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Algebra 2 Solving Systems of Equations in Three Variables

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Section 3: Systems of Equations and Inequalities: Lecture 4 | 28:53 min

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Lecture Comments (6)

1 answer

Last reply by: Dr Carleen Eaton
Sun Apr 8, 2018 2:47 PM

Post by John Stedge on March 27, 2018

At approx. 4 mins why don't you just multiply the equation by a negative two; this gets rid of having to carry the negative sign and the possibility of adding the wrong number on accident. Just a suggestion :)!

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.

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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

Solve:
(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

Solution (x,y,z) = (2,4, − 4)

Solve:
(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

Solution (x,y,z) = ( − 6,3, − 4)

Solve:
(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

Solution (x,y,z) = (2, − 1,1)

Solve:
(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

Solution (x,y,z) = ( − 4, − 6,6)

Solve:
(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

Solution (x,y,z) = (2, − 1,2)

Solve:
(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

Solution (x,y,z) = ( − 1, − 2,0)

Solve:
(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

solution (x,y,z) = ( − 1,0,0)

Solve:
(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

This is never true, therefore the system of eq. has no solution.

Solve:
(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

This is never true, therefore the system of eq. has no solution.

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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

Solving Systems in Three Variables

Number of Solutions

Example 1: Solve 3 Variables

Example 2: Solve 3 Variables

Example 3: Solve 3 Variables

Example 4: Solve 3 Variables

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

Section 1: Equations and Inequalities
	Expressions and Formulas	22:23
	Properties of Real Numbers	20:15
	Solving Equations	19:10
	Solving Absolute Value Equations	17:31
	Solving Inequalities	17:14
	Solving Compound and Absolute Value Inequalities	25:00
Section 2: Linear Relations and Functions
	Relations and Functions	32:05
	Linear Equations	14:46
	Slope	23:07
	Writing Linear Functions	23:05
	Special Functions	31:05
	Graphing Inequalities	21:42
Section 3: Systems of Equations and Inequalities
	Solving Systems of Equations by Graphing	17:13
	Solving Systems of Equations Algebraically	23:53
	Solving Systems of Inequalities By Graphing	27:12
	Solving Systems of Equations in Three Variables	28:53
Section 4: Matrices
	Basic Matrix Concepts	11:34
	Matrix Operations	21:36
	Matrix Multiplication	29:36
	Determinants	33:13
	Cramer's Rule	28:25
	Identity and Inverse Matrices	22:25
	Solving Systems of Equations Using Matrices	22:32
Section 5: Quadratic Functions and Inequalities
	Graphing Quadratic Functions	31:48
	Solving Quadratic Equations by Graphing	27:03
	Solving Quadratic Equations by Factoring	19:53
	Imaginary and Complex Numbers	35:45
	Completing the Square	27:11
	Quadratic Formula and the Discriminant	22:48
	Analyzing the Graphs of Quadratic Functions	30:07
	Graphing and Solving Quadratic Inequalities	27:05
Section 6: Polynomial Functions
	Properties of Exponents	19:29
	Operations on Polynomials	13:27
	Dividing Polynomials	31:11
	Polynomial Functions	22:30
	Analyzing Graphs of Polynomial Functions	33:29
	Solving Polynomial Functions	21:10
	Remainder and Factor Theorems	31:21
	Roots and Zeros	31:27
	Rational Zero Theorem	31:16
Section 7: Radical Expressions and Inequalities
	Operations on Functions	34:30
	Inverse Functions and Relations	22:42
	Square Root Functions and Inequalities	30:04
	nth Roots	20:46
	Operations with Radical Expressions	41:11
	Rational Exponents	30:45
	Solving Radical Equations and Inequalities	31:27
Section 8: Rational Equations and Inequalities
	Multiplying and Dividing Rational Expressions	40:54
	Adding and Subtracting Rational Expressions	55:04
	Graphing Rational Functions	57:13
	Direct, Joint, and Inverse Variation	20:21
	Solving Rational Equations and Inequalities	55:14
Section 9: Exponential and Logarithmic Relations
	Exponential Functions	35:58
	Logarithms and Logarithmic Functions	45:54
	Properties of Logarithms	28:43
	Common Logarithms	25:23
	Base e and Natural Logarithms	21:14
	Exponential Growth and Decay	24:30
Section 10: Conic Sections
	Midpoint and Distance Formulas	32:42
	Parabolas	41:27
	Circles	21:03
	Ellipses	46:51
	Hyperbolas	38:15
	Conic Sections	18:43
	Solving Quadratic Systems	47:04
Section 11: Sequences and Series
	Arithmetic Sequences	21:16
	Arithmetic Series	21:36
	Geometric Sequences	23:03
	Geometric Series	22:43
	Infinite Geometric Series	18:32
	Recursion and Special Sequences	14:34
	Binomial Theorem	48:30

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 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 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 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 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 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 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

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Dr. Carleen Eaton

Solving Systems of Equations in Three Variables

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro