Download - Algebra 2 unit 3.5
![Page 1: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/1.jpg)
UNIT 3.5 SYSTEMS OFUNIT 3.5 SYSTEMS OFTHREE VARIABLESTHREE VARIABLES
![Page 2: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/2.jpg)
Warm UpSolve each system of equations algebraically.
Classify each system and determine the number of solutions.
1. 2.x = 4y + 10
4x + 2y = 4
6x – 5y = 9
2x – y =1(2, –2) (–1,–3)
3. 4.3x – y = 8
6x – 2y = 2
x = 3y – 1
6x – 12y = –4
inconsistent; none consistent, independent; one
![Page 3: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/3.jpg)
![Page 4: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/4.jpg)
![Page 5: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/5.jpg)
Use elimination to solve the system of equations.
Example 1: Solving a Linear System in Three Variables
Step 1 Eliminate one variable.
5x – 2y – 3z = –7
2x – 3y + z = –16
3x + 4y – 2z = 7
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.
1
2
3
![Page 6: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/6.jpg)
Example 1 Continued
5x – 2y – 3z = –7
11x – 11y = –55
3(2x –3y + z = –16)
5x – 2y – 3z = –7
6x – 9y + 3z = –48
1
2
1
4
3x + 4y – 2z = 7
7x – 2y = –25
2(2x –3y + z = –16)3x + 4y – 2z = 74x – 6y + 2z = –32
3
2
Multiply equation - by 3, and add to equation .1
2
Multiply equation - by 2, and add to equation .3
2
5
Use equations and to create a second equation in x and y.
3 2
![Page 7: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/7.jpg)
11x – 11y = –55
7x – 2y = –25Step 2: You now have a 2-by-2 system. Solve for x and y.
4
5
Example 1 Continued
![Page 8: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/8.jpg)
2x – 3y + z = –16
2(–3) – 3(2) + z = –16
2
1
1
Step 3 Substitute for x and y in one of the original equations to solve for z.
z = –4
Substitute –3 for x and 2 for y.
Solve for y.
The solution is (–3, 2, –4).
Example 1 Continued
![Page 9: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/9.jpg)
Use elimination to solve the system of equations.
Step 1 Eliminate one variable.
–x + y + 2z = 7
2x + 3y + z = 1
–3x – 4y + z = 4
1
2
3
Check It Out! Example 1
In this system, z is a reasonable choice to eliminate first because the coefficient of z in the second equation is 1.
![Page 10: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/10.jpg)
–x + y + 2z = 7
–5x – 5y = 5
–2(2x + 3y + z = 1) –4x – 6y – 2z = –2
1
2
1
4
5x + 9y = –1
–2(–3x – 4y + z = 4)–x + y + 2z = 7
6x + 8y – 2z = –81
3
Multiply equation - by –2, and add to equation .1
2
Multiply equation - by –2, and add to equation .1
3
5
Check It Out! Example 1 Continued
–x + y + 2z = 7
–x + y + 2z = 7
Use equations and to create a second equation in x and y.
1 3
![Page 11: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/11.jpg)
You now have a 2-by-2 system.
Check It Out! Example 1 Continued
4
5
–5x – 5y = 5
5x + 9y = –1
![Page 12: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/12.jpg)
4y = 4
4
5
1
Add equation to equation .45
Step 2 Eliminate another variable. Then solve for the remaining variable.
You can eliminate x by using methods from Lesson 3-2.
Solve for y.
Check It Out! Example 1 Continued
–5x – 5y = 55x + 9y = –1
y = 1
![Page 13: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/13.jpg)
–5x – 5(1) = 5
4
1
1
Step 3 Use one of the equations in your 2-by-2 system to solve for x.
x = –2
Substitute 1 for y.
Solve for x.
Check It Out! Example 1
–5x – 5y = 5
–5x – 5 = 5
–5x = 10
![Page 14: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/14.jpg)
2(–2) +3(1) + z = 1
2x +3y + z = 12
1
1
Step 4 Substitute for x and y in one of the original equations to solve for z.
z = 2
Substitute –2 for x and 1 for y.
Solve for z.
The solution is (–2, 1, 2).
Check It Out! Example 1
–4 + 3 + z = 1
![Page 15: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/15.jpg)
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.
Example 2: Business Application
Orchestra Mezzanine Balcony Total Sales
Fri 200 30 40 $1470
Sat 250 60 50 $1950
Sun 150 30 0 $1050
![Page 16: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/16.jpg)
Example 2 ContinuedStep 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
250x + 60y + 50z = 1950
150x + 30y = 1050
1
2
3
Friday’s sales.
Saturday’s sales.
Sunday’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.
![Page 17: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/17.jpg)
5(200x + 30y + 40z = 1470)
–4(250x + 60y + 50z = 1950)
1
Step 2 Eliminate z.
Multiply equation by 5 and equation by –4 and add.1
2
2
1000x + 150y + 200z = 7350
–1000x – 240y – 200z = –7800
y = 5
Example 2 Continued
By eliminating z, due to the coefficients of x, you also eliminated x providing a solution for y.
![Page 18: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/18.jpg)
150x + 30y = 1050
150x + 30(5) = 1050
3 Substitute 5 for y.
x = 6
Solve for x.
Step 3 Use equation to solve for x.3
Example 2 Continued
![Page 19: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/19.jpg)
200x + 30y + 40z = 14701 Substitute 6 for x and 5 for y.
1
z = 3
Solve for x.
Step 4 Use equations or to solve for z.21
200(6) + 30(5) + 40z = 1470
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.
Example 2 Continued
![Page 20: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/20.jpg)
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?
Name
1st Place
2nd Place
3rd Place
TotalPoints
Jada 3 1 4 15
Maria 2 4 0 14
Al 2 2 3 13
Winter Fair Chili Cook-off
![Page 21: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/21.jpg)
Check It Out! Example 2 Continued
Step 1 Let x represent first-place points, y represent second-place points, and z represent third- place points.
Write a system of equations to represent the data in the table.
3x + y + 4z = 15
2x + 4y = 14
2x + 2y + 3z = 13
1
2
3
Jada’s points.
Maria’s points.
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.
![Page 22: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/22.jpg)
3(3x + y + 4z = 15)
–4(2x + 2y + 3z = 13)
1
Step 2 Eliminate z.
Multiply equation by 3 and equation by –4 and add.3
1
3
9x + 3y + 12z = 45
–8x – 8y – 12z = –52
x – 5y = –7 4
Check It Out! Example 2 Continued
2
–2(x – 5y = –7)4
2x + 4y = 14–2x + 10y = 14
2x + 4y = 14
y = 2
Multiply equation by –2 and add to equation .2
4
Solve for y.
![Page 23: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/23.jpg)
2x + 4y = 14
Step 3 Use equation to solve for x.2
2
2x + 4(2) = 14
x = 3
Solve for x.
Substitute 2 for y.
Check It Out! Example 2 Continued
![Page 24: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/24.jpg)
Step 4 Substitute for x and y in one of the original equations to solve for z.
z = 1 Solve for z.
2x + 2y + 3z = 133
2(3) + 2(2) + 3z = 13
6 + 4 + 3z = 13
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.
Check It Out! Example 2 Continued
![Page 25: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/25.jpg)
Consistent means that the system of equations has at least one solution.
Remember!
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.
![Page 26: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/26.jpg)
Classify the system as consistent or inconsistent, and determine the number of solutions.
Example 3: Classifying Systems with Infinite Many Solutions or No Solutions
2x – 6y + 4z = 2
–3x + 9y – 6z = –3
5x – 15y + 10z = 5
1
2
3
![Page 27: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/27.jpg)
Example 3 Continued
3(2x – 6y + 4z = 2)
2(–3x + 9y – 6z = –3)
First, eliminate x.
1
2
6x – 18y + 12z = 6
–6x + 18y – 12z = –6
0 = 0
Multiply equation by 3 and equation by 2 and add.2
1
The elimination method is convenient because the numbers you need to multiply the equations are small.
![Page 28: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/28.jpg)
Example 3 Continued
5(2x – 6y + 4z = 2)
–2(5x – 15y + 10z = 5)
1
3
10x – 30y + 20z = 10
–10x + 30y – 20z = –10
0 = 0
Multiply equation by 5 and equation by –2 and add.
3
1
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.
![Page 29: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/29.jpg)
Check It Out! Example 3a
Classify the system, and determine the number of solutions.
3x – y + 2z = 4
2x – y + 3z = 7
–9x + 3y – 6z = –12
1
2
3
![Page 30: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/30.jpg)
3x – y + 2z = 4–1(2x – y + 3z = 7)
First, eliminate y.
1
3
3x – y + 2z = 4
–2x + y – 3z = –7
x – z = –3
Multiply equation by –1 and add to equation . 1
2
The elimination method is convenient because the numbers you need to multiply the equations by are small.
Check It Out! Example 3a Continued
4
![Page 31: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/31.jpg)
3(2x – y + 3z = 7)–9x + 3y – 6z = –12
2
3
6x – 3y + 9z = 21
–9x + 3y – 6z = –12
–3x + 3z = 9
Multiply equation by 3 and add to equation . 3
2
Now you have a 2-by-2 system.
x – z = –3
–3x + 3z = 9 5
4
5
Check It Out! Example 3a Continued
![Page 32: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/32.jpg)
3(x – z = –3)
–3x + 3z = 9 5
4 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.
Eliminate x.
Check It Out! Example 3a Continued
![Page 33: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/33.jpg)
Check It Out! Example 3b
Classify the system, and determine the number of solutions.
2x – y + 3z = 6
2x – 4y + 6z = 10
y – z = –2
1
2
3
![Page 34: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/34.jpg)
y – z = –2y = z – 2
3Solve for y.
Use the substitution method. Solve for y in equation 3.
Check It Out! Example 3b Continued
Substitute equation in for y in equation .4 1
4
2x – y + 3z = 6
2x – (z – 2) + 3z = 6
2x – z + 2 + 3z = 6
2x + 2z = 4 5
![Page 35: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/35.jpg)
Substitute equation in for y in equation .4 2
2x – 4y + 6z = 10
2x – 4(z – 2) + 6z = 102x – 4z + 8 + 6z = 10
2x + 2z = 2 6
Now you have a 2-by-2 system.
2x + 2z = 4
2x + 2z = 2 6
5
Check It Out! Example 3b Continued
![Page 36: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/36.jpg)
2x + 2z = 4
–1(2x + 2z = 2)6
5
Eliminate z.
0 2
Check It Out! Example 3b Continued
Because 0 is never equal to 2, the equation is a contradiction. Therefore, the system is inconsistent and has no solutions.
![Page 37: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/37.jpg)
Lesson Quiz: Part I
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.
paperback: $1;
Hard-cover
Paper- back
Audio Books
Total Spent
Hal 3 4 1 $17
Ina 2 5 1 $15
Joy 3 3 2 $20
1.
hardcover: $3;
audio books: $4
![Page 38: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/38.jpg)
2.
3.
2x – y + 2z = 5
–3x +y – z = –1
x – y + 3z = 2
9x – 3y + 6z = 3
12x – 4y + 8z = 4
–6x + 2y – 4z = 5
inconsistent; none
consistent; dependent; infinite
Lesson Quiz: Part II
Classify each system and determine the number of solutions.
![Page 39: Algebra 2 unit 3.5](https://reader030.vdocuments.us/reader030/viewer/2022032618/55b7f624bb61ebcb1b8b4789/html5/thumbnails/39.jpg)
All rights belong to their respective owners.Copyright Disclaimer Under Section 107 of the Copyright Act 1976, allowance is made for "fair use" for purposes such as criticism, comment, news reporting, TEACHING, scholarship, and research. Fair use is a use permitted by copyright statute that might otherwise be infringing. Non-profit, EDUCATIONAL or personal use tips the balance in favor of fair use.