Chapter 1 Algebra: Understand Numerical And Algebraic Expressions
Vocabulary Review
Page 68 Exercise 1 Answer
The number 5 in the expression \(5^3\) is the base
Result
base
Page 68 Exercise 2 Answer
A(n) exponent describes the number of times the base is used as a factor.
Result
Exponent
Page 68 Exercise 3 Answer
There are four terms in the expression
5k + 3 − 2k + 7
Result
Terms
Page 68 Exercise 4 Answer
A variable is a quantity that can change or vary.
Result
Variable
Page 68 Exercise 5 Answer
3x + 2 is equivalent to 4x + 3 − x − 1
4x + 3 − x − 1
= 4x − x + 3 − 1 Grouping Like terms
= 3x + 2 Combine Like terms
Result
3x + 2 and 4x + 3 − x − 1 are equivalent
Page 68 Exercise 6 Answer
7x − x Evaluate
= 6x Combine Like Terms
Result
7x − x and 6x are equivalent expressions.
Page 68 Exercise 7 Answer
3(2x − 3) Evaluate
= 3(2x) + 3(−3) Use Distributive Property
= 6x − 9 Multiply
Result
3(2x − 3) and 6x − 9 are equivalent
Page 68 Exercise 8 Answer
x − 8 + 3x Evaluate
= x + 3x − 8 Group Like Term
= 4x − 8 Combine Like Terms
= 4(x) − 4(2) Use Distributive Property
= 4(x − 2) 4 is the common factor
Result
x − 8 + 3x and 4(x − 2) are equivalent
Page 68 Exercise 9 Answer
3a + 3z
The terms 3a and 3z are Not Like Terms because the terms do not have the same variables.
3a + 3z → N
The terms \(\frac{x}{3}\) and \(\frac{x}{4}\) are Like Terms because both the terms have the same variables.
\(\frac{x}{3}+\frac{x}{4} \rightarrow \underline{Y}\)4j − j + 3.8j
The terms 4j, j and 3.8j are Like Terms because all the terms have the same variables.
4j − j + 3.8j → Y
Result
3a + 3z → N
\(\frac{x}{3}+\frac{x}{4} \rightarrow \underline{Y}\)
4j − j + 3.8j → Y
Page 68 Exercise 10 Answer
One way to simplify the expression 4(3q − q):
The given algebraic expression have the same variables so we can combine the Like Terms and then simplify the expressions by Multiplying it
4(3q − q)
= 4((3 − 1)q)
= 4(2q)
= 8q
Result
4(3q − q) and 8q are equivalent
Page 69 Exercise 1 Answer
22 less than 5 times a number f = 5f − 22
Result
5f − 22
Page 69 Exercise 2 Answer
48 times a number of game markers, g = 48 × g
Result
48 × g
Page 69 Exercise 3 Answer
a number of eggs, e, divided by 12 = e ÷ 12
Result
e ÷ 12
Page 69 Exercise 4 Answer
3 times the sum of m and 7 = 3(m + 7)
Result
3(m + 7)
Page 69 Exercise 1 Answer
80 − \(4^2\) ÷ 8 Evaluate
= 80 − 16 ÷ 8 Evaluate the power
= 80 − 2 Divide
= 78 Subtract
Result
78
Page 69 Exercise 2 Answer
92.3 − (3.2 ÷ 0.4) × \(2^3\) Evaluate
= 92.3 − 8 × \(2^3\) Evaluate inside the parentheses
= 92.3 − 8 × 8 Evaluate the power
= 92.3 − 64 Multiply
= 28.3 Subtract
Result
28.3
Page 69 Exercise 3 Answer
\(\left[\left(2^3 \times 2.5\right) \div \frac{1}{2}\right]+120\) Evaluate
= \(\left[(8 \times 2.5) \div \frac{1}{2}\right]+120\) Evaluate inside the parentheses
= \(\left[20 \div \frac{1}{2}\right]+120\) Evaluate inside the parentheses
= 40 + 120 Evaluate inside the bracket
= 160 Add
Result
160
Page 69 Exercise 4 Answer
[20 + (2.5 ⋅ 3)] − \(3^3\) Evaluate
= [20 + 7.5] − \(3^3\) Evaluate inside the parentheses
= 27.5 − \(3^3\) Evaluate inside the bracket
= 27.5 − 27 Evaluate the power
= 0.5 Subtract
Result
0.5
Page 69 Exercise 5 Answer
\(\left[\left(2 \times 10^0\right) \div \frac{1}{3}\right]+8\) Evaluate
= \(\left[(2 \times 1) \div \frac{1}{3}\right]+8\) Any non zero number raised to an exponent of zero has a value of 1
= \(\left[2 \div \frac{1}{3}\right]+8\) Evaluate inside the parentheses
= 6 + 8 Evaluate inside the bracket
= 14 Add
Result
14
Page 69 Exercise 1 Answer
\(9^2\) Evaluate
= 9 × 9 Write as repeated multiplication
= 81 Multiply
Result
81
Page 69 Exercise 2 Answer
\(99^1\) Evaluate
Any number raised to an exponent of 1 has a value same as base.
Result
99
Page 69 Exercise 3 Answer
\(3105^0\) Evaluate
= 1 Any non zero number raised to an exponent of zero has a value of 1
Result
1
Page 70 Exercise 1 Answer
12x − 7 Given
= 12(4) − 7 Substitute x = 4
= 48 − 7 Multiply
= 41 Subtract
Result
41
Page 70 Exercise 2 Answer
\(4^2\) ÷ y Given
= \(4^2\)÷ 8 Substitute x = 4 and y = 8
= 16 ÷ 8 Evaluate the power
= 2 Divide
Result
2
Page 70 Exercise 3 Answer
5z + 3n − \(z^3\) Given
= 5(1) + 3(7) − \((1)^3\) Substitute z = 1 and n = 7
= 5 + 21 − 1 Multiply
= 26 − 1 Add
= 25 Subtract
Result
25
Page 70 Exercise 4 Answer
\(y^2\) ÷ 2x + 3n − z Given
= \((8)^2\) ÷ 2(4) + 3(7) − (1) Substitute the value of variables
= 64 ÷ 8 + 21 − 1 Evaluate
= 8 + 21 − 1 Divide
= 29 − 1 Add
= 28 Subtract
Result
28
Page 70 Exercise 1 Answer
12 + y
The expression 12+y has two terms.
One term is 12 and other term is y
Note: Each part of the expression that is separated by a plus or a minus sign is called a term.
Result
The expression 12 + y has two terms.
Page 70 Exercise 2 Answer
8x + (9 ÷ 3) − 4.3
The expression 8x + (9 ÷ 3) − 4.3 has three terms.
One term is 8x, second term is 9 ÷ 3 and third term is 4.3
Note: Each part of the expression that is separated by a plus or a minus sign is called a term.
Result
The expression 8x+(9÷3)−4.3 has three terms.
8 is the coefficient of x
Page 70 Exercise 3 Answer
Write an expression that has four terms and includes two variables.
2y − 3x + 3 ⋅ 2 − (8 ÷ 2)
The expression 2y − 3x + 3 ⋅ 2 − (8 ÷ 2) has four terms.
First term is 2y
Second term is 3x
Third term is 3 ⋅ 2
Fourth term is (8 ÷ 2)
Note: Each part of the expression that is separated by a plus or a minus sign is called a term.
Result
2y − 3x + 3⋅2 − (8 ÷ 2)
Page 70 Exercise 1 Answer
2(x + 4) Given
= 2(x) + 2(4) Using Distributive Property
= 2x + 8 Multiply
Result
2(x + 4) and 2x + 8 are equivalent
Page 70 Exercise 2 Answer
5x − 45 Given
= 5(x) − 5(9) Using Distributive Property
= 5(x − 9) 5 is the common factor
Result
5x − 45 and 5(x − 9) are equivalent expressions
Page 70 Exercise 3 Answer
3(x + 7) Given
= 3(x) + 3(7) Using Distributive Property
= 3x + 21 Multiply
Result
3(x + 7) and 3x + 21 are equivalent expressions
Page 71 Exercise 1 Answer
For y = 1
5(2.2y + 1) − 3 = 5(2.2(1) + 1) − 3 = 5(2.2 + 1) − 3 = 5(3.2) − 3 = 16 − 3 = 13
11y + 5 − y = 11(1) + 5 − (1) = 11 + 5 − 1 = 16 − 1 = 15
11y + 2 = 11(1) + 2 = 11 + 2 = 13
For y = 2
5(2.2y + 1)− 3 = 5(2.2(2) + 1) − 3 = 5(4.4 + 1) − 3 = 5(5.4) − 3 = 27 − 3 = 24
11y + 5 − y = 11(2) + 5 − (2) = 22 + 5 − 2 = 27 − 2 = 25
11y + 2 = 11(2) + 2 = 22 + 2 = 24
For y = 3
5(2.2y + 1) − 3 = 5(2.2(3) + 1) − 3 = 5(6.6 + 1) − 3 = 5(7.6) − 3 = 38 − 3 = 35
11y + 5 − y = 11(3) + 5 − (3) = 33 + 5 − 3 = 38 − 2 = 36
11y + 2 = 11(3) + 2 = 33 + 2 = 35
Result
The expressions 5(2.2y + 1) − 3 and 11y + 2 are equivalent.
Page 71 Exercise 2 Answer
Let x = 2
10x − 3 + 2x − 5 = 10(2) − 3 + 2(2) − 5 = 20 − 3 + 4 − 5
= 17 + 4 − 5 = 21 − 5 = 16
4(3x − 2) = 4(3(2) − 2) = 4(6 − 2) = 4(4) = 16
Let x = 3
10x − 3 + 2x − 5 = 10(3) − 3 + 2(3) − 5 = 30 − 3 + 6 − 5
= 27 + 6 − 5 = 33 − 5 = 28
4(3x − 2) = 4(3(3) − 2) = 4(9 − 2) = 4(7) = 28
The expressions 10x – 3 + 2x – 5 and 4(3x – 2) name the same number for same same value of x.
Hence, the expressions are equivalent.
Result
YES
Page 71 Exercise 3 Answer
Let y = 2
3y + 3 = 3(2) + 3 = 6 + 3 = 9
\(9\left(y+\frac{1}{3}\right)=9\left(2+\frac{1}{3}\right)=9(2)+9\left(\frac{1}{3}\right)=18+3=21\)Let y = 3
3y + 3 = 3(3) + 3 = 9 + 3 = 12
\(9\left(y+\frac{1}{3}\right)=9\left(3+\frac{1}{3}\right)=9(3)+9\left(\frac{1}{3}\right)=27+3=30\)The expressions 3y + 3 and \(9\left(y+\frac{1}{3}\right)\) does not name the same number for same value of variable y.
Hence, the expressions are not equivalent
Result
NO
Page 71 Exercise 4 Answer
6(3x + 1) Given
= 6(3x) + 6(1) Using Distributive Property
= 18x + 6 Multiply
= 9x + 9x + 6 18x can be written as 9x and 9x
Result
6(3x + 1) and 9x + 9x + 6 are equivalent
Page 71 Exercise 1 Answer
9y + 4 − 6y Given
= 9y − 6y + 4 Commutative Property of Addition
= 3y + 4 Simplify
Result
3y + 4
Page 71 Exercise 2 Answer
3x + 5 + 7x Given
= 3x + 7x + 5 Commutative Property of Addition
= 10x + 5 Simplify
Result
10x + 5
Page 71 Exercise 3 Answer
2y + 8 − y Given
= 2y − y + 8 Commutative Property of Addition
= y + 8 Simplify
Result
y + 8
Page 71 Exercise 4 Answer
8x + 13 − 3x + 9 Given
= 8x − 3x + 13 + 9 Commutative Property of Addition
= 5x + 22 Simplify
Result
5x + 22
Page 71 Exercise 5 Answer
\(y^2\) + \(3y^2\) Given
= \(1y^2\) + \(3y^2\) Identity Property of Multiplication
= \(4y^2\) Combine Like terms
Result
\(4y^2\)Page 71 Exercise 6 Answer
4x + 15 − 3x + 10 Given
= 4x − 3x + 15 + 10 Commutative Property of Addition
= x + 25 Simplify
Result
x + 25
Page 71 Exercise 7 Answer
20y − 15 − 6y Given
= 20y − 6y − 15 Commutative Property of Addition
= 14y − 15 Simplify
Result
14y − 15
Page 71 Exercise 8 Answer
10x + 2x − 12x Given
= 12x − 12x Add
= 0 Subtract
Result
0
Page 72 Exercise 1 Answer
5x → It represent the number of hours Michael practices his drums for x weeks.
3x → It represent the number of hours Michael practices his cello for x weeks.
Result
5x represent number of hours of practices of drums over x weeks
3x represent number of hours of practices of cello over x weeks
Page 72 Exercise 2 Answer
2(5x + 3x)
Write Equivalent expressions using properties
→ By Combining like terms, I can write this expression:
2(5x + 3x) = 2((5 + 3)x) = 2(8x)
→ By using the Distributive Properties, I can write this expression:
2(5x + 3x) = 2(5x) + 2(3x) = 10x + 6x
Result
By combining Like Terms → 2(8x)
By using Distributive Property → 10x + 6x
Page 72 Exercise 1 Answer
F = 1.8 × (K − 273) + 32 Given
F = 1.8 × (323 − 273) + 32 Substitute K = 323
F = 1.8 × 50 + 32 Evaluate inside parentheses
F = 90 + 32 Multiply
F = 122 Add
Result
The Temperature in Fahrenheit is 122°F
Page 72 Exercise 2 Answer
Step 1 → Identify the values in the formulas
I = unknown Interest
p = principal loan amount = $4000
r = interest rate = 4% = 0.04
t = time in years = 5
Step 2 → Substitute the values in the formula and evaluate
I = prt
I = (4000 ⋅ 0.04 ⋅ 5) Substitute the value
I = 800
The amount of interest Yolanda will pay is $800
Result
$800
Page 72 Exercise 3 Answer
Step 1 → Identify the values in the formulas
P = Perimeter of the rectangle
l = length = 6.5 cm
w = width = 5.5 cm
Step 2 → Substitute the values in the formula and evaluate
P = 2l + 2w
P = 2(6.5) + 2(5.5) Substitute the value
P = 13 + 11 Multiply
P = 24
The Perimeter of the rectangle is 24 centimeters.
Result
24 cm
Page 73 Exercise 1 Answer
(4.5 + 7.6) − 8 ÷ 2.5 Evaluate
= 12.1 − 8 ÷ 2.5 Evaluate inside the parentheses
= 12.1 − 3.2 Divide
= 8.9 Subtract
Result
8.9
Page 73 Exercise 2 Answer
Number of Large balloons in each package = 12
Number of Large Balloons packages = p
Total Number of balloons in p packages of large balloons
= 12 × p or 12p
Result
12 × p
12p
Page 73 Exercise 3 Answer
5h + 8
The expression 5h + 8 has 2 terms.
One term is 5h and other term is 8
NOTE : Each part of the expression which is separated by a plus or minus sign is called a term.
Result
The expression has two terms → 5h and 8
Page 73 Exercise 4 Answer
5h + 8
5 is the coefficient of variable h in the expression 5h + 8
Result
5
Page 73 Exercise 5 Answer
\(3^4\) → base = 3 and exponent = 4
\(3^4\)= 3 × 3 × 3 × 3
= 81
\(4^3\) → base = 4 and exponent = 3
\(4^3\)= 4 × 4 × 4
= 64
Result
\(3^4\) = 81
\(4^3\) = 64
Page 73 Exercise 6 Answer
6a) 6.5 × 4 − 7.8
= 26 − 7.8 Multiply
Yes, 6.5 × 4 − 7.8 and 26 − 7.8 are equivalent.
6b) 10.3 + (8.7 − 4.2)
= 10.3 + 8.7 − 4.2 Open Parentheses
= 19 − 4.2 Multiply
Yes, 10.3 + (8.7 − 4.2) and 19 − 4.2 are equivalent.
6c) (\(5^2\) + 3.4) ÷ 6.8
=(25 + 3.4) ÷ 6.8 \(5^2\) = 25
= 28.4 ÷ 6.8 Add
No, (\(5^2\) + 3.4) ÷ 6.8 and 13.4 ÷ 6.8 are not equivalent.
6d) 6.5 × (12.6 − 9.3)
= 6.5 × 3.3 Subtract
Yes, 6.5 × (12.6 − 9.3) and 6.5 × 3.3 are equivalent.
Result
6a) YES
6b) YES
6c) NO
6d) YES
Page 73 Exercise 7 Answer
Cost of each necklace = $3
Number of necklace sold = n
Amount spent on supplies = $15
Algebraic Expressions to show how much she earns:
Cost of each necklace × Number of necklaces sold − Supplies
= 3n − 15
Result
3n − 15
Page 74 Exercise 8 Answer
Daily fee = $50
Cost per mile driven = $0.35
Part A
Let m = the number of miles Mr.Parker drives for the day
Algebraic Expression that shows the amount he will pay for the van:
50 + 0.35m
Part B
Evaluate the expression for m = 80:
50 + 0.35(80)
= 50 + 28
= $78
Result
Part A → 50 + 0.35m
Part B → $78
Page 74 Exercise 9 Answer
Step 1 → Identify the values in the formulas
g = number of gallons of gasoline used
m = miles per gallon = 18
d = distance traveled = 315
Step 2 → Substitute the values in the formula and evaluate
g = \(\frac{d}{m}\)
g = \(\frac{315}{18}\) Substitute the value
g = 17.5
Result
C) 17.5 gallons
Page 74 Exercise 10 Answer
10a)
4(5c + 3)
= 4(5c) + 4(3) Using Distributive Property
= 20c + 12 Multiply
Properties of operations cannot be used to write 4(5c + 3) as 9c + 7
4(5c + 3) and 9c + 7 are not Equivalent expression.
10b)
2(8f − 5)
= 2(8f) − 2(5) Using Distributive Property
= 16f − 10 10 is the common factor
Properties of operations cannot be used to write 2(8f – 5) as 10f – 10
2(8f − 5) and 10f – 10 are not Equivalent expression.
10c)
3(4g + 7)
= 3(4g) + 3(7) Using Distributive Property
= 12g + 21 Multiply
12g + 21 and 3(4g + 7) are Equivalent expression.
10d)
6(4j − 6)
= 6(4j) + 6(−6) Using Distributive Property
= 24j − 36 Multiply
Properties of Operations cannot be used to write 6(4j − 6) as 24 − 36j
6(4j − 6) and 23 − 36j are not Equivalent expression.
Result
10a) NO
10b) NO
10c) YES
10d) NO
Page 74 Exercise 11 Answer
(6 + 9) + 11 = 6 + (9 + 11)
Associative Property
16k − 24 = 8(2k − 3)
Distributive Property
12 × 4 × 5 = 4 × 5 × 12
Commutative Property
Result
Associative Property → 16k − 24 = 8(2k − 3)
Distributive Property → 16k − 24 = 8(2k − 3)
Commutative Property → 12 × 4 × 5 = 4 × 5 × 12
Page 74 Exercise 12 Answer
\(3^5\)= 3 × 3 × 3 × 3 × 3
= 243
= 5 × 5 × 5
= 125
5 × 5 × 5
= 125
3 × 3 × 3 × 3 × 3
= 243
3 × 3 × 3 × 5 × 5
= 27 × 25
= 675
Result
Expression equal to 243:
\(3^5\)3 × 3 × 3 × 3 × 3
Page 75 Exercise 13 Answer
d = 65t
FOR t = 3
d = 65t = 65(3) = 195
FOR t = 4
d = 65t = 65(4) = 260
FOR t = 6
d = 65t = 65(6) = 390
FOR t = 7
d = 65t = 65(7) = 455
FOR t = 10
d = 65t = 65(10) = 650
Result
260; 455; 650
Page 75 Exercise 14 Answer
5b + 13 − 2b − 7 Evaluate
= 5b − 2b + 13 − 7 Commutative Property of Addition
= 3b + 6 Combining Like Terms
Result
A) 3b + 6
Page 75 Exercise 15 Answer
3n + 4 + 3n + 4 + 4n
= 3n + 3n + 4n + 4 + 4 Commutative Property of Addition
= 10n + 8
3n + 4 + 3n + 4 + 4n ≠ 12n − 8
11n + 4 + n − 12
= 11n + n − 12 + 4 Commutative Property of Addition
= 12n − 8
11n + 4 + n − 12 = 12n − 8
6(6n − 2)
= 6(6n) + 6(−2) Distributive Property
= 12n − 12
6(6n − 2) ≠ 12n − 8
4(3n − 2)
= 4(3n) + 4(−2) Distributive Property
= 12n − 8
4(3n − 2) = 12n − 8
4n + \(2^2\) − 12 + 8n
= 4n + 4 − 12 + 8n
= 4n + 8n − 12 + 4 Commutative Property of Addition
= 12n − 8
4n + \(2^2\) − 12 + 8n = 12n − 8
Result
12n − 8 is equivalent to the following:
→ 11n + 4 + n − 12
→ 4(3n − 2)
→ 4n + \(2^2\) − 12 + 8n
Page 75 Exercise 16 Answer
Part A:
Length = 3w + 1: Width = 4
Algebraic Expression for perimeter of the rectangle:
2l + 2w
= 2(3w + 1) + 2(4)
Part B:
2(3w + 1) + 2(4)
= 2(3w) + 2(1) + 2(8) Distributive Property
= 6w + 2 + 16
= 6w + 18
Part C:
6w + 18
= 6(8) + 18 Substitute w = 8
= 48 + 18
= 66
Result
Part A → 2(3w + 1) + 2(4)
Part B → 6w + 18
Part C → 66 units
Page 76 Exercise 17 Answer
10x − 4x + 6
= 6x + 6 Combine Like Term
3(2x + 2)
= 3(2x) + 3(2) Distributive Property
= 6x + 6
Thus, 10x − 4x + 6 and 3(2x + 2) are equivalent expressions.
Quinn is not correct.
Result
No, I do not agree
The expression 10x − 4x + 6 and 3(2x + 2) are equivalent.
Page 76 Exercise 18 Answer
Part A:
Use Distributive Property to write an expression that is equivalent to 3(3x − 5)
3(3x − 5)
= 3(3x) + 3(−5) Using Distributive Property
= 9x − 15
Part B:
Use Distributive Property to write an expression that is equivalent to 6x − 15
6x − 15
= 3(2x) + 3(−5) Using Distributive Property
= 3(2x − 5)
Part C:
Explain whether the expression are equivalent
The expressions 3(3x − 5) and 6x − 15 are not equivalent
Result
Part A → 9x − 15
Part B → 3(2x − 5)
Part C → Not Equivalent
Page 76 Exercise 19 Answer
Step 1 → Identify the values in the formulas
V = Volume of the cube
s = length of each side = 12 m
Step 2 → Substitute the values in the formula and evaluate
V = \(s^3\)
V = \((12)^3\) Substitute the value
V = 1728 \(m^3\)
Result
D) 1728 \(m^3\)
Page 76 Exercise 20 Answer
3(14x + 23 + 5x)
Write Equivalent expressions using properties
→ By Combining like terms, I can write this expression:
3(14x + 23 + 5x) = 3((14 + 5)x + 23) = 3(19x + 23)
→ By using the Distributive Properties, I can write this expression:
3(14x + 23 + 5x) = 3(14x) + 3(23) + 3(5x) = 42x + 69 + 15x
Result
3(19x + 23)
42x + 69 + 15x
Page 77 Exercise 1 Answer
Registration cost = $5
Each class = $8
Part A:
Let c = number of exercise classes a person takes.
Algebraic Expression that show the amount a person will pay to take exercise classes.
5 + 8c
Part B:
For c = 1
5 + 8c = 5 + 8(1) = 5 + 8 = 13
For c = 3
5 + 8c = 5 + 8(3) = 5 + 24 = 29
For c = 5
5 + 8c = 5 + 8(5) = 5 + 40 = 45
For c = 8
5 + 8c = 5 + 8(8) = 5 + 64 = 69
For c = 10
5 + 8c = 5 + 8(10) = 5 + 80 = 85
Part C:
New Registration = $0
New Each Class = $8 + $1 = $9
Algebraic Expression :9c
Part D:
For c = 1
9c = 9(1) = 9
For c = 3
9c = 9(3) = 27
For c = 5
9c = 9(5) = 45
For c = 8
9c = 9(8) = 72
For c = 10
9c = 9(10) = 90
Part E:
It cost less to attend 3 exercise classes after Danny changes the registration.
It cost same to attend 5 exercise classes after Danny changes the registration.
It cost more to attend 10 exercise classes after Danny changes the registration.
Result
Part A → 5 + 8c
Part B → 29; 69; 85
Part C → 9c
Part D → 27; 72; 90
Part E → Exercise 3 = less; Exercise 5 = same; Exercise 10 = more
Page 78 Exercise 2 Answer
Before Changes
After Changes
Comparing both the prices before and after changes:
If I were running Danny′s exercise class, I would prefer the plan after Danny changes the registration fee.
Result
After Danny changes the registration fee.