Solving
Linear
Systems
Solving
Linear
Systems
3-6
3-6 in Three Variables
in Three Variables
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
22
3-6
Solving Linear Systems
in Three Variables
Warm Up
Solve each system of equations algebraically.
1.
x = 4y + 10
4x + 2y = 4
(2, –2) 2.
6x – 5y = 9
2x – y =1
(–1,–3)
Classify each system and determine the
number of solutions.
3x – y = 8
x = 3y – 1
4.
3.
6x – 2y = 2
6x – 12y = –4
inconsistent; none
Holt Algebra 2
consistent, independent; one
3-6
Solving Linear Systems
in Three Variables
Objectives
Represent solutions to systems of
equations in three dimensions
graphically.
Solve systems of equations in three
dimensions algebraically.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Systems of three equations with three
variables are often called 3-by-3 systems.
In general, to find a single solution to any
system of equations, you need as many
equations as you have variables.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Recall from Lesson 3-5 that the graph of a
linear equation in three variables is a plane.
When you graph a system of three linear
equations in three dimensions, the result is
three planes that may or may not intersect.
The solution to the system is the set of points
where all three planes intersect. These
systems may have one, infinitely many, or no
solution.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Identifying the exact solution from a
graph of a 3-by-3 system can be very
difficult. However, you can use the
methods of elimination and substitution to
reduce a 3-by-3 system to a 2-by-2
system and then use the methods that
you learned in Lesson 3-2.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 1: Solving a Linear System in Three
Variables
Use elimination to solve the system of equations.
5x – 2y – 3z = –7
1
2x – 3y + z = –16
2
3x + 4y – 2z = 7
3
Step 1 Eliminate one variable.
In this system, z is a reasonable choice to eliminate
first because the coefficient of z in the second
equation is 1 and z is easy to eliminate from the
other equations.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 1 Continued
1
2
5x – 2y – 3z = –7
5x – 2y – 3z = –7
3(2x –3y + z = –16)
Multiply equation
-2 by 3, and add
6x – 9y + 3z = –48 to equation 1 .
11x – 11y
Use equations
in x and y.
3
and
2
= –55
4
to create a second equation
1
3
2
3x + 4y – 2z = 7
3x + 4y – 2z = 7
Multiply equation
2(2x –3y + z = –16) 4x – 6y + 2z = –32 -2 by 2, and add
7x – 2y
Holt Algebra 2
= –25
to equation
5
3
.
3-6
Solving Linear Systems
in Three Variables
Example 1 Continued
You now have a 2-by-2 system.
Holt Algebra 2
11x – 11y = –55
4
7x – 2y = –25
5
3-6
Solving Linear Systems
in Three Variables
Example 1 Continued
Step 2 Eliminate another variable. Then solve for
the remaining variable.
You can eliminate y by using methods from
Lesson 3-2.
4
5
equation
–2(11x – 11y = –55) –22x + 22y = 110 1 Multiply
-4 by –2, and
11(7x – 2y = –25) 77x – 22y = –275 equation -5 by 11
55x
1
= –165
x = –3
Holt Algebra 2
and add.
Solve for x.
3-6
Solving Linear Systems
in Three Variables
Example 1 Continued
Step 3 Use one of the equations in your 2-by-2
system to solve for y.
4
11x – 11y = –55
1
11(–3) – 11y = –55
Substitute –3 for x.
1
y=2
Holt Algebra 2
Solve for y.
3-6
Solving Linear Systems
in Three Variables
Example 1 Continued
Step 4 Substitute for x and y in one of the original
equations to solve for z.
2
2x – 3y + z = –16
2(–3) – 3(2) + z = –16
z = –4
The solution is (–3, 2, –4).
Holt Algebra 2
Substitute
1 –3 for x and
2 for y.
1Solve for y.
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 1
Use elimination to solve the system of equations.
–x + y + 2z = 7
1
2x + 3y + z = 1
2
–3x – 4y + z = 4
3
Step 1 Eliminate one variable.
In this system, z is a reasonable choice to eliminate
first because the coefficient of z in the second
equation is 1.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 1 Continued
1
2
–x + y + 2z = 7 Multiply equation
–x + y + 2z = 7
–2(2x + 3y + z = 1) –4x – 6y – 2z = –2 -2 by –2, and add
1
–5x – 5y
Use equations
in x and y.
1
and
3
=5
to equation
.
4
to create a second equation
1
1
3
–x + y + 2z = 7
–2(–3x – 4y + z = 4)
–x + y + 2z = 7 Multiply equation
6x + 8y – 2z = –8 -3 by –2, and add
to equation
5x + 9y
Holt Algebra 2
= –1
5
1
.
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 1 Continued
You now have a 2-by-2 system.
–5x – 5y = 5
5x + 9y = –1
Holt Algebra 2
4
5
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 1 Continued
Step 2 Eliminate another variable. Then solve for
the remaining variable.
You can eliminate x by using methods from
Lesson 3-2.
4
5
–5x – 5y = 5
5x + 9y = –1
4y = 4
y=1
Holt Algebra 2
Add equation
Solve for y.
5
1
to equation
4
.
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 1
Step 3 Use one of the equations in your 2-by-2
system to solve for x.
4
–5x – 5y = 5
Substitute 1 for y.
–5x – 5(1) = 5
1
–5x – 5 = 5
–5x = 10
x = –2
Holt Algebra 2
Solve for x.
1
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 1
Step 4 Substitute for x and y in one of the original
equations to solve for z.
2
2x +3y + z = 1
2(–2) +3(1) + z = 1
–4 + 3 + z = 1
z=2
Substitute –2 for x and
1 for y.
Solve for 1z.
1
The solution is (–2, 1, 2).
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
You can also use substitution to solve a
3-by-3 system. Again, the first step is to
reduce the 3-by-3 system to a 2-by-2
system.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 2: Business Application
The table shows the number of each type of
ticket sold and the total sales amount for each
night of the school play. Find the price of each
type of ticket.
Orchestra
Mezzanine Balcony
Total Sales
Fri
200
30
40
$1470
Sat
250
60
50
$1950
Sun
150
30
0
$1050
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 2 Continued
Step 1 Let x represent the price of an orchestra seat,
y represent the price of a mezzanine seat, and z
represent the present of a balcony seat.
Write a system of equations to represent the data in
the table.
200x + 30y + 40z = 1470
1
250x + 60y + 50z = 1950
2
Saturday’s sales.
150x + 30y = 1050
3
Sunday’s sales.
Friday’s sales.
A variable is “missing” in the last equation; however,
the same solution methods apply. Elimination is a good
choice because eliminating z is straightforward.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 2 Continued
Step 2 Eliminate z.
Multiply equation
1
2
1
by 5 and equation
5(200x + 30y + 40z = 1470)
–4(250x + 60y + 50z = 1950)
2
by –4 and add.
1000x + 150y + 200z = 7350
–1000x – 240y – 200z = –7800
y
=5
By eliminating z, due to the coefficients of x, you also
eliminated x providing a solution for y.
Holt Algebra 2
Solving Linear Systems
in Three Variables
3-6
Example 2 Continued
Step 3 Use equation
150x + 30y = 1050
150x + 30(5) = 1050
3
x=6
Holt Algebra 2
3
to solve for x.
Substitute 5 for y.
Solve for x.
3-6
Solving Linear Systems
in Three Variables
Example 2 Continued
Step 4 Use equations
1
or
2
to solve for z.
1
1
200x + 30y + 40z = 1470
200(6) + 30(5) + 40z = 1470
Substitute 6 for x and 5 for y.
Solve for x.
z=3
The solution to the system is (6, 5, 3). So, the
cost of an orchestra seat is $6, the cost of a
mezzanine seat is $5, and the cost of a balcony
seat is $3.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 2
Jada’s chili won first place at the winter fair.
The table shows the results of the voting.
How many points are first-, second-, and
third-place votes worth?
Winter Fair Chili Cook-off
Name
1st
Place
2nd
Place
3rd
Place
Total
Points
Jada
3
1
4
15
Maria
2
4
0
14
Al
2
2
3
13
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 2 Continued
Step 1 Let x represent first-place points, y represent
second-place points, and z represent thirdplace points.
Write a system of equations to represent the data in
the table.
3x + y + 4z = 15
1
Jada’s points.
2x + 4y = 14
2
Maria’s points.
2x + 2y + 3z = 13
3
Al’s points.
A variable is “missing” in one equation; however, the
same solution methods apply. Elimination is a good
choice because eliminating z is straightforward.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 2 Continued
Step 2 Eliminate z.
Multiply equation
1
by 3 and equation
3
by –4 and add.
1
3(3x + y + 4z = 15)
9x + 3y + 12z = 45
3
–4(2x + 2y + 3z = 13)
–8x – 8y – 12z = –52
x – 5y
Multiply equation
4
2
4
–2(x – 5y = –7)
2x + 4y = 14
Holt Algebra 2
= –7
by –2 and add to equation
2
4
.
–2x + 10y = 14
2x + 4y = 14
y = 2 Solve for y.
Solving Linear Systems
in Three Variables
3-6
Check It Out! Example 2 Continued
Step 3 Use equation
2
2x + 4y = 14
2x + 4(2) = 14
x=3
Holt Algebra 2
2
to solve for x.
Substitute 2 for y.
Solve for x.
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 2 Continued
Step 4 Substitute for x and y in one of the original
equations to solve for z.
3
2x + 2y + 3z = 13
2(3) + 2(2) + 3z = 13
6 + 4 + 3z = 13
z=1
Solve for z.
The solution to the system is (3, 2, 1). The points for
first-place is 3, the points for second-place is 2, and 1
point for third-place.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
The systems in Examples 1 and 2 have unique
solutions. However, 3-by-3 systems may have
no solution or an infinite number of solutions.
Remember!
Consistent means that the system of equations
has at least one solution.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 3: Classifying Systems with Infinite Many
Solutions or No Solutions
Classify the system as consistent or inconsistent,
and determine the number of solutions.
2x – 6y + 4z = 2
–3x + 9y – 6z = –3
5x – 15y + 10z = 5
Holt Algebra 2
1
2
3
3-6
Solving Linear Systems
in Three Variables
Example 3 Continued
The elimination method is convenient because the
numbers you need to multiply the equations are small.
First, eliminate x.
Multiply equation
1
2
1
by 3 and equation
3(2x – 6y + 4z = 2)
2(–3x + 9y – 6z = –3)
2
by 2 and add.
6x – 18y + 12z = 6
–6x + 18y – 12z = –6
0=0 
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Example 3 Continued
1
3
Multiply equation 1 by 5 and equation 3 by –2
and add.
10x – 30y + 20z = 10
5(2x – 6y + 4z = 2)
–10x + 30y – 20z = –10
–2(5x – 15y + 10z = 5)
0 = 0 
Because 0 is always equal to 0, the equation is
an identity. Therefore, the system is consistent,
dependent and has an infinite number of solutions.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3a
Classify the system, and determine the number
of solutions.
3x – y + 2z = 4
1
2x – y + 3z = 7
2
–9x + 3y – 6z = –12
Holt Algebra 2
3
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3a Continued
The elimination method is convenient because the
numbers you need to multiply the equations by are
small.
First, eliminate y.
Multiply equation
1
3
2
by –1 and add to equation
3x – y + 2z = 4
–1(2x – y + 3z = 7)
.
3x – y + 2z = 4
–2x + y – 3z = –7
x
Holt Algebra 2
1
– z = –3
4
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3a Continued
Multiply equation
2
3
2
by 3 and add to equation
3(2x – y + 3z = 7)
–9x + 3y – 6z = –12
–9x + 3y – 6z = –12
Now you have a 2-by-2 system.
–3x + 3z = 9
Holt Algebra 2
4
5
.
6x – 3y + 9z = 21
–3x
x – z = –3
3
+ 3z = 9
5
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3a Continued
Eliminate x.
4
5
3(x – z = –3)
–3x + 3z = 9
3x – 3z = –9
–3x + 3z = 9
0=0

Because 0 is always equal to 0, the equation is
an identity. Therefore, the system is consistent,
dependent, and has an infinite number of solutions.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3b
Classify the system, and determine the number
of solutions.
Holt Algebra 2
2x – y + 3z = 6
1
2x – 4y + 6z = 10
2
y – z = –2
3
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3b Continued
Use the substitution method. Solve for y in equation 3.
3
y – z = –2
y=z–2
Solve for y.
4
Substitute equation
4
in for y in equation
2x – y + 3z = 6
2x – (z – 2) + 3z = 6
2x – z + 2 + 3z = 6
2x + 2z = 4
Holt Algebra 2
5
1
.
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3b Continued
Substitute equation
4
in for y in equation
2x – 4y + 6z = 10
2x – 4(z – 2) + 6z = 10
2x – 4z + 8 + 6z = 10
2x + 2z = 2
6
Now you have a 2-by-2 system.
2x + 2z = 4
2x + 2z = 2
Holt Algebra 2
5
6
2
.
3-6
Solving Linear Systems
in Three Variables
Check It Out! Example 3b Continued
Eliminate z.
5
6
2x + 2z = 4
–1(2x + 2z = 2)
02

Because 0 is never equal to 2, the equation is a
contradiction. Therefore, the system is
inconsistent and has no solutions.
Holt Algebra 2
3-6
Solving Linear Systems
in Three Variables
Lesson Quiz: Part I
1. At the library book sale, each type of book is
priced differently. The table shows the number of
books Joy and her friends each bought, and the
amount each person spent. Find the price of each
type of book.
Hardcover
Paper- Audio Total
back Books Spent
Hal
3
4
1
$17
Ina
2
5
1
$15
Joy
3
3
2
$20
Holt Algebra 2
hardcover: $3;
paperback: $1;
audio books: $4
3-6
Solving Linear Systems
in Three Variables
Lesson Quiz: Part II
Classify each system and determine the number
of solutions.
2x – y + 2z = 5
2.
–3x +y – z = –1
inconsistent; none
x – y + 3z = 2
9x – 3y + 6z = 3
3.
12x – 4y + 8z = 4
–6x + 2y – 4z = 5
Holt Algebra 2
consistent; dependent;
infinite
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