Download Solving Linear Systems in Three Variables

LESSON
5.3
Name
Solving Linear Systems
in Three Variables
Class
5.3
Date
Solving Linear Systems
in Three Variables
Essential Question: How can you find the solution(s) of a system of three linear equations
in three variables?
Texas Math Standards
Resource
Locker
A2.3.B Solve systems of three linear equations in three variables by using Gaussian
elimination, technology with matrices, and substitution. Also A2.3.A, A2.4.A
The student is expected to:
A2.3.B
Explore
Solve systems of three linear equations in three variables by using
Gaussian elimination, technology with matrices, and substitution.
Also A2.3.A, A2.4.A
Recognizing Ways that Planes Can Intersect
Recall that a linear equation in two variables defines a line. Consider a linear equation in three variables. An example
is shown.
5 = 3x + 2y + 6z
Mathematical Processes
A linear equation in three variables has three distinct variables, each of which is either first degree or has a
coefficient of zero.
A2.1.B
Use a problem-solving model that incorporates analyzing given
information, formulating a plan or strategy, determining a solution,
justifying the solution, and evaluating the problem-solving process and
the reasonableness of the solution.
Just as the two numbers that satisfy a linear equation in two variables are called an ordered pair, the three numbers
that satisfy a linear equation in three variables are called an ordered triple and are written (x, y, z).
Language Objective
Three linear equations in three variables, considered together, form a system of three linear equations in three
variables. The solutions of a system like this depend on the ways three planes can intersect.
The set of all ordered pairs satisfying a linear equation in two variables forms a line. Likewise the set of all ordered
triples satisfying a linear equation in three variables forms a plane.
1.B.1, 1.B.2, 1.E.1, 1.E.3

Label the kind of solution methods shown to solve systems of three linear
equations in three variables. Explain to a partner which method is easiest
to use in a particular context and why.
Essential Question: How can you find
the solution(s) of a system of three
linear equations in three variables?
Possible answer: Solve using substitution,
elimination, or matrices.
planes?
© Houghton Mifflin Harcourt Publishing Company
ENGAGE
The diagrams show some ways three planes can intersect. How many points lie on all 3

0
The diagram shows three intersecting planes.
1
How many points lie on all 3 planes?
PREVIEW: LESSON
PERFORMANCE TASK
View the Engage section online. Discuss the photo
and how you can use a system of linear equations to
determine the number of different components of
inline skates a company can afford to purchase. Then
preview the Lesson Performance Task.
Module 5
ges must
EDIT--Chan
DO NOT Key=TX-A
Correction
be made
through “File
Lesson 3
269
info”
Date
Class
ms
ear Syste
Solving Lin riables
in Three Va
Name
5.3
ce
Resour
ions
Locker
linear equat
of three
of a system
solution(s)
find the
can you
ion: How
les?
Gaussian
in three variab
es by using
in three variabl Also A2.3.A, A2.4.A
equations
ution.
three linear
es, and substit
rsect
systems of
with matric
A2.3.B Solve
le
technology
es Can Inte
An examp
Plan
les.
elimination,
g Ways that a linear equation in three variab
er
Recognizin
line. Consid
Explore
defines a
variables
on in two
a linear equati
or has a
2y + 6z
Recall that
first degree
5 = 3x +
is shown.
which is either
les, each of
variab
t
rs
distinc
three numbe
les has three
d pair, the
an ordere
in three variab
are called
x, y, z).
equation
variables
A linear
are written (
d
on in two
of zero.
d triple and
all ordere
coefficient
a linear equati
an ordere
the set of
Likewise
rs that satisfy variables are called
forms a line.
two numbe
on in three
variables
Just as the
on in two
a linear equati
ons in three
that satisfy
a linear equati forms a plane.
ing
equati
d pairs satisfy in three variables
three linear
all ordere
system of
ct.
equation
The set of
er, form a
can interse
ing a linear
ered togeth
three planes
triples satisfy
les, consid
the ways
in three variablike this depend on
on all 3
equations
points lie
Three linear solutions of a system
many
How
The
intersect.
variables.
planes can
HARDCOVER PAGES 191204
Quest
Essential
A2_MTXESE353930_U2M05L3 269
ms show
The diagra

some ways
Turn to these pages to
find this lesson in the
hardcover student
edition.
three
0
Harcour t
Publishin
y
g Compan
planes?
n Mifflin
.
© Houghto

cting planes
three interse
m shows
The diagra
?
on all 3 planes
points lie
How many
1
Lesson 3
269
Module 5
269
Lesson 5.3
L3 269
0_U2M05
SE35393
A2_MTXE
2/22/14
5:59 AM
2/22/14 5:59 AM

The diagram shows planes intersecting in a different way.
Describe the intersection.
EXPLORE
The intersection is a line.
Recognizing Ways that Planes Can
Intersect
an infinite number
How many points lie in all 3 planes?
Reflect
1.
Discussion Give an example of three planes that intersect at exactly one point.
Sample answers: The three dimensional coordinate axes; three faces of a cube that share a
INTEGRATE TECHNOLOGY
Have students draw three intersecting planes using an
art application on their iPads.
common corner.
Solving a System of Three Linear Equations
Using Substitution
Explain 1
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Modeling
A system of three linear equations is solved in the same manner as a system of two linear equations. It just has more
steps.
Example 1

Solve the system using substitution.
⎧⎪
⎪ -2x + y + 3z = 20
1
-3x + 2y + z = 21
2
⎪
⎪ 3x - 2y + 3z = -9
⎩
3
⎨
Sharing their diagrams of three intersecting planes
with the class, students will see the possible ways
three planes can intersect. Their intersection may be
a line or a point. In some cases, three planes will not
intersect.
Choose an equation and variable to start with. The easiest equations to solve are those that have a variable
with a coefficient of 1. Solve for y.
y = 2x - 3z + 20
Now substitute for y in equations
2 and
3 and simplify.
21 = -3x + 2(2x - 3z + 20) + z
-9 = 3x - 2(2x - 3z + 20) + 3z
21 = -3x + 4x - 6z + 40 + z
-9 = 3x - 4x + 6z - 40 + 3z
21 = x - 5z + 40
-19 = x - 5z
-9 = -x + 9z - 40
31 = -x + 9z
4
EXPLAIN 1
© Houghton Mifflin Harcourt Publishing Company
-2x + y + 3z = 20
5
Solving a System of Three Linear
Equations Using Substitution
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Communication
Have students explain to a partner the process of
solving a system of three linear equations in three
variables by substitution.
Module 5
270
Lesson 3
PROFESSIONAL DEVELOPMENT
A2_MTXESE353930_U2M05L3.indd 270
Integrate Mathematical Processes
2/20/14 4:16 AM
This lesson provides an opportunity to address Mathematical Process TEKS
A2.2.B, which calls for students to “use a problem-solving model that incorporates
analyzing given information, formulating a plan or strategy, determining a
solution, justifying the solution, and evaluating the problem-solving process and
the reasonableness of the solution.” Students solve systems of linear equations in
three variables using three different methods: substitution, elimination, and
matrices. Students need to choose among these methods to solve real-world
problems, and verify that the solution is indeed reasonable.
Solving Linear Systems in Three Variables
270
This results in the following linear system in two variables:
QUESTIONING STRATEGIES
⎧
-19 = x - 5z
⎨
After solving one equation for a single
variable, what is the next step when solving a
system of three linear equations in three variables by
substitution? Substitute into the two other original
equations to get a system of two linear equations in
two variables.
4
31 = -x + 9z
⎩
5
Solve equation [4] for x.
x = 5z - 19
Substitute into equation [5] and solve for z. Then use the value for z to find the values of x.
31 = -(5z - 19) + 9z
31 = -x + 9(3)
31 = 4z + 19
31 = -x + 27
3=z
-4 = x
Finally, solve the equation for y when x = -4 and z = 3.
y = 2x - 3z + 20
y = 2(-4) - 3(3) + 20
y=3
Therefore, the solution to the system of three linear equations is the ordered triple (-4, 3, 3).
B
There is a unique parabolic function passing through any three noncollinear points in the coordinate plane
provided that no two of the points have the same x-coordinate. Find the parabola that passes through the
points (2, 1), (-1, 4), and (-2, 3).
The general form of a parabola is the quadratic equation y = ax 2 + bx + c. In order to find the
equation of the parabola, we must identify the values of a, b, and c. Since each point lies on the parabola,
substituting the coordinates of each point into the general equation produces a different equation.
⎧
1
= a(2) + b(2) + c
⎨
4
= a(-1) + b(-1) + c
⎩
3
⎪
© Houghton Mifflin Harcourt Publishing Company
⎪
2
2
( ) + b ( -2 ) + c
⇒
2
= a -2
⎧
1
⎨4=a-b+c
2
⎪ 1 = 4a + 2b + c
⎪ 3 = 4a - 2b + c
⎩
3
Choose an equation in which it is easier to isolate a variable. Solve equation [2] for c.
c = 4 - a + b [2]
Now substitute for c = in equations [1] and [3].
1 = 4a + 2b +
(
4-a+b
1 = 3a + 3b + 4
-3 = 3a + 3b
)
3 = 4a - 2b +
3 = 3a -
-1 = 3a - b
[4]
Module 5
271
( 4-a+b )
b +4
[5]
Lesson 3
COLLABORATIVE LEARNING
A2_MTXESE353930_U2M05L3 271
Peer-to-Peer Activity
Have students work in groups of three. Give each group a system of three linear
equations in three variables to solve. Explain that each student in a group should
solve the system with a different method. Then, have students compare their
solutions.
271
Lesson 5.3
2/23/14 4:51 AM
This results in the following linear system in two variables:
⎧
⎨
3a + 3b = -3
[4]
3a - b = -1
[5]
⎩
Solve equation [5] for b.
b = 3a +
1
Substitute into equation [4] and solve for a. Then use the value for a to find the values of b.
3a + 3
(
3a + 1
( _)
) = -3
3 - 1 +3b = -3
2
3
- +3b = -3
2
3
3b = - 2
1
b = -2
_
3 = -3
3a + 9a +
_
_
12a = -6
1
a = -2
_
Then use the values a and b to solve for c.
( ) ( )
1 + -_
1 =4
c=4-a+b=4- -_
2
2
So the equation of the parabola connecting (2, 1), (-1, 4), and (-2, 3) is
_1
y = -2 x 2 -
_1
2
x+
4
.
Your Turn
2.
⎧ x + 2y + z = 8
⎨ 2xx ++yy+-3zz == 47
⎩
3.
© Houghton Mifflin Harcourt Publishing Company
x + 2y + z = 8 → z = 8 - x - 2y
Substitute for z in the other equations.
x + y + 3(8 - x - 2y) = 7
2x + y - (8 - x - 2y) = 4
↓
↓
x+y=4
-2x - 5y = -17
Solving this system yields x = 1 and y = 3.
z = 8 - x - 2y → z = 8 - 1 - 2(3) = 1
So the ordered triple is (1, 3, 1).
2x - y - 3z = 1
4x + 3y + 2z = -4
⎩ -3x + 2y + 5z = -3
⎧
⎨
2x - y -3z = 1 → 2x - 3z - 1 = y
Substitute for y in the other equations.
-3x + 2(2x - 3z - 1) + 5z = -3
4x + 3(2x - 3z -1) + 2z = -4
↓
↓
10x - 7z = -1
x - z = -1
Solving this system yields x = 2 and z = 3.
2x - 3z -1 = y → 2(2) - 3(3) - 1 = y = -6
So the ordered triple is (2, -6, 3).
Module 5
272
Lesson 3
DIFFERENTIATE INSTRUCTION
A2_MTXESE353930_U2M05L3.indd 272
Visual Cues
19/03/15 11:57 AM
Show students a graph of all of the possible ways three planes can intersect and
have them identify the number and nature of the solutions in each case.
Solving Linear Systems in Three Variables 272
Solving a System of Three Linear Equations
Using Elimination
Explain 2
EXPLAIN 2
You can also solve systems of three linear equations using elimination.
Solving a System of Three Linear
Equations Using Elimination
Example 2

AVOID COMMON ERRORS
⎧ -2x + y + 3z = 20
1
⎪⎨ -3x + 2y + z = 21
⎪ 3x - 2y + 3z = -9
2
3
⎩
When solving a system of three linear equations by
elimination, students may choose two equations,
eliminate one of the variables, and then choose
another pair of the original equations and eliminate a
different variable instead of the same variable. Tell
students that in order to solve the system, they must
eliminate the same variable from both pairs of
original equations.
Begin by looking for variables with coefficients that are either the same or additive inverses of each other.
When subtracted or added, these pairs will eliminate that variable. Subtract equation 3 from equation
1
1
3
to eliminate the z variable.
-2x + y + 3z = 20
3x - 2y + 3z = -9
―――――――――
-5x + 3y + 0 = 29
Next multiply 2
4
by -3 and add it to 1 to eliminate the same variable.
1
-2x + y + 3z = 20
2
―――――――――
-2x + y + 3z = 20
-3(-3x + 2y + z = 21)
⇒
9x - 6y - 3z = -63
――――――――
7x - 5y + 0 = -43
5
This results in the system of two linear equations below.
⎧
⎨
-5x + 3y = 29
7x - 5y = -43
© Houghton Mifflin Harcourt Publishing Company
⎩
4
5
To solve this system, multiply 4
4
5
5(-5x + 3y = 29)
by 5 and add the result to the product of 5 and 3.
-25 + 15y = 145
3(7x - 5y = -43)
⇒
―――――――
21x - 15y = -129
―――――――
-4x + 0 = 16
-4x = 16
x = -4
Substitute to solve for y and z.
-5x + 3y = 29
-3x + 2y + z = 21
[4]
-5(-4) + 3y = 29
[2]
-3(-4) + 2(3) + z = 21
y=3
z=3
The solution to the system is the ordered triple (-4, 3, 3).
Module 5
273
Lesson 3
LANGUAGE SUPPORT
A2_MTXESE353930_U2M05L3.indd 273
Communicate Math
Give each pair of students a sheet with three linear equations and the beginnings
of the three different solution methods for the system. Have them label the kind of
solution method shown (elimination, substitution, matrices). The partners must
agree that the labels are accurate and then describe which method is easiest for
each of them and why.
Students then solve the systems using the chosen method, and compare solutions.
Suggest students discuss any disagreements about a method’s ease of use.
273
Lesson 5.3
2/20/14 4:16 AM
B
⎧ x + 2y + 3z = 9
1
x + 3y + 2z = 5
2
⎪
⎨
⎪
x + 4y - z = -5
⎩
QUESTIONING STRATEGIES
3
x
Begin by subtracting equation 2 from equation 1 to eliminate
1
x + 2y + 3z = 9
2
x + 3y + 2z = 5
―――――――――
0x - y + z = 4
4
Now subtract equation 3 from equation
x
1 to eliminate
.
.
x + 2y + 3z = 9
1
3
What will happen if you choose two of the
three linear equations in the system, eliminate
one of the three variables, then choose another pair
of the original equations and eliminate a different
variable? You will then have a system of two
equations in three variables that is impossible
to solve.
x + 4y - z = -5
―――――――――
0x - 2y + 4z = 14
5
This results in a system of two linear equations:
⎧
⎨
⎩
-y + z = 4
4
-2y + 4z = 14
5
_
To solve this system, multiply equation 5
-y + z = 4
4
-1
2
and add it to equation 4 .
-y + z = 4
-__12 (-2y + 4z = 14)
5
by
⇒
y - 2z = -7
―――――
0y - z = -3
z=3
© Houghton Mifflin Harcourt Publishing Company
―――――――
Substitute to solve for y and x.
-y + z = 4
-y + 3 = 4
[4]
x + 2y + 3z = 9
x + 2(-1) + 3(3) = 9
y = -1
The solution to the system is the ordered triple
Module 5
A2_MTXESE353930_U2M05L3 274
[1]
x=2
(2, -1, 3)
274
.
Lesson 3
15-01-11 3:50 AM
Solving Linear Systems in Three Variables 274
Your Turn
EXPLAIN 3
4.
Solving a System of Three Linear
Equations Using Matrices
⎧ x + 2y + z = 8
⎪⎨ 2x + y - z = 4
⎪ x + y + 3z = 7
3
⎩
x + 2y + z = 8
+ 2x + y - z = 4
――――――
3x + 3y = 12
Remember to write each equation in the form
ax + by + cz = d before writing the matrix so that
the elements of the matrix are in the correct order.
3(2x + y - z = 4)
+ 7x + 4y = 19
――――――
⇒ 6x + 3y - 3z = 12
© Houghton Mifflin Harcourt Publishing Company
⇒
11(10x - 7z = -1)
⇒
-7(17x - 11z = 1)
⇒ -7x - 7y = -28
――――――――
+ 7x + 4y = 19
――――――
3y = 9
y=3
2
3
6x - 3y - 9z = 3
8x + 6y + 4z = -8
9x - 6y - 15z = 9
―――――――
17x - 11z = 1
110x - 77z = -11
-119x + 77z = -7
――――――――
-9x = -18
x= 2
10x + 7z = - 1→ 10(2) - 7z = -1 → z = 3
2x - y - 3z = 1 → 2(2) - y - 3(3) = 1 → y = -6
So the ordered triple is (2, -6, 3).
So the ordered triple is (1, 3, 1).
Solving a System of Three Linear Equations
Using Matrices
You can represent systems of three linear equations in a matrix. A matrix is a rectangular array of numbers
enclosed in brackets. Matrices are referred to by size: an m-by-n matrix has m rows and n columns.
A system of three linear equations can be written in a 3-by-4 matrix by first rearranging the equations so all of
the variables are to the left of the equals sign and the constant term is to the right. Each row now corresponds to
an equation. Enter the coefficients of the variables in the equation as the first three numbers in the row. Enter the
constant that was on the right side of the equation as the fourth number.
The system
⎧
Module 5
A2_MTXESE353930_U2M05L3 275
2x + y + 3z = 20 is expressed as
⎪⎨ 5x + 2y + z = 21
⎪ 3x - 2y + 7z = 9
⎩
Lesson 5.3
2(4x + 3y + 2z = -4)
1
4x + 3y + 2z = -4
――――――――
10x - 7z = -1
――――――――――
x + 2y + z = 8 → 1 + 2(3) + z = 8 ⇾ z = 1
275
⇒
+ x + y + 3z = 7 -3(-3x + 2y + 5z = -3)
――――――
7x + 4y = 19
x + (3) = 4 → x = 1
Explain 3
3(2x - y - 3z = 1)
4x + 3y + 2z = -4
―――――――
⇒ x+y=4
+ x + y + 3z = 7
――――――
-7(x + y = 4)
2x - y - 3z = 1
⎪
⎨ 4x + 3y + 2z = -4
⎪ -3x + 2y + 5z = -3
2
⎩
AVOID COMMON ERRORS
⎧
5.
1
⎢
⎡
⎣
2 1
3 20
5 2
1 21
3 -2 7 -9
275
⎥
⎤
in matrix form.
⎦
Lesson 3
15-01-11 3:50 AM
Gaussian Elimination is a formalized process of using matrices to eliminate two of the variables in each equation in
the system. This results in an easy way to find the solution set. The process involves using elementary row operations
to generate equivalent matrices that lead to a solution.
QUESTIONING STRATEGIES
What operations will produce a matrix that is
row-equivalent to the original? Interchange
two rows; multiply a row by a nonzero constant;
then add a multiple of one row to another.
The elementary row operations are
(1) Multiplying a row by a constant – When performing row multiplication, the product of the original value and the
constant replaces each value in the row.
(2) Adding two rows – In row addition, each value in the second row mentioned in the addition is replaced by the
sum of the values in the equivalent column of the two rows being added. These operations can also be performed
together.
The elimination can be continued past this point to a matrix in which the solutions can be simply read directly out of
the matrix. You can use a graphing calculator to perform these operations. The commands are shown in the table.
Command
Meaning
Syntax
*row(
replace each value in the row indicated with the product
of the current value and the given number
*row(value,matrix,row)
row+(
replace rowB with the sum of rowA and the current rowB
row+(matrix,rowA,rowB)
*row+(
replace rowB with the product of the given value and
rowA added to the current value of rowB
*row+(value,matrix,rowA,rowB)
Example 3
Solve the system of three linear equations using matrices.

⎧ -2x + y + 3z = 20
⎪⎨ -3x + 2y + z = 21
⎪ 3x - 2y + 3z = -9
⎩
Input the system as a 3-by-4 matrix. Multiply the first row by –0.5. Enter the command into
your calculator. Press enter to view the result.
© Houghton Mifflin Harcourt Publishing Company
Module 5
A2_MTXESE353930_U2M05L3 276
276
Lesson 3
15-01-11 3:50 AM
Solving Linear Systems in Three Variables 276
© Houghton Mifflin Harcourt Publishing Company
To reuse the resulting matrix, store it into Matrix B. Add 3 times row 1 to row 2. Press enter
to view the result. Remember to store the result into a new matrix.
Multiply row 2 by 2.
Add -3 times row 1 to row 3.
Add 0.5 times row 2 to row 3.
Multiply row 3 by 0.25.
Add 7 times row 3 to row 2.
Add 1.5 times row 3 to row 1.
Add 0.5 times row 2 to row 1.
The first row tells us that x = -4, the second row tells us that y = 3, and the third row tells us that z = 3.
So the solution is the ordered triple (-4, 3, 3).
Module 5
A2_MTXESE353930_U2M05L3.indd 277
277
Lesson 5.3
277
Lesson 3
1/14/15 9:41 PM
⎧ x + 2y + 3z = 9
x + 3y + 2z = 5
B ⎨
⎩ x + 4y - z = -5
⎡
⎢
⎢
⎢
Write as a matrix.
⎢
⎢
⎣
1
2
3
9
1
3
2
5
1
4
-1
-5
Perform row operations.
-r1 + r2
⎡1
⎢0
⎣1
2
3
1 -1
4 -1
⎢ 0
⎣
⎥
⎥
⎥
⎥
⎥
⎦
-r1 + r3
⎡1
9 ⎤
-4
-5 ⎦
⎥
⎢0
⎣0
-0.5r3
⎡
⎢1
⎢0
⎢
⎤
-2r2 + r3
2
3
1 -1
2 -4
⎡1
9 ⎤
-4
-14 ⎦
⎥
⎢0
⎣0
r3 + r2
2
1
3
-1
0
1
⎤
9 ⎥
-4 ⎥
⎥
3 ⎥
⎡1
⎢
9 ⎤
-4
-6 ⎦
⎥
-3r3 + r1
3
9 ⎤
0 -1
1 3 ⎦
2
0 1
⎣0 0
2
3
1 -1
0 -2
⎥
⎦
⎡
⎢ 1
⎢
0
⎢⎣ 0
0
2
⎢
1
0
0
1
⎤
0 ⎥⎥
-1
3
⎥
⎥⎦
-2r2 + r1
⎡1
0
1
⎣0 0
⎢0
0
2⎤
0 -1
1
3⎦
⎥
The solution is the ordered triple
(2, –1, 3)
.
Your Turn
⎨ 2xx ++yy+-3zz == 47
⎡1 2
1 8⎤
Input the system as a 3-by-4 matrix. 2 1 -1 4
3 7⎦
⎣1 1
⎩
⎢
⎥
Multiply row 1 by -2 and add it to row 2. Multiply row 1 by -1 and add it to row 3. Multiply
1
1
row 2 by -_
. Add row 2 to row 3. Multiply row 3 by _
. Multiply row 3 by -1 and add it to
3
3
row 1. Multiply row 3 by -1 and add it to row 2. Multiply row 2 by -1 and add it to row 1.
© Houghton Mifflin Harcourt Publishing Company
⎧ x + 2y + z = 8
6.
⎡1 0 0 1⎤
0 1 0 3 So, the solution of the system is (1, 3, 1).
⎣0 0 1 1⎦
⎢
Module 5
A2_MTXESE353930_U2M05L3 278
⎥
278
Lesson 3
1/12/15 9:31 PM
Solving Linear Systems in Three Variables 278
Explain 4
EXPLAIN 4
Example 4
Solving a Real-World Problem

INTEGRATE MATHEMATICAL
PROCESSES
Focus on Communication
A child has $6.17 in change in her piggy bank. The change
consists of 113 coins in a mix of pennies, nickels, and
quarters. If there are 8 times as many nickels as pennies,
how many of each coin does the child have? Solve using
substitution.
Begin by setting up a system of equations, and use p for the
number of pennies, n for the number of nickels, and q for
the number of quarters. Use the relationships in the problem
statement to write the equations.
Have students work in pairs or small groups to write
and solve another real-world problem that can be
modeled with a system of three linear equations in
three variables.
The total number of coins is the sum of the number of each coin. So, the first
equation is p + n + q = 113.
The total value of the coins is $6.17 or 617 cents (converting the value to cents will allow all coefficients to
be integers). The second equation will be p + 5n + 25q = 617.
The third relationship given is that there are eight times as many nickels as pennies or, n = 8p. This gives
the following system of equations:
QUESTIONING STRATEGIES
⎧ p + n + q = 113
⎨
1
p + 5n + 25q = 617
⎩ n = 8p
2
3
Equation 3 is already solved for n. Substitute for n in equations 1 and 2 and simplify.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©val
lawless/Shutterstock
Will there always be a solution to a system
of three linear equations in three variables
written to model a real-world problem? No. There
will not be a solution if all of the given parameters
cannot be met.
Solving a Real-World Problem
p + (8p) + q = 113
p + 5(8p) + 25q = 617
9p + q = 113 4
p + 40p + 25q = 617
41p + 25q = 617
This results in the following linear
system in two variables:
Solve equation 4 for q.
⎧ 9p + q = 113
9p + q = 113
⎨
4
⎩ 41p + 25q = 617
5
q = 113 - 9p
5
Substitute for q in equation 5 and solve for p.
Use at p = 12 to find q and n.
41p + 25(113 - 9p) = 617
q = 113 - 9p
n = 8p
q = 113 - 9(12)
n = 8(12)
q=5
n = 96
41p + 2825 - 225p = 617
12 = p
The child’s piggy bank contains 12 pennies, 96 nickels, and 5 quarters.
Module 5
A2_MTXESE353930_U2M05L3 279
279
Lesson 5.3
279
Lesson 3
15-01-11 3:50 AM
B
A student is shopping for clothes. The student needs to buy an equal number of shirts and
ties. He also needs to buy four times as many shirts as pants. Shirts cost $35, ties cost $25,
and pants cost $40. If the student spends $560, how many shirts, pants, and ties did he get?
CONNECT VOCABULARY
Compare and contrast a system of three linear
equations in three variables to the linear-quadratic
systems discussed in the previous lesson. Have
students complete a chart showing the similarities
and differences between these two kinds of systems.
Begin by setting up a system of equations, using s for the number of shirts, t for the number of ties, and p for
the number pairs of pants. Use the relationships in the problem statement to write the equations.
The number of shirts is equal to the number of ties. So, the first equation is s = t.
The number of shirts is equal to 4 times the number of pairs of pants, so a second equation is s = 4p .
The total the student spent is the sum of the cost of the shirts, the ties, and the pairs of pants.
35s + 25t + 40p = 560
The system of equations is below.
⎧s=t
⎨
1
s = 4p
⎩ 35s + 25t + 40p = 560
Equation
4p = s
2
3
1 is already solved for t. Solve equation 2
for p.
_1
p = 4s
Substitute for p and t in equation 3
( )
and solve for s.
1 s = 560
35s + 25(s) + 40 _
4
35s + 25s + 10s = 560
70 s = 560
s= 8
Evaluate the equation solved for p above at s = 8 to find p.
© Houghton Mifflin Harcourt Publishing Company
1s
p=_
4
1 (8) = 2
p=_
4
Recall that s = t, so t = 8.
The student bought 8 shirts, 8 ties, and 2 pairs of pants.
Module 5
A2_MTXESE353930_U2M05L3.indd 280
280
Lesson 3
2/20/14 4:16 AM
Solving Linear Systems in Three Variables 280
Your Turn
ELABORATE
7.
AVOID COMMON ERRORS
Students may assume that systems have a single
solution, namely, that the equations are unique and
their graphs intersect at a point. Remind them that a
system of three linear equations may be dependent;
that is, all three equations may describe the same
plane, or the system may be inconsistent, meaning
there is no point or line in which all three planes
intersect.
Louie Dampier is the leading scorer in the history of the American Basketball Association (ABA). His
13,726 points were scored on two-point baskets, three-point baskets, and one-point free throws. In his
ABA career, Dampier made 2144 more two-point baskets than free throws and 1558 more free throws than
three-point baskets. How many three-point baskets, two-point baskets, and free throws did Dampier make?
r = f + 2144
f = t + 1558
⇒ r = (t + 1558) + 2144
⇒
r = t + 3702
2r + 3t + f = 13726
2(t + 3702) + 3t + (t + 1558) = 13726
f = 794 + 1558 = 2352
t = 794
↓
r = 2352 + 2144 = 4496
So, t = 794, f = 2352 and r = 4496.
Elaborate
SUMMARIZE THE LESSON
What are the principal methods of solving
a real-world problem that can be modeled
by a system of three linear equations in three
variables? Translate to a system, then use
substitution, elimination, or matrices to solve
it and apply the solution to the problem.
8.
If you are given a system of linear equations in three variables, but the system only has two equations, what
happens when you try to solve it?
The solution will be a line because it will be in terms of two variables.
9.
Discussion Why does a system need to have at least as many equations as unknowns to have a unique
solution?
If there is not an equation for each variable, the solution processes outlined above
© Houghton Mifflin Harcourt Publishing Company
cannot progress to the end, and you will be left with one variable defined only in
terms of another.
10. Essential Question Check-In How can you find the solution to a system of three linear equations
in three variables?
The system can be solved by substitution, elimination, or by using matrices.
Module 5
A2_MTXESE353930_U2M05L3.indd 281
281
Lesson 5.3
281
Lesson 3
1/24/15 9:12 PM
Evaluate: Homework and Practice
Solve the system using substitution.
1.
⎧
⎪ 4x + y - 2z = -6 1
⎪
2x - 3y + 3z = 9 2
⎨
⎪
x - 2y = 0 3
⎪
⎩
2.
x - 2y = 0 → x = 2y
ASSIGNMENT GUIDE
↓
x + 17y = 13
↓
6x - 2y + 4(4y - 3) = 0
9y - 2z = -6
↓
2(2y) - 3y + 3z = 9
3x + 7y = 6
↓
Solving this system yields x = 0.25
and y = 0.75.
y + 3z = 9
z = 4y - 3 = 4(0.75) - 3 = 0
Solving this system yields y = 0 and z = 3
So the ordered triple is (0.25, 0.75, 0).
x = 2(0) = 0
So the ordered triple is (0, 0, 3).
Solve the system using elimination.
3.
⎧
⎪ x + 5y + 3z = 4 1
⎪
4y - z = 3 2
⎨
⎪ 6x - 2y + 4z = 0 3
⎪
⎩
x + 5y + 3(4y - 3) = 4
4(2y) + y - 2z = -6
⎪ 4x + y - 2z = -6
• Online Homework
• Hints and Help
• Extra Practice
4y - z = 3 → 4y - 3 = z
Now substitute for x in the first and
second equations and simplify.
⎧
EVALUATE
1
⎪
⎩
12x + 3y - 6z = -18
2 (2x - 3y + 3z = 9)
4x - 6y + 6z = 18
2x - 3y + 3z = 9
――――――――
-3 (x - 2y = 0)
――――――――
2 (14x - 3z = -9)
x + 6z = 18
―――――――
⇒
2x - 3y + 3z = 9
――――――――
14x - 3z = -9
© Houghton Mifflin Harcourt Publishing Company
⎪
⎨ 2x - 3y + 3z = 9 2
⎪
x - 2y = 0 3
3 (4x + y - 2z = -6) ⇒
- 3x + 6y = 0
―――――――
x + 6z = 18
⇒
28x - 6z = -18
x + 6z = 18
―――――――
29x = 0
x=0
14x - 3z = -9 → 14 (0) - 3z = -9 → z = 3
Depth of Knowledge (D.O.K.)
Exercise 14
Example 1
Solving a System of Three Linear
Equations Using Substitution
Exercises 1–2
Example 2
Solving a System of Three Linear
Equations Using Elimination
Exercises 3–5
Example 3
Solving a System of Three Linear
Equations Using Matrices
Exercises 6–7
Example 4
Solving a Real-World Problem
Exercises 11–13
Lesson 3
282
Exercise
Explore
Recognizing Ways that Planes Can
Intersect
Have students work in small groups. Have each group
use pieces of paper to model systems of three
equations in three variables. Have each group model
a consistent system with a single point as a solution,
a consistent system with a line as a solution, an
inconsistent system, and a dependent system.
So the ordered triple is (0, 0, 3).
A2_MTXESE353930_U2M05L3 282
Practice
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Modeling
4x + y - 2z = -6 → 4 (0) + y - 2(3) = -6 → y = 0
Module 5
Concepts and Skills
Mathematical Processes
1–5
1 Recall of Information
1.F Analyze relationships
6–7
1 Recall of Information
1.C Select tools
8–9
1 Recall of Information
1.F Analyze relationships
10
2 Skills/Concepts
1.D Multiple representations
11–13
2 Skills/Concepts
1.A Everyday life
1/12/15 9:40 PM
Solving Linear Systems in Three Variables 282
⎧
AVOID COMMON ERRORS
4.
Remind students that a system of three linear
equations may have no solution (when all three
planes do not intersect), a single solution (when the
planes intersect at a single point), or an infinite
number of solutions (when the planes intersect in a
line and when the system is dependent).
⎪ x + 5y + 3z = 4 1
⎪
4y - z = 3 2
⎨
⎪ 6x - 2y + 4z = 0 3
⎪
⎩
3 (4y - z = 3)
⇒
x + 5y + 3z = 4
―――――――
12y - 3z = 9
x + 5y + 3z = 4
―――――――
x + 17y = 13
4 (4y - z = 3) ⇒
6x - 2y + 4z = 0
―――――――
16y - 4z = 12
6x - 2y + 4z = 0
―――――――
6x + 14y = 12
↓
3x + 7y = 6
-3 (x + 17y = 13) ⇒
3x + 7y = 6
―――――――
-3x - 51y = -39
3x + 7y = 6
―――――――
-44y = -33
y = 0.75
x + 17y = 13 → x + 17 (0.75) = 13 → x = 0.25
x + 5y + 3z = 4 → 0.25 + 5(0.75) + 3z = 4 → z = 0
So the ordered triple is (0.25, 0.75, 0).
⎧
5.
⎪ 2x - y + 3z = -12 1
⎪ -x + 2y - 3z = 15 2
⎨
⎪
y + 5z = -6 3
⎪
⎩
2x - y + 3z = -12
-x + 2y - 3z = 15
―――――――――
x+ y=3
5 (-x + 2y - 3z = 15) ⇒
© Houghton Mifflin Harcourt Publishing Company
3 (y + 5z = -6)
―――――――――
5 (x + y = 3)
-5x + 13y = 57
――――――――
⇒
-5x + 10y - 15z = 75
3y + 15z = -18
―――――――――
-5x + 13y = 57
5x + 5y = 15
-5x + 13y = 57
――――――
―――――――
18y = 72
x + y = 3 → x + (4) = 3 → x = -1
y=4
2x - y + 3z = -12 → 2 (-1) - 4 + 3z = -12
3z = -6 → z = -2
So the ordered triple is (-1, 4, -2).
Module 5
Exercise
A2_MTXESE353930_U2M05L3 283
283
Lesson 5.3
Lesson 3
283
Depth of Knowledge (D.O.K.)
Mathematical Processes
14–15
3 Strategic Thinking
1.G Explain and justify arguments
16
3 Strategic Thinking
1.D Multiple representations
1/12/15 9:44 PM
Solve the system of three linear equations using matrices.
⎧
6.
⎪ 4x + y - 2z = -6 1
⎪
⎨ 2x - 3y + 3z = 9
⎪
x - 2y = 0
⎪
2
3
⎩
QUESTIONING STRATEGIES
Input the system as a 3-by-4 matrix.
[A] ⎡
1
4
2 -3
⎣ 1 -2
⎢
Can a system of three linear equations in three
variables be solved by graphing on the
coordinate plane? Why or why not? No; graphing in
three variables requires a dimension for each of the
variables, so a third variable would require a third
axis, and thus a three-dimensional coordinate
system.
-2 -6 ⎤
3
9
0
0⎦
⎥
Switch row 3 with row 1 to make the matrix easier to solve.
Multiply row 1 by -4 and add it to row 3.
Multiply row 1 by -2 and add it to row 2.
Multiply row 2 by -9 and add it to row 3.
Multiply row 3 by - 1 .
29
Multiply row 3 by -3 and add it to row 2.
_
Multiply row 2 by 2 and add it to row 1.
⎡1 0 0 0⎤
0 1 0 0
⎣0 0 1 3⎦
⎢
⎥
So, the solution is (0, 0, 3).
⎧
7.
⎪ x + 5y + 3z = 4 1
⎪
⎨
⎪
4y - z = 3 2
⎪ 6x - 2y + 4z = 0
⎩
3
Input the system as a 3-by-4 matrix.
[A] ⎡
3 4⎤
5
1
4 -1 3
0
4 0⎦
⎣ 6 -2
⎢
⎥
Multiply row 1 by -6 and add it to row 3.
© Houghton Mifflin Harcourt Publishing Company
Multiply row 2 by 8 and add it to row 3.
Multiply row 3 by - 1 .
22
Add row 3 to row 2.
_
Multiply row 2 by 0.25.
Multiply row 2 by -5 and add it to row 1.
Multiply row 3 by -3 and add it to row 1.
⎡ 1 0 0 .25 ⎤
0 1 0 .75
⎣0 0 1 0 ⎦
⎢
⎥
So, the solution is (0.25, 0.75, 0).
Module 5
A2_MTXESE353930_U2M05L3 284
284
Lesson 3
1/12/15 9:48 PM
Solving Linear Systems in Three Variables 284
Solve the system of linear equations using your method of choice.
MULTIPLE REPRESENTATIONS
⎧
Show students that there are other ways to solve
systems that involve determinants or augmented
matrices with row reduction. Show students how a
two-by-two system can be solved by using these
methods, then explain how the methods can be
applied to three-by-three systems.
8.
⎪
2x - y + 3z = 5 1
⎪
⎨ -6x + 3y - 9z = -15 2
⎪
⎪
4x - 2y + 6z = 10 3
⎩
2x - y + 3z = 5
→
2x + 3z - 5 = y
-6x + 3 (2x + 3z - 5) - 9z = -15
4x - 2 (2x + 3z - 5) + 6z = 10
-15 = -15
10 = 10
-6x + 6x + 9z - 15 - 9z = -15
4x - 4x - 6z + 10 + 6z = 10
Both equations are true, so the system has infinitely many solutions.
9.
⎧
⎪3x + 4y - z = -7 1
⎪
⎨ x - 5y + 2z = 19 2
⎪
⎪ 5x + y - 2z = 5
3
⎩
3x + 4y - z = -7
x - 5y + 2 (3x + 4y + 7) = 19
→
3x + 4y + 7 = z
5x + y - 2 (3x + 4y + 7) = 5
⇒
x - 5y + 6x + 8y + 14 = 19
5x + y - 6x - 8y - 14 = 5
7x + 3y = 5
-x - 7y = 19
Solving this system yields x = 2 and y = -3
© Houghton Mifflin Harcourt Publishing Company
3x + 4y + 7 = z → 3(2) + 4(-3) + 7 = z → 1 = z
So the ordered triple is (2, -3, 1)
10. Find the equation of the parabola passing through the points (3, 7), (30, -11), and (0, -1).
Identify the values of a, b, and c for the general form of a parabola
(7) = a(3) 2 + b (3) + c
(−11) = a (30) + b (30) + c
2
2
(−1) = a (0) + b(0) + c
⎧ 8 = 9a + 3b
⇒⎨
⎩ -1 = 90a + 3b
7 = 9a + 3b + c
⇒
−11 = 900a + 30b + c
7 = 9a + 3b − 1
⇒
−11 = 900a + 30b − 1
−1 = c
1
Solving this systems yields b = 3 and a = -_
9
1 2
x + 3x - 1
So, the equation of the parabola connecting (3, 7), (30, -11), and (0, -1) is y = -_
9
Module 5
A2_MTXESE353930_U2M05L3 285
285
Lesson 5.3
285
Lesson 3
1/12/15 9:47 PM
11. Geometry In triangle ABC, the measure of angle X is eight times the sum of the
measures of angles Y and Z. The measure of angle Y is three times the
measure of angle Z. What are the measures of the angles?
Y = 3Z
X = 8(Y + Z) = 8(3Z) + 8Z = 32Z
Substitute the equations for X and Y in terms of Z into the first equation.
X + Y + Z = 180 → (32Z) + (3Z) + Z = 180 → Z = 5
Y = 3(5)
X = 8(15) + 8(5)
Y = 15
X = 160
So, angle X measures 160°, angle Y measures 15°, and angle Z measures 5°.
12. The combined age of three relatives is 120 years. James is
three times the age of Dan, and Paul is two times the sum of
the ages of James and Dan. How old is each person?
Choose an equation and variable to start with. The
third equation is already solved for J: J = 3D
Now substitute for J in the second equation and
simplify.
P = 2J + 2D = 2(3D) + 2D = 8D
J + D + P = 120 → (3D) + D + (8D) = 120 → D = 10
J = 3D = 3(10) = 30
P = 2J + 2D = 2(30) + 2(10) = 80
© Houghton Mifflin Harcourt Publishing Company • Image Credits: (t) ©Steve
Hix/Somos Images/Corbis; (b) ©xPACIFICA/Corbis
So, Dan is 10 years old, James is 30 years old, and
Paul is 80 years old.
13. Economics At a stock exchange there were a total of
10,000 shares sold in one day. Stock A had four times as
many shares sold as Stock B. The number of shares sold for
Stock C was equal to the sum of the numbers of shares sold
for Stock A and Stock B. How many shares of each stock
were sold?
A = 4B
C = A + B = (4B) + B = 5B
A + B + C = 10,000 → (4B) + B + (5B) = 10,000 → B = 1000
A = 4B = 4(1000) = 4000
C = A + B = 4000 + 1000 = 5000
4000 shares of Stock A, 1000 shares of Stock B, and 5000 shares of Stock C were sold.
Module 5
A2_MTXESE353930_U2M05L3 286
286
Lesson 3
15-01-11 3:50 AM
Solving Linear Systems in Three Variables 286
JOURNAL
H.O.T. Focus on Higher Order Thinking
Have students list the methods they have learned for
solving systems of three linear equations in three
variables, and include an example problem for each
method.
14. Communicate Mathematical Ideas Explain how you know when a system has infinitely many
solutions or when it has no solutions.
If when solving the system you get a true statement (such as 0 = 0), then that system
has infinitely many solutions. If when solving the system you get a false statement
(such as 1 = 3), then that system will have no solution.
15. Explain the Error A student was asked to solve this system of equations using matrices. Find and correct
the student’s error.
⎧ 5x +7y + 9x = 0
⎨
⎩
x - y + z = -3
8x + y = 12
7 9⎤
-1 1
1 0⎦
⎣8
⎡5
⎢1
⎥
The student did not include the last column of numbers
representing the values to the right of the equals sign.
The correct matrix set up should be as follows.
⎡5
0⎤
7 9
1 -1 1 -3
⎣8
1 0 12⎦
⎢
⎥
16. Critical Thinking Explain why the following system of equations cannot be solved.
⎧ 7x + y + 6z = 1
⎨
When solving a system of equations, there must be at least as many
equations as there are variables.
© Houghton Mifflin Harcourt Publishing Company
⎩ -x - 4y + 8z = 9
Module 5
A2_MTXESE353930_U2M05L3 287
287
Lesson 5.3
287
Lesson 3
15-01-11 3:49 AM
Lesson Performance Task
INTEGRATE MATHEMATICAL
PROCESSES
Focus on Modeling
A company that manufactures inline skates needs to order three
parts—part A, part B, and part C. For one shipping order the company
needs to buy a total of 6000 parts. There are four times as many B
parts as C parts. The total number of A parts is one-fifth the sum of
the B and C parts. On previous orders, the costs had been $0.25 for
part A, $0.50 for part B, and $0.75 for part C, resulting in a cost of
$3000 for all the parts in one order. When filling out an order for
new parts, the company sees that it now costs $0.60 for part A, $0.40
for part B, and $0.60 for part C. Will the company be able to buy the
same quantity of parts at the same price as before with the new prices?
Because the information is listed in a long paragraph,
students may have difficulty setting up the system.
First have them read the statement of the problem
and decide what the variables are. Have them
consider each sentence of the paragraph separately
and write it in simplified form, if necessary. Lightly
crossing off each item of information as it is turned
into an equation will make it easier to see what
remains to be modeled.
QUESTIONING STRATEGIES
⎧A + B + C = 6000
1
A = __
(B + C)
5
⇒
⎩B=C
⎧(C) + 5C = 6000
⎨
⇒ A=C
⎩ B = 4C
⎨
1 (4C) + C
A=_
)
(
5
B
=
4C
⎩
⇒
⎧C = 1000
A=C
⎩ B = 4C
⎨
⇒
⎧A + 5C = 6000
⇒ A=C
⎩ B = 4C
⎨
⎧C = 1000
A=C
⎩ B = 4C
⎨
⇒
⎧ C = 1000
A = 1000
⎩ B = 4(1000) = 4000
⎨
Determine the new costs of each part with the price change.
Part A = 1000 · 0.60 = 600
Part B = 4000 · 0.40 = 1600
Part C = 1000 · 0.60 = 600
The total cost will now be $2800, so the company will be able to afford the parts it needs and can
actually buy more.
Module 5
288
What quantities do each of the equations
represent? One equation represents the total
number of parts, one represents the relationship
between the numbers of units of A to the numbers
of units of B and C, and one represents the
relationship between the numbers of units of B to
the numbers of units of C.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Henry
Westheim Photography/Alamy
⎨
⎧A + (4C) + C = 6000
Lesson 3
EXTENSION ACTIVITY
A2_MTXESE353930_U2M05L3.indd 288
The techniques for solving three equations in three unknowns can be extended to
solve systems with four or more equations. Ask students to explore solution
methods for solving systems of four or more equations, and to examine their
graphing calculators’ matrix-solving capabilities as well. Have students report on
the methods used, as well as on the differences they found. They should note that
graphing is not an option, because a fourth dimension, or greater, would be
required.
2/20/14 4:15 AM
Scoring Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain his/her reasoning.
0 points: Student does not demonstrate understanding of the problem.
Solving Linear Systems in Three Variables 288