 6z = − 20
8( − 1) − 6z = − 20
z = 2  Using y = − 1 and z = 2, solve for x using (a), (b), or (c)
 − 3x + 2y + 3z = − 2
− 3x + 2( − 1) + 3(2) = − 2
x = 2
(a)x + y + z = − 3
(b)3x − 4y = 5
(c)2x + 3y − z = − 8
 The strategy to solve a system of equations in three variables is to divide and conquer.
 Break down 3x3 system into 2x2 systems, solve for one variable, then work backwards to solve for all three variables. This is done in several steps.
 Choose the easiest variable to eliminate.
 The Easiest variable to eliminate is z because we can add (a) + (c) to eliminate z,
 (a) + (c):
(x + y + z = − 3) + (2x + 3y − z = − 8) = (3x + 4y = − 11)  Combine (3x + 4y =  11) and equation (b) and solve for one of the variables.
 (3x + 4y = − 11) + (3x − 4y = 5) =
6x = − 6
x = − 1  Use x = − 1 to solve for y using (3x + 4y = − 11)
 3x + 4y = − 11
3( − 1) + 4y = − 11
− 3 + 4y = − 11
y = − 2  Using x = − 1 and y = − 2, solve for z using (a), (b), or (c)
 x + y + z = − 3
− 1 + − 2 + z = − 3
z = 0
(a)2x − 2y + 3z = − 2
(b) − y + 3z = 0
(c) − 2x + y − 3z = 2
 The strategy to solve a system of equations in three variables is to divide and conquer.
 Break down 3x3 system into 2x2 systems, solve for one variable, then work backwards to solve for all three variables. This is done in several steps.
 Choose the easiest variable to eliminate.
 The Easiest variable to eliminate is x because we can add (a) + (c) to eliminate x,
 (a) + (c):
(2x − 2y + 3z = − 2) + ( −2x + y − 3z = 2 ) = (− y = 0)  y = 0
 Solve for z using y=0 and (b)
 − y + 3z = 0
− 0 + 3z = 0
z = 0  Use y = 0 and z = 0 to solve for x using (a) or (b) or (c)
 2x − 2y + 3z = − 2
2x − 2(0) + 3(0) = − 2
x = − 1
(a)x + y + 2z = 4
(b) − 2y − z = − 3
(c) − x + y − z = − 4
 The strategy to solve a system of equations in three variables is to divide and conquer.
 Break down 3x3 system into 2x2 systems, solve for one variable, then work backwards to solve for all three variables. This is done in several steps.
 Choose the easiest variable to eliminate.
 The Easiest variable to eliminate is x because we can add (a) + (c) to eliminate x,
 (a) + (c):
(x + y + 2z = 4) + (− x + y − z = − 4) = (2y + z = 0)  Combine (2y + z = 0) and equation (b) and solve for one of the variables.
 (2y + z = 0) + (− 2y − z = − 3)
0 = −3
(a)x + 4y + 2z = 6
(b)3x + 4y − 6z = − 12
(c) − 2x − 4y + 2z = − 18
 The strategy to solve a system of equations in three variables is to divide and conquer.
 Break down 3x3 system into 2x2 systems, solve for one variable, then work backwards to solve for all three variables. This is done in several steps.
 Choose the easiest variable to eliminate.
 The Easiest variable to eliminate is y because we can add (a) + (c) and (b) and (c),
 Derive the two Systems of Equations Sys.
 (a) + (c):
(x + 4y + 2z = 6) + ( − 2x − 4y + 2z = − 18 ) = ( − x + 4z = − 12)  (b) + (c):
(3x + 4y − 6z = − 12) + (− 2x − 4y + 2z = − 18 ) = (x − 4z = − 30 )  Combine and solve for one of the variables.
 (− x + 4z = − 12) + (x − 4z = − 30)
0 = −42
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Solving Systems of Equations in Three Variables
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro 0:00
 Solving Systems in Three Variables 0:17
 Triple of Values
 Example: Three Variables
 Number of Solutions 5:55
 One Solution
 No Solution
 Infinite Solutions
 Example 1: Solve 3 Variables 7:59
 Example 2: Solve 3 Variables 13:50
 Example 3: Solve 3 Variables 19:54
 Example 4: Solve 3 Variables 25:50
Algebra 2
Transcription: Solving Systems of Equations in Three Variables
Welcome to Educator.com.0000
Today we are going to be going on to talk about solving systems of equations in three variables.0002
In previous lessons, we talked about how to solve systems with two variables.0007
So now, we are going to go up to systems involving three variables.0012
In order to solve these systems, you are actually going to use the same strategies you used for solving systems with two variables.0017
Recall those techniques: substitution, elimination, and multiplication.0027
This time, though, a solution is an ordered triple of values.0031
So, you will end up having three variables: for example, (x,y,z).0035
And the solution would be something like (5,3,2), where x is 5, y is 3, and z is 2.0040
Now, just to work out an example to show you how to approach these using the same techniques that you already know:0056
looking at this system of equations, I have three equations, and I have three total variables.0082
And the idea is to work with the three equations so that you get one of the variables to drop out.0088
Once you get one of the variables to drop out, you will be left with a system of two equations0096
with two variables, and you already know how to work with that.0100
So, the technique would be to first just consider two of the equations together; I am going to call these equations 1, 2, and 3, so I can keep track of them.0103
I am going to first consider equations 1 and 2.0113
And when I look at these, I see that the two z's have opposite coefficients.0116
And you will recall that elimination works really well in that situation.0125
So, I am going to take 2x + 3y  z = 5 (that is equation 1) and equation 2: 3x  2y + z = 4, and I am going to add those.0129
This will give me 5x, and then 3y  2y is going to give me y; the z's drop out; 5 and 4 is 9.0146
So now, I have a new equation, and I will just mark this out so I can keep track of it.0158
So, once you have gotten a variable to drop out, work with two different equations to get the same variable to drop out.0167
So, I got z to drop out; and what I want to do is work with two different equationsI worked with 1 and 2.0175
I could work with 1 and 2, or I could work with 2 and 3.0182
And I am actually going to work with 2 and 3; and I want to get z to drop out.0187
Equations 2 and 3: looking at this, how am I going to get z to drop out?0193
Well, in order to do that, I could use elimination; but I am first going to have to multiply this second equation by 2.0197
So, this is equation 2; and it is 3x  2y + z = 4; and I am going to multiply that by 2 to give me 6x  4y + 2z = 8.0207
Now, I am going to take equation 3 (this is equation 2, and I am going to take equation 3): I want to make sure that I am working with two different equations.0229
So, I have 1 and 2, and 2 and 3 (or I could have done 1 and 3either way).0237
4x + y + 2z = 3: my goal is to get the z's to drop out.0243
In order to do that, I am going to have to subtract: I need to subtract 3 from 2.0256
I want to be very careful with my signs here, so I am going to change this to adding the opposite.0267
This is going to give me 6x + 4x, which is 10x; 4y and y is 5y; 2z and 2z drops out 8 minus 3 is 5; OK.0276
At this point, what I am left with is a system of two equations with two unknowns.0297
Once you get that far, you proceed using the techniques that we learned previously (again, substitution, elimination, and multiplication).0303
But since you are only working with two equations with two unknowns, you are on familiar territory.0312
And you can solve for one of the variables and then find that value; substitute in for the other variable0316
and find that value; and then you can go back and find z.0325
And we are going to work more examples on this; but the basic technique is to work with two equations0329
to eliminate a variable, using either elimination or substitution, then work with two other equations0335
to eliminate that same variable, resulting in two equations with two variables,0342
allowing you to solve for one variable, then the other, and then the third.0349
Just as in systems with two variables, a system with three variables may have one solution, no solutions, or an infinite number of solutions.0357
So, recall: the solution set here, if I had three variables (x, y, and z) would be a value for x0368
(such as 2), a value for y (such as 4), and a value for z (such as 1) that would satisfy all three equations.0375
The other possibility is that there may be no solutions; recall from working with systems of equations with two variables0384
you know that you are in this situation when you are using elimination, or you are using substitution;0393
you are going along; and then you see variables start to drop out, and you end up with an equation0399
where you have a constant equaling another constant, which is never true.0406
So, if you start seeing variables drop out, and you end up with something such as 4 = 7 (which is never true),0413
this tells you that there is no solution to this system of equations.0420
There can be an infinite number of solutions; when you are working with your equations;0426
you are eliminating; you are substituting; you are using your techniques; you are being careful;0434
you are doing everything right, and then you see variables drop out, and you get a constant equaling a constant, like 2 = 2.0438
Well, that is always true; and this means that that system of equations has an infinite number of solutions.0447
So, typically, you will get one solution: a value for x, y, and z that is the set of values that makes the equations truethat satisfies the equations.0457
You may end up, though, with no solutions (there are no solutions to this equation) or an infinite number of solutions for this system of equations.0467
OK, in the first example, we are given a system of three equations with three variables.0479
So again, I am going to work with two of the equations; I will number them 1, 2, and 3.0486
And I will work with two of them, and then a different two.0493
So, first, I am going to look at the second two equations; and they are very easy to work with, because I have opposite coefficients.0500
I could also use substitution, because I have coefficients of 1; but I am just going to use elimination.0509
So first, I am going to work with 2 and 3: y  z = 2, x (let me move that, so it doesn't create confusion) + 2y + z = 2.0515
OK, here I end up with x; all I am doing is adding these together using elimination.0533
x + 3y; the z's drop out; 2 + 2 is 4.0543
OK, so I eliminated z from this first set of equations; now, I need to work with a different set of equations.0550
And there are actually multiple different ways to approach this, and I am going to work with 1 and 2.0567
I already worked with 2 and 3, so I eliminated z; and I want to...actually, I already have...this does not have z in it, so I don't even need to proceed.0579
That way, I already have two equations with two unknowns; this is a particularly easy situation, compared with when all three have 3 variables.0598
OK, so I look up here, and I have my new equation; I can call it equation 4.0607
And then, I have equation 1; and these just have x and y.0613
So, I am just going to proceed, like I usually do, with two equations with two unknowns.0616
Let me rewrite these right here: 2x + y = 3, and this is x + 3y = 4.0620
I can use substitution; that would be fine, because I have a coefficient of 1.0634
So here, I am going to solve for y; and this will give me y = 2x + 3.0640
And then, I am going to substitute into this equation; so I have x, and I am going to substitute for y: plus 3, times 2x, plus 3, equals 4.0651
Working this out: x...3 times 2x is 6x; 3 times 3 is 9; it equals 4.0668
Here, now, I just have an equation with a single variable, so I can solve that.0679
First, I am combining like terms: x  6x is 5x; plus 9 equals 4.0684
Subtract 9 from both sides: 5x = 5; x = 1.0692
OK, so the first thing I wanted to do is just get rid of one of the variables; I am just working with two variables.0700
And I did that by just adding these two; in the second and third equations, the z dropped out.0705
I was lucky, because the first equation already didn't have a z; so I had two equations with two unknowns.0711
And then, I just used those two; I solved by substitution, and I came up with x = 1.0717
Since I know that x equals 1, I can go ahead and substitute this into this equation to find y.0726
So, looking at equation 1: 2x + y = 3; I know that x equals 1, so that is 2(1) + y = 3, or 2 + y = 3.0734
Subtracting 2 from both sides, I get y = 1; so now, I have x, and I have y.0754
I need to find z: well, this will easily tell me what z is.0759
That is y  z = 2, and I know y: y equals 1, so 1  z = 2; z = 1, therefore z = 1.0765
So, the solution to this set of equations is that x equals 1, y equals 1, and z equals 1.0780
The hardest step was just getting rid of that third unknown (the third variable).0791
I did that by adding these two together: then, working with two equations with two variables, I was able to solve for x.0797
Once I am there, all I have to do is start substituting.0805
Here, I substituted x into the first equation and solved for y.0807
Once I got y, then I was able to substitute y into the second equation to solve for z.0813
So, 1, 1, 1 is the solution for this system of equations with three variables.0822
OK, Example 2: again, my goal is going to be to get a variable to drop out, so I am just left with two variables.0833
Looking at this first and second equation, considering these together, y and y...if I add those together,0843
the y's will drop out, because they have opposite coefficients (1 and 1).0852
So, I am going to start off by adding the first two equations: this is 1 and 2.0857
x + y + z = 2; and then, I am going to add x  y + 2z = 1; and I am going to come up with a new equation.0861
This is 2x; the y's drop out; z + 2z is 3z; 2  1 is 1; OK, I have this.0874
And I worked with these first two; I now need to work with two different equations to get the same variable to drop out.0886
This time, I am going to pick equations 1 and 3, and I want y to drop out.0894
So, I have equations 1 and 3: that is x + y + z = 2, and (equation 3) that is 2x + y + 2z = 2.0900
Now, I need to subtract in order to get the y to drop out.0919
To keep everything straight, as far as my signs go, I am going to keep the first equation the same;0925
but for the second one, I am going to change it to adding the opposite: add 2x, y, 2z, and 2.0931
OK, x  2x is x; the y's drop out, which is just what I wanted; z  2z is z; 2  2 is 0.0944
Now, I have two equations and two variables; I am rewriting these two up here to see what I have to work with.0956
2x + 3z = 1; now, I just use my usual methods of solving a system of equations with two variables.0965
Since I have coefficients here of 1, it is pretty easy to use substitution, so I am going to solve for x in this second equation,0976
and then substitute that value up in the first equation.0984
I have x  z = 0, which would give me x = z, or x = z.0989
So, I am going to take this z and substitute it in right here; OK, that gives me 2x + 3z = 1, and let x equal z.0996
So, 2 times z, plus 3z, equals 1; that is 2z + 3z = 1.1014
Combine these two like terms to get z = 1; now, I have my first value.1025
OK, so I know that z equals 1, so I am on my way.1034
And I look up here, and I see, "Well, I know that x equals z, so that makes it very easy to solve for x."1038
If x equals z, and z equals 1, then x equals 1.1047
So now, I have x = 1, z = 1; I am just missing y.1059
Well, look at that first equation: it tells me that x + y + z = 2.1063
The x is 1; I don't know y; and I know that z is 1; these two cancel, and that gives me y = 2.1074
So, 1 + 1 is 0, so I end up with y = 2.1088
Putting all this together up here as my solution, I end up with x = 1, y = 2, and z = 1.1092
That was a lot of steps; it is really important to keep track of what you are working withespecially, in the beginning,1106
that you work with two equations (I worked with 1 and 2) to get a variable to drop out.1112
I added those, and the y's dropped out; then I want to work with either 1 and 3 or 2 and 3 (two different equations) to get the y to drop out.1118
I chose 1 and 3; and I saw that I could get the y to drop out of 1 and 3 if I just subtracted 3 from 1; that is what I did right here.1128
At that point, I clearly mark out what I ended up with, which is two equations with two variables.1140
We wrote those up here; and I decided I was going to use substitution.1146
I solved for x in this second equation: x equals z; I substituted that in right here, into the first equation.1151
That allowed me to have one equation with one variable, z; and I determined that z equals 1.1162
From there, it was much easier, because I saw that x equals z, and I knew z; so x equals 1.1170
I had x; I had z; and I had three equations that I could have used,1180
but I picked the easiest one to substitute in x and z and solve for y to get my set of solutions.1184
Again, this is a set of three equations with three variables that I need to approach systematically.1198
And my first goal is to eliminate the same variable, so I am working with two equations with two variables.1204
And I see several possibilities; you could approach it differently, and you will come up with the same answer, as long as you follow the rules and the steps.1216
I am seeing that y and y are opposite in terms of coefficients (1 and 1), so I am going to add those two.1227
I am going to add 1 and 3; that is going to give me 5x; the y's drop out, so 5x  z; 1  3 is 2.1235
I just marked that, so I can keep track of it, because I am going to need to use it in a minute,1260
once I generate another equation in which y has been eliminated.1265
OK, I worked with the first and the third; now, I need to work with two different equations;1270
and I am going to work with the first and the second, and I want to eliminate y.1276
In order to eliminate y, I need to multiply the first equation by 2; so I will do that up here.1282
This is going to give me 4x  2y + 2z = 2; that is the first equation.1294
Now, I am going to add it to the second equation: + x + 2y  z = 0.1305
Adding these together, I am going to get 5x; the y's drop out; 2z  z is z; 2 and 0 is 2.1319
OK, I first worked with the first and the third, and then I worked with the first and the second, to get the y's to drop out.1330
So now, I have two equations and two variables: x and z.1338
So, put these together so I can see what is going on with them: 5x  2 = z; 5x + z = 2.1345
Well, I can see that, if I add these, z will drop out; and I am just back to my usual two equations with two variablesusual techniques.1355
5x and 5x is 10x; the z's drop out; 2 and 2 is 0.1365
Divide both sides by 10; it gives me x = 0.1373
So, I have my first value, which makes things much easier.1377
Since I know that x equals 0, I can substitute into either of these to solve for z.1384
So, 5x  z = 2; so 5(0)  z = 2; 0  z = 2; z = 2; divide both sides by 1 to get z = 2.1389
I have my second value; OK, so I know x; I know z; I just need y.1411
I am going to solve for y; I could use any of theseI am going to pick the top one and solve for y, knowing that x equals 0 and z equals 2.1417
OK, that is 0  y + 2 = 1, which gives me y + 2 = 1.1437
Subtract 2 from both sides to get y; if I say 1  2, that is going to give me 1.1447
Multiply all of that by 1 to give me y = 1.1455
OK, so the solution here is x = 0, y = 1, and z = 2.1461
And I approached that by seeing that I could add 1 and 3 because of y and y, and those would drop out.1470
I could have added the first two and had the z drop out, and had that be my variable to eliminate; I happened to choose y.1479
I added those together and got this equation.1486
Then, I worked with the first and the second equationa little more complicated, because to get opposite coefficients,1490
I had to multiply the first equation by 2.1496
I did that to generate this equation, which I added to the second; this is the resulting equation.1500
I then had two equations with two variables; I looked at those two and saw I had opposite coefficients with z.1507
So, when I added them together, z dropped out, and I could solve for x.1514
Once I determined that x is 0, I substituted 0 into this top equation for x to solve for z, and determined that z equals 2.1519
At that point, I just needed to solve for y, so I took this equation and substituted my value for x and my value for z, and determined that y equals 1.1529
So again, we are using the same techniques that we used previously, only you are working with more equations, and there is more to keep track of.1543
OK, in this system of equations (three equations with three unknowns), it is a little bit more complicated.1551
I do have one equation that has a coefficient of 1, but the rest of them have larger coefficients.1559
So, I am going to work with the first and second equations; and what I want to do is eliminate the z.1566
So, in order to do that, I am going to need to multiply this first equation by 2.1577
I want to work with the first and second equations; I want to get the z to drop out.1583
But I need to multiply this by 2 first; so let me do that right over here.1589
That is going to give me 6x  4y + 2z = 8; so I am going to rewrite that right here: 6x  4y + 2z = 8.1600
And that came from that first equation, multiplied by 2; and I am going to add it to the second equation.1615
So, + 6x + 4y  2z = 2: now, you might have already seen what happened.1621
I was really just focusing on "OK, I want to get the z's to be the same or opposite coefficients," so that they would cancel out.1635
But what happened is: everything ended up with opposite coefficients: 6 and 6; 4 and 4; 2 and 2; OK.1640
So, I have 6x  6x; that is 0; 4y and 4y0; 2z and 2z0; 8 and 210.1654
0 = 10: well, we know that 0 does not equal 10, so this is not true; it is never true that 0 equals 10.1669
So, there is no solution to this set of equations.1678
I could have had a situation where I got a solution.1682
I could have also had a situation where maybe I got 10 = 10 with a different system of equations.1686
If I had come up with something like 0 = 0 (that is always true) or 10 = 10, then I would have had an infinite number of solutions.1694
But instead, what happened is that my variables drop out; I got c = d, which is saying1701
that I have a constant that is equal to another constant, which is never true; so there is no solution.1706
So, this one turned out to actually be less work than the others.1712
But when this happens, you just want to be really careful that you were doing everything correctly;1714
you didn't make a mistake; but actually, it can end up that there simply is no solution to the system of equations.1718
That concludes this lesson on Educator.com on solving systems of equations with three variables; and I will see you next lesson!1725
1 answer
Last reply by: Dr Carleen Eaton
Sat Feb 23, 2013 6:35 PM
Post by bo young lee on February 19, 2013
i dont get the example 1 of the slove for y. can you explain more easy?
1 answer
Last reply by: Dr Carleen Eaton
Sun Jan 29, 2012 4:41 PM
Post by Ken Mullin on January 25, 2012
Very concise explanations accompanying videogood review of each step.
I sometimes accompany an explanation of this type systems of equationstripleswith a quick solution using matrices on the TI84.
Students become skillful in obtaining a quick answer for x, y, and z when the identity matrix appears.