DO NOT EDIT--Changes must be made through “File … Solve systems of three linear equations in...

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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

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Explore Recognizing Ways that Planes Can IntersectRecall 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

A linear equation in three variables has three distinct variables, each of which is either first degree or has a coefficient of zero.

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) .

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.

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 diagrams show some ways three planes can intersect. How many points lie on all 3

planes?

The diagram shows three intersecting planes.

How many points lie on all 3 planes?

0

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Module 5 269 Lesson 3

5 . 3 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?

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A2_MTXESE353930_U2M05L3 269 2/22/14 5:59 AM

Texas Math StandardsThe student is expected to:

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

Mathematical Processes

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.

Language Objective

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.

HARDCOVER PAGES 191204

Turn to these pages to find this lesson in the hardcover student edition.

Solving Linear Systems in Three Variables

ENGAGE 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.

PREVIEW: LESSON PERFORMANCE TASKView 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.

269

HARDCOVER PAGES

Turn to these pages to find this lesson in the hardcover student edition.

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

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Explore 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

A linear equation in three variables has three distinct variables, each of which is either first degree or has a

coefficient of zero.

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) .

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.

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 diagrams show some ways three planes can intersect. How many points lie on all 3

planes?

The diagram shows three intersecting planes.

How many points lie on all 3 planes?

0

1

Module 5

269

Lesson 3

5 . 3 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?

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CorrectionKey=TX-A

A2_MTXESE353930_U2M05L3 269

2/22/14 5:59 AM

269 Lesson 5 . 3

L E S S O N 5 . 3

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The diagram shows planes intersecting in a different way.

Describe the intersection.

How many points lie in all 3 planes?

Reflect

1. Discussion Give an example of three planes that intersect at exactly one point.

Explain 1 Solving a System of Three Linear Equations Using Substitution

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

-3x + 2y + z = 21

3x - 2y + 3z = -9

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.

-2x + y + 3z = 20

y = 2x - 3z + 20

Now substitute for y in equations and 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 -9 = -x + 9z - 40

-19 = x - 5z 31 = -x + 9z

1

2

2

4 5

3

3

The intersection is a line.

an infinite number

Sample answers: The three dimensional coordinate axes; three faces of a cube that share a

common corner.




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Integrate Mathematical ProcessesThis 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.

EXPLORE Recognizing Ways that Planes Can Intersect

INTEGRATE TECHNOLOGYHave students draw three intersecting planes using an art application on their iPads.

INTEGRATE MATHEMATICAL PROCESSESFocus on ModelingSharing 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.

EXPLAIN 1 Solving a System of Three Linear Equations Using Substitution

INTEGRATE MATHEMATICAL PROCESSESFocus on CommunicationHave students explain to a partner the process of solving a system of three linear equations in three variables by substitution.

PROFESSIONAL DEVELOPMENT

Solving Linear Systems in Three Variables 270


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This results in the following linear system in two variables:

⎧

⎨

⎩

-19 = x - 5z 31 = -x + 9z

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) 2 + b (2) + c

⇒

⎧ ⎪

⎨

⎪

⎩

1 = 4a + 2b + c4 = a (-1) 2 + b (-1) + c 4 = a - b + c

3 = 4a - 2b + c = a

( )

2

+ b ( )

+ c

Choose an equation in which it is easier to isolate a variable. Solve equation [2] for c.

c = 4 - a + [2]

Now substitute for c = in equations [1] and [3].

1 = 4a + 2b + ( ) 3 = 4a - 2b + ( ) 1 = + 3b + 4 3 = 3a - + 4

= 3a + 3b [4] = 3a - b [5]

4

5

1

2

33 -2 -2

b

3a

-3

b

-1

4 - a + b 4 - a + b


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COLLABORATIVE LEARNING

Peer-to-Peer ActivityHave 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.

QUESTIONING STRATEGIESAfter 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.

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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 +

Substitute into equation [4] and solve for a. Then use the value for a to find the values of b.

3a + 3 ( ) = -3 3 (- 1 _ 2 ) +3b = -3

3a + 9a + = -3 +3b = -3

12a = 3b =

a = b =

Then use the values a and b to solve for c.

c = 4 - a + b = 4 - (- 1 _ 2 ) + (- 1 _ 2 ) = 4

So the equation of the parabola connecting (2, 1) , (-1, 4) , and (-2, 3) is

y = x 2 - x + .

Your Turn

2.

⎧

⎨

⎩

x + 2y + z = 82x + y - z = 4x + y + 3z = 7

3.

⎧

⎨ ⎩

2x - y - 3z = 14x + 3y + 2z = -4

-3x + 2y + 5z = -3

1

3a + 1

3

- 1 _ 2

- 1 _ 2

- 3 _ 2

- 3 _ 2

-6

x + 2y + z = 8 → z = 8 - x - 2ySubstitute for z in the other equations. 2x + y - (8 - x - 2y) = 4 x + y + 3 (8 - x - 2y) = 7 ↓ ↓ x + y = 4 -2x - 5y = -17Solving this system yields x = 1 and y = 3.z = 8 - x - 2y → z = 8 - 1 - 2 (3) = 1So the ordered triple is (1, 3, 1) .

2x - y -3z = 1 → 2x - 3z - 1 = ySubstitute for y in the other equations. 4x + 3 (2x - 3z -1) + 2z = -4 -3x + 2 (2x - 3z - 1) + 5z = -3 ↓ ↓ 10x - 7z = -1 x - z = -1Solving this system yields x = 2 and z = 3.2x - 3z -1 = y → 2 (2) - 3 (3) - 1 = y = -6So the ordered triple is (2, -6, 3) .

- 1 _ 2 1 _ 2 4


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DIFFERENTIATE INSTRUCTION

Visual CuesShow 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.



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Explain 2 Solving a System of Three Linear Equations Using Elimination

You can also solve systems of three linear equations using elimination.

Example 2

⎧ ⎪ ⎨ ⎪

⎩

-2x + y + 3z = 20

-3x + 2y + z = 21

3x - 2y + 3z = -9

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 from equation

to eliminate the z variable.

-2x + y + 3z = 20

――――――――― 3x - 2y + 3z = -9

-5x + 3y + 0 = 29

Next multiply by -3 and add it to to eliminate the same variable.

-2x + y + 3z = 20 -2x + y + 3z = 20

――――――――― -3 (-3x + 2y + z = 21) ⇒ ―――――――― 9x - 6y - 3z = -63

7x - 5y + 0 = -43

This results in the system of two linear equations below.

⎧

⎨ ⎩

-5x + 3y = 29

7x - 5y = -43

To solve this system, multiply by 5 and add the result to the product of and 3.

5 (-5x + 3y = 29) -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 [4] -3x + 2y + z = 21 [2]

-5 (-4) + 3y = 29 -3 (-4) + 2 (3) + z = 21

y = 3 z = 3

The solution to the system is the ordered triple (-4, 3, 3) .

3

3

1

1

2

4

2

4

4

4

5

5

5

1

1

2

3

5

1




LANGUAGE SUPPORT

Communicate MathGive 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.

EXPLAIN 2 Solving a System of Three Linear Equations Using Elimination

AVOID COMMON ERRORSWhen 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.

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B

⎧ ⎪ ⎨

⎪

⎩

x + 2y + 3z = 9

x + 3y + 2z = 5

x + 4y - z = -5

Begin by subtracting equation from equation to eliminate .

x + 2y + 3z = 9

――――――――― x + 3y + 2z = 5

0x - y + z = 4

Now subtract equation from equation to eliminate .

x + 2y + 3z = 9

――――――――― x + 4y - z = -5

0x - 2y + 4z = 14

This results in a system of two linear equations:

⎧

⎨

⎩

To solve this system, multiply equation by and add it to equation .

-y + z = 4 -y + z = 4

――――――― - 1 __ 2 (-2y + 4z = 14) ⇒ ――――― y - 2z = -7

0y - z = -3

z = 3

Substitute to solve for y and x.

-y + z = 4 [4] x + 2y + 3z = 9 [1]

-y + 3 = 4 x + 2 (-1) + 3 (3) = 9

y = -1 x = 2

The solution to the system is the ordered triple .

4

5

45

1

1

1

1

4

4

2

3

2

3

5

5

1

3

2

x

x

- 1 _ 2

-2y + 4z = 14

-y + z = 4

(2, -1, 3)




A2_MTXESE353930_U2M05L3 274 15-01-11 3:50 AM

QUESTIONING STRATEGIESWhat 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.



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Your Turn

4.

⎧ ⎪ ⎨

⎪

⎩

x + 2y + z = 8

2x + y - z = 4

x + y + 3z = 7

5.

⎧ ⎪ ⎨

⎪

⎩

2x - y - 3z = 1

4x + 3y + 2z = -4

-3x + 2y + 5z = -3

1

2

3

Explain 3 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 is expressed as in matrix form.

⎧ ⎪ ⎨

⎪

⎩

2x + y + 3z = 20

5x + 2y + z = 21

3x - 2y + 7z = 9

⎡

⎢

⎣

2 1 3 20

⎤

⎥

⎦

5 2 1 21

3 -2 7 -9

1

2

3

x + 2y + z = 8

+ ―――――― 2x + y - z = 4

3x + 3y = 12 ⇒ x + y = 4

3 (2x + y - z = 4) ⇒ 6x + 3y - 3z = 12

+ ―――――― x + y + 3z = 7 + ―――――― x + y + 3z = 7

7x + 4y = 19

-7 (x + y = 4) ⇒ -7x - 7y = -28

+ ―――――― 7x + 4y = 19 + ―――――― 7x + 4y = 19

3y = 9

y = 3

x + (3) = 4 → x = 1

x + 2y + z = 8 → 1 + 2 (3) + z = 8 ⇾ z = 1

So the ordered triple is (1, 3, 1) .

3 (2x - y - 3z = 1) ⇒ 6x - 3y - 9z = 3

――――――― 4x + 3y + 2z = -4 ―――――――― 4x + 3y + 2z = -4

10x - 7z = -1

2 (4x + 3y + 2z = -4) ⇒ 8x + 6y + 4z = -8

―――――――――― -3 (-3x + 2y + 5z = -3) ――――――― 9x - 6y - 15z = 9

17x - 11z = 1

11 (10x - 7z = -1) ⇒ 110x - 77z = -11

―――――――― -7 (17x - 11z = 1) ―――――――― -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) .



A2_MTXESE353930_U2M05L3 275 15-01-11 3:50 AM

EXPLAIN 3 Solving a System of Three Linear Equations Using Matrices

AVOID COMMON ERRORSRemember 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.

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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.

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.




A2_MTXESE353930_U2M05L3 276 15-01-11 3:50 AM

QUESTIONING STRATEGIESWhat 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.



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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).



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277 Lesson 5 . 3


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B ⎧

⎨

⎩

x + 2y + 3z = 9

x + 3y + 2z = 5 x + 4y - z = -5

Write as a matrix.

⎡

⎢

⎢

⎢

⎢

⎢

⎣

⎤

⎥

⎥

⎥

⎥

⎥

⎦

Perform row operations. -r1 + r2 -r1 + r3 -2r2 + r3

⎡

⎢

⎣ 1 2 3 9

0 1 -1 -4 1 4 -1 -5

⎤

⎥

⎦

⎡

⎢

⎣ 1 2 3 9

0 1 -1 -4 0 2 -4 -14

⎤

⎥

⎦

⎡

⎢

⎣ 1 2 3 9

0 1 -1 -4 0 0 -2 -6

⎤

⎥

⎦

-0.5r3 r3 + r2 -3r3 + r1

⎡ ⎢

⎢

⎢

⎢

⎣

1 2 3 9

0 1 -1 -4

⎤ ⎥

⎥

⎥

⎥

⎦

⎡

⎢

⎣ 1 2 3 9

0 1 0 -1 0 0 1 3

⎤

⎥

⎦

-2r2 + r1

⎡

⎢

⎣ 1 0 0 2

0 1 0 -1 0 0 1 3

⎤

⎥

⎦ The solution is the ordered triple .

Your Turn

6.

⎧

⎨

⎩

x + 2y + z = 8

2x + y - z = 4 x + y + 3z = 7

⎡

⎢

⎢

⎢

⎢ ⎣

0 1 0 -1

0 0 1 3

⎤

⎥

⎥

⎥

⎥ ⎦

1

1

1

2

3

4

3

2

-1

9

5

-5

1 2 0 0

0 0 1 3

(2, –1, 3)

Input the system as a 3-by-4 matrix.

⎡

⎢

⎣ 1

2 1

2

1 1

1

-1

3

8

4

7

⎤

⎥

⎦

Multiply row 1 by -2 and add it to row 2. Multiply row 1 by -1 and add it to row 3. Multiply

row 2 by - 1 _ 3 . Add row 2 to row 3. Multiply row 3 by 1 _ 3 . Multiply row 3 by -1 and add it to

row 1. Multiply row 3 by -1 and add it to row 2. Multiply row 2 by -1 and add it to row 1.

⎡

⎢

⎣ 1 0 0 1

0 1 0 3 0 0 1 1

⎤

⎥

⎦ So, the solution of the system is (1, 3, 1) .




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Explain 4 Solving a Real-World Problem

Example 4

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.

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:

⎧

⎨

⎩

p + n + q = 113

p + 5n + 25q = 617

n = 8p

Equation is already solved for n. Substitute for n in equations and and simplify.

p + (8p) + q = 113 p + 5 (8p) + 25q = 617

9p + q = 113 p + 40p + 25q = 617

41p + 25q = 617

This results in the following linear Solve equation for q. system in two variables:

⎧

⎨

⎩ 9p + q = 113

41p + 25q = 617

9p + q = 113

q = 113 - 9p

Substitute for q in equation and solve for p. Use at p = 12 to find q and n.

41p + 25 (113 - 9p) = 617 q = 113 - 9p n = 8p

41p + 2825 - 225p = 617 q = 113 - 9 (12) n = 8 (12)

12 = p q = 5 n = 96

The child’s piggy bank contains 12 pennies, 96 nickels, and 5 quarters.

1

1

4

5

4

4

5

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2

2

3

3



A2_MTXESE353930_U2M05L3 279 15-01-11 3:50 AM

EXPLAIN 4 Solving a Real-World Problem

INTEGRATE MATHEMATICAL PROCESSESFocus on CommunicationHave 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.

QUESTIONING STRATEGIESWill 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.

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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?

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 = .

The total the student spent is the sum of the cost of the shirts, the ties, and the pairs of pants.

35s + 25t + 40p =

The system of equations is below.

⎧

⎨

⎩

s = t

s = 4p 35s + 25t + 40p = 560

Equation is already solved for t. Solve equation for p.

4p = s

p =

Substitute for p and t in equation and solve for s.

35s + 25 (s) + 40 ( 1 _ 4 s) = 560

35s + 25s + 10s = 560

s = 560

s =

Evaluate the equation solved for p above at s = 8 to find p.

p = 1 _ 4 s

p = 1 _ 4 (8) = 2

Recall that s = t, so t = 8.

The student bought shirts, ties, and pairs of pants.

21

3

1

2

3

4p

70

8

560

1 _ 4 s

8 8 2





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.



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Your Turn

7. 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?

Elaborate

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?

9. Discussion Why does a system need to have at least as many equations as unknowns to have a unique solution?

10. Essential Question Check-In How can you find the solution to a system of three linear equations in three variables?

r = f + 2144 ⇒ r = (t + 1558) + 2144 ⇒ r = t + 3702

f = t + 1558

2r + 3t + f = 13726

2 (t + 3702) + 3t + (t + 1558) = 13726

t = 794

f = 794 + 1558 = 2352

↓r = 2352 + 2144 = 4496

So, t = 794, f = 2352 and r = 4496.

If there is not an equation for each variable, the solution processes outlined above

cannot progress to the end, and you will be left with one variable defined only in

terms of another.

The solution will be a line because it will be in terms of two variables.

The system can be solved by substitution, elimination, or by using matrices.



A2_MTXESE353930_U2M05L3.indd 281 1/24/15 9:12 PM

ELABORATE AVOID COMMON ERRORSStudents 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.

SUMMARIZE THE LESSONWhat 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.

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• Online Homework• Hints and Help• Extra Practice

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 + 5y + 3z = 4 1

4y - z = 3 2 6x - 2y + 4z = 0 3

Solve the system using elimination.

3.

⎧ ⎪ ⎪

⎨ ⎪ ⎪

⎩

4x + y - 2z = -6 1

2x - 3y + 3z = 9 2 x - 2y = 0 3

x - 2y = 0 → x = 2y

Now substitute for x in the first and second equations and simplify.

4 (2y) + y - 2z = -6

↓ 9y - 2z = -6

2 (2y) - 3y + 3z = 9

↓ y + 3z = 9

Solving this system yields y = 0 and z = 3

x = 2 (0) = 0


4y - z = 3 → 4y - 3 = z

x + 5y + 3 (4y - 3) = 4

↓ x + 17y = 13

6x - 2y + 4 (4y - 3) = 0

↓3x + 7y = 6

Solving this system yields x = 0.25 and y = 0.75.

z = 4y - 3 = 4 (0.75) - 3 = 0

So the ordered triple is (0.25, 0.75, 0) .

3 (4x + y - 2z = -6) ⇒ 12x + 3y - 6z = -18

―――――――― 2x - 3y + 3z = 9 ―――――――― 2x - 3y + 3z = 9

14x - 3z = -9

2 (2x - 3y + 3z = 9) ⇒ 4x - 6y + 6z = 18

―――――――― -3 (x - 2y = 0) ――――――― - 3x + 6y = 0

x + 6z = 18

2 (14x - 3z = -9) ⇒ 28x - 6z = -18

――――――― x + 6z = 18 ――――――― x + 6z = 18

29x = 0

x = 0

14x - 3z = -9 → 14 (0) - 3z = -9 → z = 3

4x + y - 2z = -6 → 4 (0) + y - 2 (3) = -6 → y = 0





A2_MTXESE353930_U2M05L3 282 1/12/15 9:40 PMExercise Depth of Knowledge (D.O.K.) 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

EVALUATE

ASSIGNMENT GUIDE

Concepts and Skills Practice

ExploreRecognizing Ways that Planes Can Intersect

Exercise 14

Example 1Solving a System of Three Linear Equations Using Substitution

Exercises 1–2

Example 2Solving a System of Three Linear Equations Using Elimination

Exercises 3–5

Example 3Solving a System of Three Linear Equations Using Matrices

Exercises 6–7

Example 4Solving a Real-World Problem

Exercises 11–13

INTEGRATE MATHEMATICAL PROCESSESFocus on ModelingHave 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.



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4.

⎧ ⎪ ⎪

⎨

⎪

⎪

⎩

x + 5y + 3z = 4 1

4y - z = 3 2 6x - 2y + 4z = 0 3

5.

⎧ ⎪ ⎪

⎨

⎪ ⎪

⎩

2x - y + 3z = -12 1

-x + 2y - 3z = 15 2 y + 5z = -6 3

3 (4y - z = 3) ⇒ 12y - 3z = 9

――――――― x + 5y + 3z = 4 ――――――― x + 5y + 3z = 4

x + 17y = 13

4 (4y - z = 3) ⇒ 16y - 4z = 12

――――――― 6x - 2y + 4z = 0 ――――――― 6x - 2y + 4z = 0

6x + 14y = 12

↓ 3x + 7y = 6

-3 (x + 17y = 13) ⇒ -3x - 51y = -39

――――――― 3x + 7y = 6 ――――――― 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) .

2x - y + 3z = -12

――――――――― -x + 2y - 3z = 15

x + y = 3

5 (-x + 2y - 3z = 15) ⇒ -5x + 10y - 15z = 75

――――――――― 3 (y + 5z = -6) ――――――――― 3y + 15z = -18

-5x + 13y = 57

5 (x + y = 3) ⇒ 5x + 5y = 15

―――――――― -5x + 13y = 57 ――――――― ―――――― -5x + 13y = 57

18y = 72

y = 4

x + y = 3 → x + (4) = 3 → x = -1

2x - y + 3z = -12 → 2 (-1) - 4 + 3z = -12

3z = -6 → z = -2

So the ordered triple is (-1, 4, -2) .



A2_MTXESE353930_U2M05L3 283 1/12/15 9:44 PMExercise 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

AVOID COMMON ERRORSRemind 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).

283 Lesson 5 . 3


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Solve the system of three linear equations using matrices.

6.

⎧ ⎪ ⎪

⎨

⎪ ⎪

⎩

4x + y - 2z = -6 1

2x - 3y + 3z = 9 2 x - 2y = 0 3

7. ⎧

⎪

⎪

⎨

⎪

⎪

⎩

x + 5y + 3z = 4 1

4y - z = 3 2 6x - 2y + 4z = 0 3


[A]

⎡

⎢

⎣ 4

2 1

1

-3

-2

-2

3

0

-6

9

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 3 by - 1 _ 29

.


Multiply row 2 by 2 and add it to row 1.

⎡

⎢

⎣ 1

0 0

0

1 0

0

0 1

0

0 3

⎤

⎥

⎦

So, the solution is (0, 0, 3).


[A]

⎡

⎢

⎣ 1

0 6

5

4

-2

3

-1

4

4

3

0 ⎤

⎥

⎦


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.



⎡

⎢

⎣ 1

0 0

0

1 0

0

0 1

.25

.75 0

⎤

⎥

⎦

So, the solution is (0.25, 0.75, 0).




A2_MTXESE353930_U2M05L3 284 1/12/15 9:48 PM

QUESTIONING STRATEGIESCan 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.



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Solve the system of linear equations using your method of choice.

8. ⎧

⎪

⎪

⎨

⎪

⎪

⎩

2x - y + 3z = 5 1

-6x + 3y - 9z = -15 2 4x - 2y + 6z = 10 3

9. ⎧

⎪

⎪

⎨

⎪

⎪

⎩

3x + 4y - z = -7 1

x - 5y + 2z = 19 2 5x + y - 2z = 5 3

10. Find the equation of the parabola passing through the points (3, 7) , (30, -11) , and (0, -1) .

2x - y + 3z = 5 → 2x + 3z - 5 = y

-6x + 3 (2x + 3z - 5) - 9z = -15 4x - 2 (2x + 3z - 5) + 6z = 10

-6x + 6x + 9z - 15 - 9z = -15 4x - 4x - 6z + 10 + 6z = 10

-15 = -15 10 = 10

Both equations are true, so the system has infinitely many solutions.

3x + 4y - z = -7 → 3x + 4y + 7 = z

x - 5y + 2 (3x + 4y + 7) = 19 ⇒ 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

3x + 4y + 7 = z → 3 (2) + 4 (-3) + 7 = z → 1 = z

So the ordered triple is (2, -3, 1)

Identify the values of a, b, and c for the general form of a parabola

(7) = a (3) 2 + b (3) + c 7 = 9a + 3b + c 7 = 9a + 3b − 1

(−11) = a (30) 2 + b (30) + c ⇒ −11 = 900a + 30b + c ⇒ −11 = 900a + 30b − 1

(−1) = a (0) 2 + b (0) + c −1 = c

⇒ ⎧

⎨ ⎩ 8 = 9a + 3b

-1 = 90a + 3b

Solving this systems yields b = 3 and a = - 1 _ 9

So, the equation of the parabola connecting (3, 7) , (30, -11) , and (0, -1) is y = - 1 _ 9 x 2 + 3x - 1



A2_MTXESE353930_U2M05L3 285 1/12/15 9:47 PM

MULTIPLE REPRESENTATIONSShow 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.

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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?

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?

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?

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°.

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

So, Dan is 10 years old, James is 30 years old, and Paul is 80 years old.

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.




A2_MTXESE353930_U2M05L3 286 15-01-11 3:50 AM



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H.O.T. Focus on Higher Order Thinking

14. Communicate Mathematical Ideas Explain how you know when a system has infinitely many solutions or when it has no solutions.

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 = -38x + y = 12

⎡

⎢

⎣ 5

1 8

7

-1 1

9

1 0

⎤

⎥

⎦

16. Critical Thinking Explain why the following system of equations cannot be solved.

⎧

⎨

⎩

7x + y + 6z = 1

-x - 4y + 8z = 9

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.

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

1 8

7

-1 1

9

1 0

0

-3 12

⎤

⎥

⎦

When solving a system of equations, there must be at least as many equations as there are variables.



A2_MTXESE353930_U2M05L3 287 15-01-11 3:49 AM

JOURNALHave 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.

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Lesson Performance Task

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?

⎧

⎨

⎩

A + B + C = 6000

A = 1 __ 5 (B + C)

B = C

⇒

⎧

⎨

⎩

A + (4C) + C = 6000

A = 1 _

5 ( (4C) + C)

B = 4C

⇒

⎧

⎨

⎩

A + 5C = 6000

A = C

B = 4C

⇒

⎧

⎨

⎩

(C) + 5C = 6000

A = C B = 4C

⇒ ⎧

⎨

⎩ C = 1000

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.




EXTENSION ACTIVITY

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.

INTEGRATE MATHEMATICAL PROCESSESFocus on ModelingBecause 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 STRATEGIESWhat 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.

Scoring Rubric2 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.



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