correctionkey=nl-d;ca-d name class date 4.4 solving linear systems … · explain 3 solving 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.

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

planes?

B The diagram shows three intersecting planes.

How many points lie on all 3 planes?

Resource Locker

Module 4 203 Lesson 4

4.4 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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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.

A

⎧ ⎪

⎪

⎨

⎪ ⎪ ⎩

-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

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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 vertical parabola passing through any three noncollinear points in the coordinate plane provided that no two of the points have the same x-coordinate. Find the vertical parabola that passes through the points (2, 1) , (-1, 4) , and (-2, 3) .

The general form of a vertical 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 relatively easy 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]

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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



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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

A

⎧ ⎪ ⎨ ⎪

⎩

-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) -25x + 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) .

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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 .

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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

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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.

A

⎧ ⎪ ⎨ ⎪

⎩

-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.



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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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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

⎤

⎥

⎥

⎥

⎥ ⎦



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Explain 4 Solving a Real-World Problem

Example 4

A 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 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.

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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.

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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?



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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

2x - 3y + 3z = 9

x - 2y = 0

2.

⎧

⎪ ⎪

⎨

⎪

⎪ ⎩

x + 5y + 3z = 4

4y - z = 3

6x - 2y + 4z = 0

Solve the system using elimination.

3.

⎧ ⎪ ⎪

⎨ ⎪ ⎪

⎩

4x + y - 2z = -6

2x - 3y + 3z = 9

x - 2y = 0

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3 3

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y4.

⎧ ⎪ ⎪

⎨

⎪

⎪

⎩

x + 5y + 3z = 4

4y - z = 3

6x - 2y + 4z = 0

5.

⎧ ⎪ ⎪

⎨

⎪ ⎪

⎩

2x - y + 3z = -12

-x + 2y - 3z = 15

y + 5z = -6

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Solve the system of three linear equations using matrices.

6.

⎧ ⎪ ⎪

⎨

⎪ ⎪

⎩

4x + y - 2z = -6

2x - 3y + 3z = 9

x - 2y = 0

7. ⎧

⎪

⎪

⎨

⎪

⎪

⎩

x + 5y + 3z = 4

4y - z = 3

6x - 2y + 4z = 0

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

8. ⎧

⎪

⎪

⎨

⎪

⎪

⎩

2x - y + 3z = 5

-6x + 3y - 9z = -15

4x - 2y + 6z = 10

9. ⎧

⎪

⎪

⎨

⎪

⎪

⎩

3x + 4y - z = -7

x - 5y + 2z = 19

5x + y - 2z = 5

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

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11. Geometry In triangle XYZ, 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?



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yH.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 When given this system of equations, a student was asked to solve using matrices. Find and correct the student’s error.

⎧

⎨

⎩

5x +7y + 9 = 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



z

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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?



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