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Holt Algebra 2
3-6
Solving Linear Systems in Three Variables 3-6
Solving Linear Systems in Three Variables
Holt Algebra 2
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
Warm Up Solve 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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
Represent solutions to systems of equations in three dimensions graphically.
Solve systems of equations in three dimensions algebraically.
Objectives
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
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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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 = 7 4x – 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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
11x – 11y = –55
7x – 2y = –25 You now have a 2-by-2 system.
4
5
Example 1 Continued
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
–2(11x – 11y = –55)
55x = –165
11(7x – 2y = –25) –22x + 22y = 110 77x – 22y = –275
4
5
1
1 Multiply equation - by –2, and equation - by 11 and add.
4
5
Step 2 Eliminate another variable. Then solve for the remaining variable.
You can eliminate y by using methods from Lesson 3-2.
x = –3 Solve for x.
Example 1 Continued
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
11x – 11y = –55
11(–3) – 11y = –55
4
1
1
Step 3 Use one of the equations in your 2-by-2 system to solve for y.
y = 2
Substitute –3 for x.
Solve for y.
Example 1 Continued
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
2x – 3y + z = –16
2(–3) – 3(2) + z = –16
2
1
1
Step 4 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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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.
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
–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 = –8
1
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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
You now have a 2-by-2 system.
Check It Out! Example 1 Continued
4
5
–5x – 5y = 5
5x + 9y = –1
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
4y = 4
4
5
1
Add equation to equation . 4 5
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 = 5 5x + 9y = –1
y = 1
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
–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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
2(–2) +3(1) + z = 1
2x +3y + z = 1 2
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
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
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
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 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.
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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.
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
200x + 30y + 40z = 1470 1 Substitute 6 for x and 5 for y.
1
z = 3 Solve for x.
Step 4 Use equations or to solve for z. 2 1
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
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?
Name
1st Place
2nd Place
3rd Place
Total Points
Jada 3 1 4 15
Maria 2 4 0 14
Al 2 2 3 13
Winter Fair Chili Cook-off
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 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.
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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.
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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 = 13 3
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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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.
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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.
ü
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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.
ü
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 2x – y + 3z = 7
–9x + 3y – 6z = –12
1
2
3
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
3(x – z = –3) –3x + 3z = 9 5
4 3x – 3z = –9 –3x + 3z = 9
0 = 0 ü
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
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.
2x – y + 3z = 6 2x – 4y + 6z = 10 y – z = –2
1
2
3
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
y – z = –2 y = z – 2
3 Solve 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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
Substitute equation in for y in equation . 4 2
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 6
5
Check It Out! Example 3b Continued
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
2x + 2z = 4 –1(2x + 2z = 2) 6
5
Eliminate z.
0 ≠ 2 û
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.
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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
Holt Algebra 2
3-6
Solving Linear Systems in Three Variables
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.
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