Transcript Algebra 2

Algebra 2
Systems With Three variables
Lesson 3-5
Goals
Goal
• To solve systems in three
variables using elimination.
Rubric
Level 1 – Know the goals.
Level 2 – Fully understand the
goals.
Level 3 – Use the goals to
solve simple problems.
Level 4 – Use the goals to
solve more advanced problems.
Level 5 – Adapts and applies
the goals to different and more
complex problems.
Vocabulary
• None
Essential Question
Big Idea: Solving Equations and
Inequalities
• How is solving a system of three equations in
three variables similar to solving a system of two
equations in two variables?
Three-Dimensional Space
• A Global Positioning System
(GPS) gives locations using the
three coordinates of latitude,
longitude, and elevation.
• You can represent any location in
three-dimensional space using
a three-dimensional coordinate
system, sometimes called
coordinate space.
Coordinate Space
• Each point in coordinate space
can be represented by an
ordered triple of the form (x,
y, z).
• The system is similar to the
coordinate plane but has an
additional coordinate based on
the z-axis.
• Notice that the axes form
three planes that intersect at
the origin.
Graphing Points in Three
Dimensions
Graph the point in three-dimensional space.
A(3, –2, 1)
From the origin,
move 3 units
forward along the
x-axis, 2 units left,
and 1 unit up.
z
y
A(3, –2, 1)

x
Graphing Points in Three
Dimensions
Graph the point in three-dimensional space.
C(–1, 0, 2)
z
From the origin,
move 1 unit back
along the x-axis, 2
units up. Notice that
this point lies in the
xz-plane because the
y-coordinate is 0.
C(–1,0, 2)

y
x
Linear Equation in Three
Dimensions
• Recall that the graph of a linear equation in
two dimensions is a straight line. In threedimensional space, the graph of a linear
equation is a plane.
• Because a plane is defined by three points,
you can graph linear equations in three
dimensions by graphing the three intercepts.
Linear Equation in Three
Dimensions
Graph the linear equation 2x – 3y + z = –6 in
three-dimensional space.
Step 1 Find the intercepts:
x-intercept: 2x – 3(0) + (0) = –6
x = –3
y-intercept: 2(0) – 3y + (0) = –6
y=2
z-intercept: 2(0) – 3(0) + z = –6
z = –6
Linear Equation in Three
Dimensions
z
Step 2 Plot the
points (–3, 0, 0),
(0, 2, 0), and
(0, 0, –6). Sketch a
plane through the
three points.
 (–3, 0, 0)

y
(0, 2, 0)
x
 (0, 0, –6)
System of Three Equations
• 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.
System of Three Equations
• 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.
System of Three Equations
• There are 4 ways in which 3 planes could
intersect.
1.
2.
3.
4.
No common intersection (no solution)
Intersection in a single point (one solution)
Intersection in a line (infinite solutions)
Intersection in a plane (all three are identical
planes – infinite solutions)
1. No Common Intersection
Each of these systems are inconsistent.
No solution.
2. Intersection in a Point
Consistent, Independent system.
One solution.
3. Intersection in a Line
Consistent, Dependent system.
Infinite solutions.
4. Intersection in a Plane
All three equations are the same plane.
Consistent, Dependent system.
Infinite solutions.
Solving a System of
Three Equations
• Identifying the exact solution from a graph
of a 3-by-3 system can be very difficult.
• However, you can use the method of
elimination to reduce a 3-by-3 system to a
2-by-2 system and then use the methods that
you learned earlier to complete the solution.
Solving a System of
Three Equations
• To solve a linear system with three unknowns, first
eliminate a variable from any two of the equations.
• Then eliminate the same variable from a different pair of
equations.
• Eliminate a second variable using the resulting two
equations in two variables to get an equation with just one
variable whose value you can now determine.
• Find the values of the remaining variables by substitution.
• The solution of the system is written as an ordered triple.
Example:
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.
Example: continued
1
2
5x – 2y – 3z = –7
5x – 2y – 3z = –7
3(2x –3y + z = –16)
6x – 9y + 3z = –48
11x – 11y
Use equations
Multiply equation
-2 by 3, and add
to equation 1 .
3
and
2
= –55
4
to create a second equation in x and y.
1
3
2
3x + 4y – 2z = 7
2(2x –3y + z = –16)
3x + 4y – 2z = 7
4x – 6y + 2z = –32
7x – 2y
= –25
Multiply equation
-2 by 2, and add
to equation 3 .
5
Example: continued
11x – 11y = –55
4
7x – 2y = –25
5
You now have a 2-by-2 system.
Example: 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
–2(11x – 11y = –55)
11(7x – 2y = –25)
–22x + 22y = 110
77x – 22y = –275
55x
= 1–165
x = –3
1
Multiply equation
-4 by –2, and
equation -5 by 11
and add.
Solve for x.
Example: 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
Solve for y.
Example: 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).
1
Substitute –3 for x and 2 for y.
1Solve
for y.
Example:
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.
Example: continued
1
2
–x + y + 2z = 7
–2(2x + 3y + z = 1)
–x + y + 2z = 7
–4x – 6y – 2z = –2
–5x – 5y
Use equations
Multiply equation
-2 by –2, and add
to equation 1 .
1
and
3
=5
4
to create a second equation in x and y.
1
1
3
–x + y + 2z = 7
–2(–3x – 4y + z = 4)
–x + y + 2z = 7
6x + 8y – 2z = –8
5x + 9y
= –1
Multiply equation
-3 by –2, and add
to equation 1 .
5
Example: continued
You now have a 2-by-2 system.
–5x – 5y = 5
5x + 9y = –1
4
5
Example: 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
Add equation
Solve for y.
5
1
to equation
4
.
Example: continued
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
Solve for x.
1
Example: continued
Step 4 Substitute for x and y in one of the original
equations to solve for z.
2
2x +3y + z = 1
Substitute –2 for x and 1 for y.
2(–2) +3(1) + z = 1
–4 + 3 + z = 1
Solve for z.
z=2
1
The solution is (–2, 1, 2).
1
Your Turn:
Solve the system.
3 x  9 y  6z  3
(1)
2x  y  z  2
xy z2
(2)
(3)
Solution
Eliminate z by adding equations (2) and (3) to get
3 x  2y  4
(4)
5.1 - 33
Your Turn:
To eliminate z from another pair of
equations, multiply both sides of equation
(2) by 6 and add the result to equation (1).
12x  6y  6z  12
3 x  9 y  6z  3
15 x  15y
 15
Multiply (2) by 6.
(1)
(5)
Make sure
equation (5) has
the same two
variables as
equation (4).
5.1 - 34
Your Turn:
To eliminate x from equations (4) and (5),
multiply both sides of equation (4) by 5 and add
the result to equation (5). Solve the resulting
equation for y.
15 x  10y  20
15 x  15y  15
5y   5
y  1
Multiply (4) by –5.
(5)
Add.
Divide by 5.
5.1 - 35
Your Turn:
Using y = – 1 , find x from equation (4)
by substitution.
3 x  2( 1)  4
(4) with y = –1.
x 2
5.1 - 36
Your Turn:
Substitute 2 for x and –1 for y in equation (3) to
find z.
2  ( 1)  z  2
(3) with x = 2, y = –1.
z 1
The solution set is {(2, –1,1)}.
5.1 - 37
Your Turn:
Solve the following system by the elimination method.
2x + 2y + z = 1
– x + y + 2z = 3
x + 2y + 4z = 0
Solution: Leave the first equation alone, and multiply
the second equation by 2, since combining these two
equations will eliminate the variable x.
2x + 2y + z = 1
– 2x + 2y + 4z = 6
Continued.
Your Turn: continued
Combine the two equations together.
4y + 5z = 7
Now combine the 2nd and 3rd equations together (no
need to multiply by any number).
3y +6z = 3
Continued.
Your Turn: continued
Multiply the first equation by 3 and the second
equation by – 4 to combine them.
12y + 15z = 21
– 12y – 24z = – 12
– 9z = 9
z = –1
Continued.
Your Turn: continued
Substitute this variable into one of the two equations
with only two variables.
4y + 5(– 1) = 7
4y – 5 = 7
4y = 12
y=3
Continued.
Your Turn: continued
Now substitute both of the values into one of the
original equations.
2x + 2(3) + (– 1) = 1
2x + 6 – 1 = 1
2x = – 4
x = –2
So the solution is (– 2, 3, – 1).
Your Turn:
Solve the system of equations.
5x + 3y + 2z = 2
2x + y – z = 5
x + 4y + 2z = 16
Step 1
Use elimination to make a system of two equations in
two variables.
5x + 3y + 2z = 2
5x + 3y + 2z = 2
2x + y – z = 5
(+)4x + 2y – 2z = 10
Multiply by 2.
9x + 5y
= 12
First equation
Second equation
Add to
eliminate z.
Your Turn: continued
5x + 3y + 2z = 2
(–) x + 4y + 2z = 16
4x – y
= –14
First equation
Third equation
Subtract to eliminate z.
Notice that the z terms in each equation have been eliminated.
The result is two equations with the same two variables x and
y.
Your Turn: continued
Step 2
Solve the system of two equations.
9x + 5y = 12
4x – y = –14
Multiply by 5
9x + 5y = 12
(+) 20x – 5y = –70
29x
= –58 Add to
eliminate y.
x = –2
Divide by 29.
Your Turn: continued
Substitute –2 for x in one of the two equations
with two variables and solve for y.
4x – y = –14
Equation with two variables
4(–2) – y = –14
–8 – y = –14
y=6
The result is x = –2 and y = 6.
Replace x with –2.
Multiply.
Simplify.
Your Turn: continued
Step 3
Solve for z using one of the original equations with
three variables.
2x + y – z = 5
Original equation with
three variables
2(–2) + 6 – z = 5
–4 + 6 – z = 5
z = –3
Replace x with –2 and y with 6.
Multiply.
Simplify.
Answer: The solution is (–2, 6, –3). You can check
this solution in the other two original
equations.
Example: 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
Example: 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.
Friday’s sales.
200x + 30y + 40z = 1470
1
250x + 60y + 50z = 1950
2
Saturday’s sales.
150x + 30y = 1050
3
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.
Example: continued
Step 2 Eliminate z.
Multiply equation
1
2
1
by 5 and equation
5(200x + 30y + 40z = 1470)
2
by –4 and add.
1000x + 150y + 200z = 7350
–4(250x + 60y + 50z = 1950) –1000x – 240y – 200z = –7800
y
=5
By eliminating z, due to the coefficients of x, you also eliminated x
providing a solution for y.
Example: continued
Step 3 Use equation
150x + 30y = 1050
150x + 30(5) = 1050
3
x=6
3
to solve for x.
Substitute 5 for y.
Solve for x.
Example: 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.
Your Turn:
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
Solution:
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.
Solution:
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
–2(x – 5y = –7)
2x + 4y = 14
4
= –7
by –2 and add to equation
–2x + 10y = 14
2x + 4y = 14
y=2
2
.
Solve for y.
4
Solution:
Step 3 Use equation
2
2x + 4y = 14
2x + 4(2) = 14
x=3
2
to solve for x.
Substitute 2 for y.
Solve for x.
Solution:
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.
Other Solutions
• The systems so far have had unique
solutions (one solution).
• However, 3-by-3 systems may have no
solution or an infinite number of solutions.
Example:
Classify the system and determine the number of solutions.
2x – 6y + 4z = 2
1
–3x + 9y – 6z = –3
2
5x – 15y + 10z = 5
3
Example: continued
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

Example: continued
Multiply equation
1
3
5(2x – 6y + 4z = 2)
–2(5x – 15y + 10z = 5)
1
by 5 and equation
3
by –2 and add.
10x – 30y + 20z = 10
–10x + 30y – 20z = –10
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.
Your Turn:
Classify each system and determine the number of
solutions.
2x – y + 2z = 5
1.
–3x +y – z = –1
inconsistent; none
x – y + 3z = 2
9x – 3y + 6z = 3
2.
12x – 4y + 8z = 4
–6x + 2y – 4z = 5
consistent; dependent;
infinite
EXTRA PRACTICE
1. What is the solution to the system of
equations shown below?
3x + y – z = 5
–15x – 5y + 5z = 11
x+y+z=2
A. (0, 6, 1)
B. (1, 0, –2)
C. infinite number of solutions
D. no solution
2. What is the solution to the system of
equations shown below?
2x + 3y – 3z = 16
x + y + z = –3
x – 2y – z = –1
A.
(–1, 2, –4)
B. (–3, –2, 2)
C.
infinite number of solutions
D.
no solution
3. What is the solution to the system of
equations shown below?
x + y – 2z = 3
–3x – 3y + 6z = –9
2x + y – z = 6
A. (1, 2, 0)
B. (2, 2, 0)
C. infinite number of solutions
D. no solution
Essential Question
Big Idea: Solving Equations and
Inequalities
• How is solving a system of three equations in
three variables similar to solving a system of two
equations in two variables?
• Use elimination, the same method you used for a
system of two equations in two variables.
Assignment
• Section 3-5, Pg. 181 – 183; #1 – 3 all, 6 12 even, 16 – 20 even (use elimination), 24
– 38 even.