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Five-Minute Check (over Lesson 3–3)
CCSS
Then/Now
New Vocabulary
Example 1: A System with One Solution
Example 2: No Solution and Infinite Solutions
Example 3: Real-World Example: Write and Solve a System
of Equations
Over Lesson 3–3
Graph the system of inequalities. Name the
coordinates of the vertices of the feasible
region. Find the maximum an minimum
values of the given function for this region.
1 ≤ x ≤ 4, y ≥ x, y ≥ 2x + 3; f(x, y) = 3x – 2y
A.
maximum: f(4, 4) = 4
minimum: f(4, 11) = –10
B.
maximum: f(4, 11) = 10
minimum: f(1, 1) = 1
C.
maximum: f(4, 4) = 4
minimum: f(1, 5) = –7
D.
maximum: f(4, 11) = 10
minimum: f(4, 4) = 4
Over Lesson 3–3
Graph the system of inequalities. Name the
coordinates of the vertices of the feasible
region. Find the maximum an minimum
values of the given function for this region.
1 ≤ x ≤ 4, y ≥ x, y ≥ 2x + 3; f(x, y) = 3x – 2y
A.
maximum: f(4, 4) = 4
minimum: f(4, 11) = –10
B.
maximum: f(4, 11) = 10
minimum: f(1, 1) = 1
C.
maximum: f(4, 4) = 4
minimum: f(1, 5) = –7
D.
maximum: f(4, 11) = 10
minimum: f(4, 4) = 4
Over Lesson 3–3
A company profits $3 for every widget it
manufactures and $2 for every plinket it
manufactures. It must make at least one widget and
one plinket each hour, but cannot make more than
7 total widgets and plinkets in any hour. What is the
most profit the company can make in any hour?
A. $21
B. $20
C. $18
D. $5
Over Lesson 3–3
A company profits $3 for every widget it
manufactures and $2 for every plinket it
manufactures. It must make at least one widget and
one plinket each hour, but cannot make more than
7 total widgets and plinkets in any hour. What is the
most profit the company can make in any hour?
A. $21
B. $20
C. $18
D. $5
Content Standards
A.CED.3 Represent constraints by equations
or inequalities, and by systems of equations
and/or inequalities, and interpret solutions as
viable or nonviable options in a modeling
context.
Mathematical Practices
3 Construct viable arguments and critique
the reasoning of others.
You solved linear equations with two variables.
• Solve systems of linear equations in three
variables.
• Solve real-world problems using systems of
linear equations in three variables.
• ordered triple
A System with One Solution
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
2x + y – z = 5
Multiply by 2.
5x + 3y + 2z = 2 First equation
(+)4x + 2y – 2z = 10 Second equation
9x + 5y
= 12 Add to
eliminate z.
A System with One Solution
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.
A System with One Solution
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.
A System with One Solution
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
Replace x with –2.
Multiply.
Simplify.
The result is x = –2 and y = 6.
A System with One Solution
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
Answer:
Replace x with –2 and y with 6.
Multiply.
Simplify.
A System with One Solution
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.
What is the solution to the system of equations
shown below?
2x + 3y – 3z = 16
x + y + z = –3
x – 2y – z = –1
A.
B. (–3, –2, 2)
C. (1, 2, –6)
D. (–1, 2, –4)
What is the solution to the system of equations
shown below?
2x + 3y – 3z = 16
x + y + z = –3
x – 2y – z = –1
A.
B. (–3, –2, 2)
C. (1, 2, –6)
D. (–1, 2, –4)
No Solution and Infinite Solutions
A. Solve the system of equations.
2x + y – 3z = 5
x + 2y – 4z = 7
6x + 3y – 9z = 15
Eliminate y in the first and third equations.
Multiply by 3.
2x + y – 3z = 5
6x + 3y – 9z = 15
6x + 3y – 9z = 15
(–)6x + 3y – 9z = 15
0= 0
No Solution and Infinite Solutions
The equation 0 = 0 is always true. This indicates that
the first and third equations represent the same plane.
Check to see if this plane intersects the second plane.
Multiply by 6.
x + 2y – 4z = 7
6x + 3y – 9z = 15
6x + 12y – 24z = 42
(–)6x + 3y – 9z = 15
9y – 15z = 27
Divide by the GCF, 3.
Answer:
3y – 5z = 9
No Solution and Infinite Solutions
The equation 0 = 0 is always true. This indicates that
the first and third equations represent the same plane.
Check to see if this plane intersects the second plane.
Multiply by 6.
x + 2y – 4z = 7
6x + 3y – 9z = 15
6x + 12y – 24z = 42
(–)6x + 3y – 9z = 15
9y – 15z = 27
Divide by the GCF, 3.
3y – 5z = 9
Answer: The planes intersect in a line. So, there is an
infinite number of solutions.
No Solution and Infinite Solutions
B. Solve the system of equations.
3x – y – 2z = 4
6x – 2y – 4z = 11
9x – 3y – 6z = 12
Eliminate x in the first two equations.
Multiply by 2.
3x – y – 2z = 4
6x – 2y – 4z = 11
6x – 2y + 4z = 8
(–) 6x – 2y – 4z = 11
0 = –3
Answer:
No Solution and Infinite Solutions
B. Solve the system of equations.
3x – y – 2z = 4
6x – 2y – 4z = 11
9x – 3y – 6z = 12
Eliminate x in the first two equations.
Multiply by 2.
3x – y – 2z = 4
6x – 2y – 4z = 11
6x – 2y + 4z = 8
(–) 6x – 2y – 4z = 11
0 = –3
Answer: The equation 0 = –3 is never true.
So, there is no solution of this system.
A. 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
A. 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
B. 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
B. 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
Write and Solve a System of
Equations
SPORTS There are 49,000 seats in a sports stadium.
Tickets for the seats in the upper level sell for $25,
the ones in the middle level cost $30, and the ones in
the bottom level are $35 each. The number of seats
in the middle and bottom levels together equals the
number of seats in the upper level. When all of the
seats are sold for an event, the total revenue is
$1,419,500. How many seats are there in each level?
Explore Read the problem and define the variables.
u = number of seats in the upper level
m = number of seats in the middle level
b = number of seats in the bottom level
Write and Solve a System of
Equations
Plan
There are 49,000 seats.
u + m + b = 49,000
When all the seats are sold, the revenue is
1,419,500. Seats cost $25, $30, and $35.
25u + 30m + 35b = 1,419,500
The number of seats in the middle and bottom
levels together equal the number of seats in the
upper level.
m+b=u
Write and Solve a System of
Equations
Solve Substitute u = m + b in each of the first
two equations.
(m + b) + m + b = 49,000
2m + 2b = 49,000
m + b = 24,500
Replace u with m + b.
Simplify.
Divide by 2.
25(m + b) + 30m + 35b = 1,419,500 Replace u with
m + b.
25m + 25b + 30m + 35b = 1,419,500 Distributive
Property
55m + 60b = 1,419,500 Simplify.
Write and Solve a System of
Equations
Now, solve the system of two equations in two variables.
Multiply by 55.
m + b = 24,500
55m + 60b = 1,419,500
55m + 55b = 1,347,500
(–) 55m + 60b = 1,419,500
–5b =
–72,000
b = 14,400
Write and Solve a System of
Equations
Substitute 14,400 for b in one of the equations with two
variables and solve for m.
m + b = 24,500
m + 14,400 = 24,500
m = 10,100
Equation with two variables
b = 14,400
Subtract 14,400 from
each side.
Write and Solve a System of
Equations
Substitute 14,400 for b and 10,100 for m in one of the
original equations with three variables.
m+b =u
10,100 + 14,400 = u
24,500 = u
Answer:
Equation with three
variables
m = 10,100, b = 14,400
Add.
Write and Solve a System of
Equations
Substitute 14,400 for b and 10,100 for m in one of the
original equations with three variables.
m+b =u
10,100 + 14,400 = u
24,500 = u
Answer:
Equation with three
variables
m = 10,100, b = 14,400
Add.
There are 24,500 upper level, 10,100
middle level, and 14,400 bottom level
seats.
Write and Solve a System of
Equations
Check Check to see if all the criteria are met.
24,500 + 10,100 + 14,400 = 49,000
The number of seats in the middle and bottom
levels equals the number of seats in the
upper level.
10,100 + 14,400 = 24,500
When all of the seats are sold, the revenue
is $1,419,500.
24,500($25) + 10,100($30) +
14,400($35) = $1,419,500
BUSINESS The school store sells pens, pencils, and
paper. The pens are $1.25 each, the pencils are $0.50
each, and the paper is $2 per pack. Yesterday the
store sold 25 items and earned $32. The number of
pens sold equaled the number of pencils sold plus
the number of packs of paper sold minus 5. How
many of each item did the store sell?
A. pens: 5; pencils: 10; paper: 10
B. pens: 8; pencils: 7; paper: 10
C. pens: 10; pencils: 7; paper: 8
D. pens: 11; pencils: 2; paper: 12
BUSINESS The school store sells pens, pencils, and
paper. The pens are $1.25 each, the pencils are $0.50
each, and the paper is $2 per pack. Yesterday the
store sold 25 items and earned $32. The number of
pens sold equaled the number of pencils sold plus
the number of packs of paper sold minus 5. How
many of each item did the store sell?
A. pens: 5; pencils: 10; paper: 10
B. pens: 8; pencils: 7; paper: 10
C. pens: 10; pencils: 7; paper: 8
D. pens: 11; pencils: 2; paper: 12