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### 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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 screen-captured 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 equations*0096

*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 equations--I 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 3--either 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 variable*0316

*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 equations*0329

*to eliminate a variable, using either elimination or substitution, then work with two other equations*0335

*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 x*0368

*(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 variables--*0384

*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 equation*0399

*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 true--that 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 with--especially, 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 variables--usual 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 these--I 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 equation--a 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 4y--0; 2z and -2z--0; 8 and 2--10.*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 saying*1701

*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 video---good review of each step.

I sometimes accompany an explanation of this type systems of equations--triples--with a quick solution using matrices on the TI-84.

Students become skillful in obtaining a quick answer for x, y, and z when the identity matrix appears.