CHAPTER 1. FACTORS AND MULTIPLES

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7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 1. FACTORS AND MULTIPLES
CHAPTER 1-1. EXPONENTIAL EXPRESSION
EXPRESSION
EXPONENTIAL NOTATION:
1 1 1
1 3
× × = ( ) or
2 2 2
2
1 1 1
1
× × = 3
2 2 2 2
3 × 3 = 3𝟐
⏟
two 3
3 × 3 × 3 × 3 = 3πŸ’
⏟
four 3
2 × 2 × 2 × πŸ• × πŸ• = 23 × 7𝟐
1
1
= 3,
2×2×2 2
BASE AND EXPONENT
WARNING:
2 + 2 + 2 ≠ 23 ,
2 ∢𝐞𝐱𝐩𝐨𝐧𝐞𝐧𝐭
⏟
3
1
1
= 2
2 × 2 × 3 × 3 2 × 32
23 ≠ 6
π›πšπ¬πž
π‘Ž0 = 1 (π‘Ž ≠ 0), 00 = 0
Example 1: Find the base and the exponent of the exponential expression.
(A) 24
Base:_____ Exponent:____
(B) 57
Base:_____ Exponent:____
(A) B : 2, E: 4
(B) B : 5, E: 7
1
(C) B : , E: 13
(C)
1 13
( )
2
Base:____ Exponent:____
(D)
4 9
( )
3
Base:_____ Exponent:____
Example 2: Express each of the following in exponential notation.
(A) 3 × 3 × 3 × 3 × 3
(B) 2 × 3 × 2
2
4
(D) B : , E: 9
3
(A) 35
(B) 22 × 3
(C) 2 × 3 × 113
(D)
1
(C) 3 × 2 × 11 × 11 × 11
1
1
1
(D) 10 × 10 × 10 × 10
(E) 2 × 33
(F) 32 × 54
(G)
(H)
(E) 2 × 2 × 3 × 3 × 3
1
1
1
1
1
(F) 3 × 5 × 5 × 3 × 5 × 5
1
1
(G) 2 × 2 × 2 × 2 × 7 × 7 × 7
1
(H) 3×5×5×5×7×7
1
104
2
1
24 ×73
1
3×53 ×72
1
7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1. Find the value of the expression.
(A) 24
1.
3 3
(A) 16
(B) (2)
(B)
27
8
2.
2. Simplify and express each of the following in exponential notation.
(A) 7 × 7 × 9
(B) 3 × 3 × 3 × 3 × 3
(A) 72 × 9
(B) 35
(C) 2 × 33 × 72
(D) 23 × 32 × 5
(E) 52 × 73 × 11
(F)
(C) 2 × 3 × 3 × 3 × 7 × 7
(D) 2 × 2 × 3 × 3 × 5
π‘Ž2 × π‘3
(G) π‘Ž2 × π‘4
(H) π‘₯ 4 × π‘¦ 2
(I)
(J)
(E) 5 × 5 × 7 × 7 × 7 × 11
(F) π‘Ž × π‘ × π‘Ž × π‘ × π‘
(K)
(L)
1 3
1
2
3
( ) ×
2 2
1 2
3
5
( ) ×( )
1
32 ×53
2×33 ×5
74 ×112
(M) 77
(N) 29
(G) π‘Ž × π‘ × π‘ × π‘ × π‘Ž × π‘
1
2
1
1
1
(H) π‘₯ × π‘₯ × π‘₯ × π‘₯ × π‘¦ × π‘¦
(J)
2
3
(K) 3×3×5×5×5
(L)
2×3×3×3×5
7×7×7×7×11×11
(M) 73 × 74
(N) 23 × 26
(O) π‘Ž5 × π‘Ž7
(P) π‘₯ 4 × π‘₯ 9
(I)
×3×2×2
1
2
3
1
5
× × ×
1
5
(O) π‘Ž12
(P)
π‘₯ 13
2
7 GRADE-I: DR. EUNKYUNG YOU
3. Find the base and the exponent of the exponential expression.
1 6
(A) (−3)4 Base:____Exponent:_____
(B) ( )
Base:_____Exponent:___
3
3.
(A) B:-3 E:4
1
(B) B:
3
E:6
4. Find the exponent of the base 5 in exponential notation : 2 × 3 × 2 × 3 × 5 × 2 × 5 × 5
5. A square has a side of length 7 cm. Find the area of the square.
6. A cube has a side of length 8 cm. Find its volume of the cube.
7. Which of the following is true?
(A) 33 = 9
(B) 2 × 2 × 2 = 32
(C) 3 × 3 × 3 × 5 × 5 = 33 × 52
(D) 3 + 3 + 3 + 3 = 34
2
2
2
2
(E) 3 × 3 × 3 × 3 =
24
3
8. Which of the following is true?
(A) 7 × 7 × 7 × 7 = 4 × 7
(B) 3 × 3 × 2 × 2 = 33 × 22
(C) π‘Ž × π‘Ž × π‘ × π‘ × π‘ = π‘Ž2 × π‘ 3
(D) π‘Ž + π‘Ž + π‘Ž + π‘Ž + π‘Ž = π‘Ž5
1
1
1
3
(E) 4 × 4 × 4 = 4
9. Sierpinski Triangle: We will create Sierpinski Triangle in the following way:
 Start with a filled-in equilateral triangle.
 Connect the midpoints of each side, dividing the triangle into four smaller equilateral
triangles.
 Remove the middle triangle.
 Repeat on all filled-in triangles infinitely.
Find the number of the shaded equilateral triangles in 𝑆9
𝑆1
𝑆2
𝑆3
First four stages of the Sierpinski triangle.
𝑆4
4.
3
5.
72 = 49 cm2
6.
73 cm3
7.
C
8.
D
9.
38
3
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 1-2. FACTORS AND MULTIPLES
DEFINITION: A number is a mathematical object used to
DEFINITION:
β–‘ Factors are numbers you can multiply together
count, label, and measure.
Natural numbers :
1, 2, 3, β‹―
Whole numbers:
0, 1, 2, 3, β‹―
Integers :
β‹― − 3, −2, −1, 0, 1, 2, 3, β‹―
to get another number:
β–‘ A number that can be divided by another number
π‘Ž without a remainder is called a multiple of the
number π‘Ž
𝟏𝟐 = πŸ‘ × πŸ’ :
DEFINITION:
3 and 4 are factors of 12 and
β–‘ A prime number is a natural number greater than 1
that has only two factors, 1 and itself.
12 is the multiple of 3 (or 4)
β–‘ A composite number is a natural number greater
𝟏𝟐 = 𝟏 × πŸπŸ = 𝟐 × πŸ” = πŸ‘ × πŸ’
than 1 that has more than two factors.
1, 2, 3, 4, 6, 12 are all factors of 12
2,3,5,7,11, 13, β‹― are prime numbers
DEFINITION: A multiple is the result of multiplying a
6 = 2 × 3 is a composite number
number by an integer
WARNING:
1 is not a prime, nor a composite number.
Multiples
3
6
9
12
β‹―
of 3:
3×1
3×2
3×3
3×4
β‹―
Example 1: Find all factors of the following number and determine whether it is a prime or a composite number
(A) 23
(B) 60
Solution:
(A) 23 is not divisible by any of the prime numbers less than 23. Therefore, 23 is a prime number.
So 1, 23 are all factors of 23
(B) Since 60 ÷ 2 = 30, 60 = 2 × 30 is a composite number.
60 = 1 × 60 = 2 × 30 = 3 × 20 = 4 × 15 = 5 × 12 = 6 × 10
The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60
Example 2: Find all factors of the following number and determine whether it is a prime or a composite number
(A) 6
(B) 11
(A) 1,2,3,6: composite
(B) 1,11: Prime
(C) 1,13: Prime
(C) 13
(D) 15
(D) 1,3,5,15: composite
(E) 1,2,5,10,25,50:
(E) 50
composite
(F) 32
(F)
1,2,4,8,16,32: composite
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7 GRADE-I: DR. EUNKYUNG YOU
Example 3: Find five multiple of 12
Solution:
Multiples of 12:
12
12 × 1
24
36
48
60
β‹―
12 × 2
12 × 3
12 × 4
12 × 5
β‹―
Example 4: Write down the first four multiples of each of the following numbers.
(A) 2
(A) 2,4,6,8
(B) 5
(B) 5,10,15,20
(C) 7,14,21,28
(D) 6,12,18,24
(C) 7
(D) 6
(E) 11,22,33,44
(F)
(E) 11
9,18,27,36
(F) 9
Example 5: Fine all the prime numbers between 0 and 50. Do the following.
Step 1: Cross out 1 since 1 is not a prime
Step 2: Cross out all multiples of 2 except 2.
Step 3: Cross out all multiples of 3 except 3.
Step 5: Continue doing this until you have visited all the numbers in the table.
1
11
21
31
41
2
12
22
32
42
3
13
23
33
43
4
14
24
34
44
5
15
25
35
45
6
16
26
36
46
7
17
27
37
47
8
18
28
38
48
9
19
29
39
49
2, 3, 5, 7, 11, 13, 17, 19, 23,
29, 31, 37, 41, 43, 47
10
20
30
40
50
Example 6: Which of the following number is a multiple of 7?
(A) 106
(B) 400
(D) 2395
(E) 1078
(C) 257
(E)
(C) 27
(D)
Example 7: Which of the following numbers have 216 as a multiple?
(A) 9
(B) 12
(D) 32
(E) 36
Example 8: Find the largest multiple of
(A) 19 which is less than 1,000
(B) 7 which is less than 1,00
(A) 988
(B) 98
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7 GRADE-I: DR. EUNKYUNG YOU
Example 9: Find the following.
(A) Factors of 54
(C) Find common Factors of 54 and 72
Example 10: Find the following.
(A) Factors of 6
(B) Factors of 72
(A) 1,2,3,6,9,18,27,54
(B) 1,2,3,4,6,8,9,12,18,
24,36,72
(C) 1,2,3,6,9,18
(D) 1,2,3,6,9,18
(D) Find factors of 18
(B) Factors of 9
(C) Find first five multiple of 6
(D) Find first five multiple of 9
(E) Find common Factors of 6 and 9
(F) Find first two common multiples of
6 and 9
Example 11: Determine whether each statement below is TRUE or FALSE
(A) If 25 is a factor of a number, then 5 is also a factor of the number.
(B) If 2 and 3 are factors of a numbers, then 6 is a factor of the number.
(C) If two numbers are multiple of 7, then their sum is a multiple of 7.
(D) The sum of a multiple of 3 and a multiple of 5 is a multiple of 15.
Example 12: There are 21 square blocks. Find the number of possible rectangular designs by
using all blocks.
(A)
(B)
(C)
(D)
(E)
(F)
1,2,3,6
1,3,9
6,12,18,24,30
9,18,27,36,45
1,3
18,36
(A)
(B)
(C)
(D)
T
T
T
F
2 designs
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7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
(A)
(B)
(C)
(D)
(E)
(F)
Composite
Prime
Composite
Prime
Composite
Composite
(C) 28
(A)
(B)
(C)
(D)
(F) 84
(E)
1,3,5,9,15,45
1,2,3,4,6,8,12,16,24,48
1,2,4,7,14,28
1,2,3,4,6,8,9,12,18,24,
36,72
1,2,3,4,6,8,12,16,24,32
43,96
1,2,3,4,6,14,21,28,42,
82
2,4,6, 8
5, 10, 15, 20
7, `4, 21, 28
6, 12, 18, 24
11, 22, 33, 44
9, 18, 27, 36
1. Determine whether the following numbers are prime numbers or composite numbers
(A) 15
(B) 23
(C) 27
(D) 53
(E) 357
2. Find all factors of the following numbers
(A) 45
(B) 48
(D) 72
(E) 96
(F) 143
(F)
3. Find the first four multiples of the following numbers
(A) 2
(B) 5
(D) 6
(E) 11
4. Which of the following numbers have 16 as a factor?
(A) 96
(B) 207
(D) 175
(E) 304
5. Which of the following numbers are not factors of 108?
(A) 6
(B) 8
(D) 18
(E) 36
6. Find the common factors to
(A) 56 and 84
(B) 96 and 120
7. Find the first three common multiples to
(A) 2 and 5
(B) 3 and 5
(C) 7
(F) 9
(A)
(B)
(C)
(D)
(E)
(F)
A, C, E
(C) 336
B
(C) 12
(C) 168 and 360
(A) 1,2,4,7,14,28
(B) 1,2,3,4,6,8,12,24
(C) 1,2,3,4,6,8,12,24
(C) 4 and 6
(A) 10,20,30
(B) 15,30,45
(C) 12,24,36
D
8. Which of the following is true?
(A) 3 is the smallest prime number
(B) All composite numbers are even
(C) The product of two different prime numbers is odd
(D) All even numbers except 2 are not prime numbers
(E) All odd numbers are prime numbers.
9. Determine whether the following statement is true or false
(A) If 9 is a factor of a number, then 3 is a factor of the number.
(B) If 2 and 7 are factors of a numbers, then 14 is a factor of the number.
(C) If 2 and 12 are factors of a numbers, then 24 is a factor of the number.
(D) If two numbers are multiple of 7, then their sum is a multiple of 7
(E) If a number is a multiple of 2 and another number is a multiple of 5, then their sum is a
multiple of 7.
(A)
(B)
(C)
(D)
(E)
True
True
False
True
False
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7 GRADE-I: DR. EUNKYUNG YOU
10. Determine whether each statement below is True or False.
(A) A prime number has only 1 and itself as its factors.
(B) 1 is a prime number.
(C) The number of factors of a prime number is 2.
(D) There is 3 prime numbers which are less than 10.
(E) If 4 is a factor of a number, then 2 is a factor of the number.
(A)
(B)
(C)
(D)
(E)
True
False
True
False
True
6
11. Find the number of multiples of 15 which is less than 100.
12. Find the biggest multiple of 11 which is less than 600.
594
13. Find the biggest multiple of 7 which is less than 200.
196
14. Find all multiples of 4 which are not multiple of 6 and less than 40.
4,8,16,20,28,32
510
15. Find the smallest multiple of 17 which is greater than 500.
16. Find the number of possible rectangular designs by using all blocks if we have
(A) 42 square blocks
(B) 72 square blocks
(A) 4
(B) 6
17. When π‘₯ is divided by 6, its quotient and remainder are 7 and 1, respectively. Find the
remainder if π‘₯ is divided by 5.
3
18. When π‘₯ is divided by 12, its quotient and remainder is 6 and 7, respectively. Find the
remainder and quotient when π‘₯ is divided by 16
Remainder 15, quotient: 4
19. Let π‘Ž, 𝑏 be two natural numbers. When π‘Ž is divided by 𝑏, its quotient and remainder are 10
and 8, respectively. Find the remainder when π‘Ž is divided by 5.
20. There are 30 male students and 42 female students in Tift High School. School decides to
divide them into classes such that each class has same numbers of boys and girls. Find all
possible number of classes.
21. There are 252 students. The teacher wants to put students into groups to do a project. Each
group should consists of the same number of students. There must be at least 5 groups but
not more than 20 groups. Find all possible number of groups and the number of people in
each group.
3
1,2,3,6
6 groups-42 people
7 groups-36 people
9 groups-28 people
12 groups-21 people
14 groups-18 people
18 groups-14 people
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7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 1-3. PRIME FACTORIZATION
DEFINITION: The process to express a composite
number as a product of prime factors is called prime
factorization.
TREE METHOD
90 = 2 × 5 × 32 since
90
β‹° β‹±
FACTOR METHOD
𝟐
90 = 2 × 45
45
β‹°
=2×5×9
β‹±
πŸ“
=2×5×3×3
9
β‹° β‹±
= 2 × 5 × 32
πŸ‘
πŸ‘
DIVISION METHOD
WARNING:
90 = 2 × 5 × 32 since
𝟐) 90
30 = 2 × πŸπŸ“ : Not prime factorization
πŸ“) 45
30 = 2 × 3 × 5: Prime factorization
πŸ‘)
9
πŸ‘
Example 1: Find the prime factorization of 120
Solution:
2 ) 120
120 = 23 × 3 × 5
2 ) 60
2 ) 30
3 ) 15
5
Example 2: Find the prime factorization of each number and express your answer in exponential notation.
(A) 42
(B) 72
(C) 100
(A) 2 × 3 × 7
(B) 23 × 32
(C) 22 × 52
(D) 27
(D) 128
(E) 144
(F) 150
(E) 24 × 32
(F)
2 × 3 × 52
(G) 23 × 5
(H) 3 × 52
(G) 40
(H) 75
(I) 108
(I)
22 × 33
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7 GRADE-I: DR. EUNKYUNG YOU
Example 3: Express 18 × 6 in exponential notation, where all the bases are prime numbers.
Solution:
18 × 6 = (2 × 9) × (2 × 3) = (2 × 3 × 3) × (2 × 3) = 22 × 33
Example 4: Express each of the following in exponential notation, where all the bases are prime numbers.
(A) 12 × 15
(B) 54 × 96
(A) 22 × 32 × 5
(B) 26 × 34
(C) 22 × 33 × 53
(D) 27 × 3 × 7 × 11
(E) 311
(F)
210
(G) 72
(H) 38
(C) 18 × 10 × 75
(D) 16 × 56 × 33
(E) 34 × 37
(F) 23 × 27
(G) 75 ÷ 73
(H) (34 )2
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7 GRADE-I: DR. EUNKYUNG YOU
Example 5: Fine all factors of 22 × 53
Solution
×
1 = 20
21
22
1 = 50
1=1
21 × 1 = 2
22 × 1 = 4
51
1 × 51 = 5
21 × 51 = 10
22 × 51 = 20
52
1 × 52 = 25
21 × 52 = 50
22 × 52 = 100
53
1 × 53 = 125 21 × 53 = 250 22 × 53 = 500
The factors are 1,2,4,5,10,20,25,50,100,125,250 and 500 and there are (3 + 1) × (2 + 1) = 12 factors of 120.
Example 6: Find all factors of the following numbers.
(B) 22 × 34
(A) 35
(C) 2 × 32 × 5
(A) 1,3,9,27,81,243
(B) 1,2,3,4,6,9,12,18,27,
36,54,81,162,216,324
(C) 1,2,3,5,6,9,10,15,18,
30,45,90
(D) 1,2,4,8,11,22,44,88
(D) 88
Example 7: Find the smallest natural number π‘Ž such that 200 × π‘Ž is a perfect square
Solution:
Step1: Find a prime factorization of 200
200 = 2 × 100 = 22 × 50 = 23 × 25 = 2πŸ‘ × 52
Step 2: If it is a perfect square, all exponents of prime factors are even
Since the exponent of 2 is odd,
200 × 2 = 24 × 52 = (20)2
Step 3: Since 200 × 2 = (20)2 , π‘Ž = 2
Example 8: Find the smallest natural number π‘Ž such that 26 × 5 × π‘Ž is a perfect square
5
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7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1. Find the prime factorization and express it in the exponential notation.
(A) 98
(B) 196
(C) 104
(A) 72 × 2
(B) 72 × 22
(C) 33 × 22
(D) 43 × 5
(D) 215
(E) 135
(F) 204
(E) 33 × 5
(F)
17 × 3 × 22
(G) 32 × 5 × 7
(G) 315
(H) 360
(I) 420
(H) 23 × 32 × 5
(I)
22 × 3 × 5 × 7
(J)
35
(K) 22 × 32 × 11
(J) 243
(K) 792
(L) 252
(L) 22 × 32 × 7
(M) 22 × 32 × 5
(N) 22 × 33 × 53
(M) 12 × 15
(N) 18 × 10 × 75
(O) 54 × 98 × 16
2. Find the prime factorization and express it as a single number in the exponential notation
(A) 2 × 2 × 3 × 3 × 5
(B) 3 × 3 × 3 × 3 × 3
(C) 7 × 7 × 9
(O) 26 × 33 × 72
(A) 22 × 32 × 5
(B) 35
(C) 72 × 5
(D) 28
(D) 8 × 8 × 4
(E) 32 × 36
(F) 24 × 23
(E) 39
(F)
27
(G) 25
(G) 28 ÷ 23
(H) (52 )3
(I) 16 × 23
(H) 56
(I)
3. Find all the factors of the following numbers
(A) 42
(B) 54
27
(A) 1,2,3,6,7,14,21,42
(C) 162
(B) 1,2,3,6,9,18,27,54
(C) 1,2,3,6,9,18,27,
54,81,162
(D) 1,2,4,8,16,32,64,128
(E) 1,3,5,7,9,15,21,35,
(D) 128
(E) 315
(F) 196
45,63,105,315
(F)
1,2,4,7,14,28,49,
98,196
(G) 1,2,3,4,6,8,9,12,16,
(G) 144
(H) 140
(I) 243
18,24,36,72,144
(H) 1,2,4,5,7,10,14,20,
28,35,70,140
(J) 25
(K) 33 × 52
(L) 22 × 3 × 52
(I)
1,3,9,27,81,243
(J)
1,2,4,8,16,32
(K) 1,3,9,27,5,15,45,
135,25,75,225,675
(L) 1,2,4,3,6,12,5,10,20
15,30,60,25,50,100,
75,150,300
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7 GRADE-I: DR. EUNKYUNG YOU
4. Find the exponent of the base 2 in the prime factorization of
1 × 2 × 3 × 4 × 5 × 6 × 7 × 8.
7
5. Find the side length of a square whose area is 36 cm2.
6 cm
6. Find the side of length of a cube whose volume is 64 cm2 .
4 cm
7. Find three common factors of the following numbers
2 × 33 × 52 × 7 and 3 × 52 × 73 × 11
8. Find the smallest natural number π‘Ž if 28 × π‘Ž is a perfect square.
9. Find the smallest natural number 𝑏 if
360
𝑏
is a perfect square.
3, 5, 7, 15,21, β‹―
π‘Ž=7
𝑏 = 10
10. Find a natural number π‘₯ such that 60 × π‘₯ become a perfect square.
π‘₯ = 15
11. Find the smallest natural number 𝑐 if 27 × π‘ is a perfect square and a multiple of 4
𝑐 = 12
12. Find the number of all natural numbers 𝑛 if
180
𝑛
is a natural number.
13. Find the smallest natural number π‘Ž if 23 × π‘Ž has exactly 10 factors
14. Find the natural number π‘₯ if the number of factors of 8 × 32 × 5π‘₯ is 36
15. When 42 is divisible by a natural number π‘₯, find all possible π‘₯.
16. Find the constants π‘Ž and 𝑏 if 2π‘Ž = 64 and 33 = 𝑏
18
6
π‘₯=2
1,2,3,6,7,14,21,42
π‘Ž = 6, 𝑏 = 27
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7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 1-4. GREATEST COMMON DIVISOR/ LEAST COMMON MULTIPLE
DEFINITION: The greatest common factor (GCF) of
DEFINITION: The least common multiple (LCM) of
two or more non-zero integers is the largest positive
two integers π‘Ž and 𝑏, usually denoted by LCM(π‘Ž, 𝑏), is the
integer that divides the numbers without a remainder.
smallest positive integer that is divisible by both π‘Ž and 𝑏.
If either π‘Ž or 𝑏 is 0, LCM(π‘Ž, 𝑏) is defined to be zero.
Factors of 16 : 1, 2, πŸ’ , 8, 16
Multiples of 2 : 2, 4, πŸ” , 8, 10, 12 , 14, β‹―
Factors of 20 : 1, 2, πŸ’ , 5, 10, 20
Multiples of 3 : 3, πŸ” , 9, 12 , 15, β‹―
Common Factors of 16, 20
GCF of 16 and 20
Common Multiples of 2, 3
LCM of 2 and 3
1, 2, 4
4
6, 12, β‹―
6
(Multiples of LCM)
(factors of GCF)
DEFINITION: two integers a and b are said to be
REMARK:
relatively prime, or coprime (also spelled co-prime) if the
only common positive factor of the two numbers is 1.
LCM of coprime numbers is the product of two numbers.
Factors of 4 : 1 , 2, 4
LCM (4,9) = 36
Factors of 9 : 1 , 3, 9
4 and 9 are coprime.
Example 1: Find the following.
(A) Factors of 35
(C) Find common Factors of 35 and 42
Example 2: Find the following.
(A) Factors of 18
(C) Find common Factors of 18 and 45
(B) Factors of 42
(A)
(B)
(C)
(D)
1,5,7,35
1,2,3,6,7,14,21,41
1,7
7
(A)
(B)
(C)
(D)
1,2,3,6,9,18
1,3,5,9,15,45
1,3,9
9
(D) Find GCF of 35 and 42
(B) Factors of 45
(D) Find GCF of 18 and 45
14
7 GRADE-I: DR. EUNKYUNG YOU
Example 3: Decide whether the following numbers are coprime or not.
(A) 2,11
(B) 3,15
(C) 12, 27
(A)
(B)
(C)
(D)
Coprime
Not
Not
Coprime
(A)
(B)
(C)
(D)
6,12,18,…
9,18,27,..
18,36,54,…
18
(A)
(B)
(C)
(D)
12,24,36,48,60,…
15,30,45,60,75,…
60,120,…
60
(D) 16, 27
Example 4: Find the following.
(A) Multiples of 6
(B) Multiples of 9
(C) Find common Multiples of 6 and 9
Example 5: Find the following.
(A) Multiples of 12
(D) Find LCM of 6 and 9
(B) Multiples of 15
(C) Find common Multiples of 12, 15
(D) Find LCM of 12 and 15
Example 6: List all the factors common to 16 and 28
Solution:
Factors of 16 : 1, 2, 4, 8, 16
Factors of 28 : 1, 2, 4, 7, 14, 28
Therefore, the greatest common factor 4: list of common factors are 1, 2, 4
Example 7: List all the factors common to
(A) 12 and 30
(B) 24 and 32.
(C) 56 and 84
(A) 1,2,3,6
(B) 1,2,4,8
(C) 1,2,4,7,14,28
15
7 GRADE-I: DR. EUNKYUNG YOU
GCF OF TWO NUMBERS
54 = 𝟐
GCF OF TWO NUMBERS
2 ) 54
72
×πŸ‘×πŸ‘×3
72 = 𝟐 × 2 × 2 × πŸ‘ × πŸ‘
GCF = 𝟐 ×
×πŸ‘×πŸ‘
= 18
36
3 ) 9
12
3
4
GCF(54,72) = 2 × 3 × 3 = 18
GCF OF THREE NUMBERS
GCF OF THREE NUMBERS
54 = 𝟐
3 ) 27
×πŸ‘×πŸ‘×3
72 = 𝟐 × 2 × 2 × πŸ‘ × πŸ‘
90 = 𝟐 ×
×πŸ‘×πŸ‘
GCF = 𝟐 ×
×πŸ‘×πŸ‘
×5
= 18
2 ) 54
72
90
3 ) 27
36
45
3 ) 9
12
15
3
4
5
GCF(54,72,90) = 2 × 3 × 3 = 18
FIND THE GCF USING EXPONENT
54 = 21 × 33
90 = 21 × 32 × 51
GCF = 𝟐𝟏 × πŸ‘πŸ :
common primes with smallest exponent
Example 8: Find the GCF of the following numbers.
(A) 16,40
(B) 18, 24
(A) 8
(B) 6
(C) 4
(D) 27
(E) 12
(C) 12, 40
(D) 108, 135
(F)
4
(G) 18
(H) 12
(E) 36, 60, 72
(F) 20, 24, 32
(G) 54, 72, 90
(H) 24, 96, 132
16
7 GRADE-I: DR. EUNKYUNG YOU
Example 9: Find the GCF of the following numbers.
(A) 23 × 3, 2 × 32
(B) 22 × 53 , 24 × 5
(A) 6
(B) 20
(C) 15
(D) 75
(E) 150
(C) 34 × 5, 2 × 3 × 52
(D) 33 × 52 , 3 × 54 × 72
(F)
392
(G) 2
(H) 70
(E) 23 × 3 × 54 , 2 × 35 × 52
(F) 23 × 5 × 74 , 23 × 3 × 72
(G) 23 × 5 × 72 , 24 × 53 , 2 × 73
(H) 22 × 5 × 7,2 × 52 × 7, 3 × 52 × 73
Example 10: Tom is covering a bottom of his bathroom with identical square tiles (which cannot be cut into smaller
pieces). The area he want to cover is 20 ft by 24 ft. Find the largest possible length of a side of the tile.
Solution: Since the number of square tiles is an integer, the length of a side of the tile is an integer which is factor to 20
and 24. That is, it is a common factor of 20 and 24. Therefore the largest possible length of a side of the tile is GCF.
20 = 22 × 5
24 = 23 × 3
The GCF of 20 and 24 is 4.
The largest possible length of a side of the tile is 4
Example 11: Tom cut a 35 cm by 28 cm rectangle paper to obtain identical squares. Find the
7 cm
largest possible length of a side of each square.
Example 12: There are 18 apples and 24 pears. We want to distribute those to students equally.
Find the maximum number of students.
6 students
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7 GRADE-I: DR. EUNKYUNG YOU
LCM OF TWO NUMBERS
54 = 𝟐
LCM OF TWO NUMBERS
2 ) 54
72
×πŸ‘×πŸ‘×πŸ‘
72 = 𝟐 × πŸ × πŸ × πŸ‘ × πŸ‘
LCM = 𝟐 × πŸ × πŸ × πŸ‘ × πŸ‘ × πŸ‘
= 216
3 ) 27
36
3 ) 9
12
3
4
LCM (54,72) = 2 × 3 × 3 × 4 = 216
LCM OF THREE NUMBERS
54 = 𝟐
LCM OF THREE NUMBERS
×πŸ‘×πŸ‘×πŸ‘
72 = 𝟐 × πŸ × πŸ × πŸ‘ × πŸ‘
90 = 𝟐 ×
×πŸ‘×πŸ‘× ×πŸ“
LCM= 𝟐 × πŸ × πŸ × πŸ‘ × πŸ‘ × πŸ‘ × πŸ“
= 1080
2 ) 54
72
90
3 ) 27
36
45
3 ) 9
12
15
3
4
5
LCM (54,72,90) = 2 × 3 × 3 × 3 × 4 × 5 = 1080
FIND THE LCM USING EXPONENT
54 = 21 × 33
90 = 21 × 32 × 51
GCF = 𝟐𝟏 × πŸ‘πŸ‘ × 51 :
all prime numbers with greatest exponent
Example 13: Find the GCF of the following numbers.
(A) 10, 40
(B) 12,45
(A) 40
(B) 180
(C) 180
(D) 450
(C) 9,12, 15
(D) 45, 50, 75
Example 14: Find the GCF of the following numbers.
(A) 23 × 3, 22 × 3 × 52
(B) 22 × 33 × 5, 2 × 34 × 7
(A) 600 = 23 × 3 × 52
(B) 11340 = 22 × 34 × 5 ×
7
(C) 11340 = 23 × 34 × 5 ×
(C) 23 × 3 × 5, 2 × 34 , 22 × 33 × 7
(D) 22 × 32 × 5, 2 × 33 × 52 , 23 × 3 × 7
7
(D) 37800 = 23 × 33 ×
52 × 7
18
7 GRADE-I: DR. EUNKYUNG YOU
RELATION BETWEEN GCF AND LCM
2 ) 6
8
3
4
6 = 2 × 3, 8 = 2 × 4
(2 × 3 × 4)
⇒6×8= ⏟
2 ×⏟
GCF(6,8) = 2
𝐆𝐂𝐅
LCM (6,8) = 2 × 3 × 4 = 24
π‹π‚πŒ
Example 15: The GCF and LCM of two integers are 12 and 180, respectively. Find the all possible two integers.
Solution: Two numbers are 12 × π‘₯ and 12 × π‘¦ where π‘₯, 𝑦 are relatively prime.
12 × π‘₯ × 12 × π‘¦ = 12 × 180 ⇒ π‘₯𝑦 = 15
So π‘₯ = 1, 𝑦 = 15 or π‘₯ = 3, 𝑦 = 5.
Therefore the two numbers are 12, 180 or 36, 60
Example 14: The GCF of a natural number 𝐴 and 15 is 3 and the LCM of these is 60. Find the
12
number A.
Example 15: The GCD of two numbers is 24 and the LCM of these is 360. Find all possible
24,360 or 72,120
two numbers.
Example 16: Tom has 60 apples and 108 tomatoes. He wants to distribute to students equally.
Find the maximum number of students.
12 students
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7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1. Find the GCF and LCM of the following numbers
(A) 12, 30
(B) 24, 32
(A)
(B)
(C)
(D)
(E)
(F)
(C) 18, 27
(D) 21, 84
(E) 25, 75
(F) 108, 135
(G)
(G) 72, 84, 108
(H) 48, 72, 120
(I) 28, 63, 91
(J) 60, 75, 200
(K) 48, 84, 144
(L) 140, 210, 350
(H)
(I)
(J)
(K)
(L)
2. Find the GCF and LCM of the following numbers
(A) 22 × 3, 2 × 33
(B) 2 × 3,
(C) 23 × 3 × 52 , 2 × 32 × 5
(A)
2
2 ×3×5
(D) 22 × 32 , 23 × 3 × 7
(E) 2 × 32 × 53 , 22 × 3 × 52 , 24 × 3 × 53
(B)
(C)
(D)
(E)
2
2
2
3
2
(F) 2 × 3 , 2 × 3 × 7, 2 × 3 × 7
2
3. Find two different pair numbers such that their GCF is 20.
(F)
GCF: 6 LCM: 60
GCF:8 LCM: 96
GCF:9 LCM: 54
GCF:21 LCM:84
GCF:25 LCM:75
GCF:27
LCM:540
GCF:12 LCM:
1512
GCF:24
LCM:720
GCF:7
LCM:3276
GCF: 5
LCM:3000
GCF:12
LCM:1008
GCF: 70
LCM:2100
GCF: 6 LCM:
22 × 33
GCF: 6 LCM:
22 × 3 × 5
GCF: 30 LCM:
23 × 32 × 52
GCF: 12 LCM:
23 × 32 × 7
GCF: 150 LCM:
24 × 32 × 53
GCF: 12 LCM:
23 × 32 × 72
20,40 or 40, 60
4. There are two wood bars of lengths 72 in and 96 in. Short bars of equal length are cut from
both bars. Find the largest possible length of each short bar.
24 in
5. There are 120 candy and 48 cupcake. We want to distribute those to students equally. Find
the maximum number of students.
24 students
6. There are 42 bananas, 36 apples, and 18 oranges. We want to distribute those to students
equally without left over. Find the maximum number of students.
6 students
7. Say you have 60 pencils, 90 pens and 120 tablets and you want to make packages of
pencils, pens and tablets to donate to your school for students who cannot afford these
supplies. What is the maximum number of packages you can make using all items, and
how many pencils, pens and tablets will be in each package?
30 packages which consist of
2 pencils, 3 pen, and 4 tablets
8. A florist has 36 roses, 27 tulips, and 18 carnations she must use to create bouquets. What is
the largest number of bouquets she can make without having any flowers left over
9
9. The prices of apple, orange, and banana are $1, $0.8, and $0.5 respectively. There are 24
apples, 30 oranges, and 48 bananas. If we want to make same size basket which consists of
all three, what is the maximum number of baskets we can make using all fruits? What is
the price of a basket?
6 baskets : $12
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7 GRADE-I: DR. EUNKYUNG YOU
10. In a department store, the price (in dollars) of a toy car is a whole number. The sales of this
toy cars on two days are $1,518 and $2,346. Find all possible numbers of cars which is
sold on each day.
2,3,6,23,46,69,138
11. You want to make two garden plots next to each other with a fence completely around each
one. One plot is 180 square feet and the other is 204 square feet. If the fence comes in 1
foot lengths, what is the greatest length of the fence you can make that is shared by both
garden plots?
12 ft
12. Identical square tiles are arranged to form a rectangular wall whose dimension is 120 cm ×
150 cm.
(A) Find the length of one side of the square tiles (length must be an integer).
(B) Find the number of tiles.
13. Using rectangular tiles whose dimension is 9 in × 15 in, we want to make a square
bathroom which has smallest area.
(A) Find the dimension of the bathroom.
(B) How many rectangular tiles do we need to make the bathroom?
1 cm : 18,000
2 cm : 4,500
3 cm : 2000
5 cm: 720
6 cm : 500
10 cm : 180
15 cm :80
30 cm : 15
(A) 45 in by 45 in
(B) 15 tiles
14. The circumference of the front wheel and the rear wheel of a bicycle are 80 cm and 60 cm
respectively. When Tom begins to ride the bicycle, 𝑃 and 𝑄 are the points of the front
wheel and the rear wheel respectively which touch the ground. Find the distance traveled
and the numbers of revolutions when P and Q next touch the ground at the same time.
240 cm :
Front- 3 revolution
Rear: 4 revolution
15. A flashing red bulb on a tree flashes once every 8 seconds. A flashing blue bulb flashes
once every 12 seconds. If they are flashing together now, how long will it take for the two
bulbs to next flash together?
24 seconds
16. The diagram shows a gear system in which the numbers of teeth on the big and small
wheels are 24 and 30 respectively. The two wheels are engaged at the start.
(A) Find the number of tooth contacts before two wheels are next engaged.
(B) Find the number of revolutions that each wheel will have made by then.
120
Big: 4 revolutions
Small: 5
revolutions
17. Find the two numbers such that its great common divisor is 3 and its least common
multiple is 60.
3,60 or 12,15
18. Find the two numbers such that its product is 300 and its least common multiple is 60.
5, 60 or 15, 20
19. Find LCM of two natural numbers such that its product is 80 and their GCF is 2.
40
20. Find GCF of two natural numbers such that its product is 576 and their LCM is 48.
12
21. Find two natural numbers such that their GCF is 9 and their LCM is 270.
22. The GCF of a number A and 23 × 3 × 72 is 22 × 3 and LCM of these is 23 × 32 × 5 × 72 .
Find the natural number A.
23. Find the smallest value of
70
𝑛
+
28
𝑛
where 𝑛,
70 28
,
𝑛 𝑛
are integers.
9, 270 or 18,135
or27,90 or 45,54
22 × 32 × 5
7
21
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 1-5. SQUARE ROOTS AND CUBIC ROOTS
SQUARE ROOTS :
CUBIC ROOTS
The (positive) square root of a number is a value that,
The cubic root of a number is a value that, when
when multiplied by itself, gives the number. Its symbol is
multiplied by itself, gives the number. Its symbol looks
called a radical and looks like this : √
like this : √
3
3
32 = 9 ∢ 3 is a square root of 9 or √9 = 3
23 = 8 ∢ 2 is a cubic root of 8 or √8 = 2
Example 1: Evaluate the number by prime factorization
(A) √49
(B) √144
(C) √26 × 32
Solution:
(A) Since 49 = 72 , √49 = 7
(B) Since 144 = 24 × 32 = (22 × 3)2 , √144 = 22 × 3 = 12
(C) Since 26 × 32 = (23 × 3)2 , √26 × 32 = 23 × 3 = 24
3
(D) Since 27 = 33 , √27 = 3
Example 2: Evaluate the number by prime factorization
(A) √16
(B) √36
(C) √225
(D) √24 × 36
(E) √64
3
(F) √216
3
Example 3: Find the length of one side of a square whose area is 1296 ft 2
Example 4: Find the length of one side of a cube whose volume is 1728 ft 3
3
(D) √27
22
7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1. Evaluate the number by prime factorization
(A) √64
(B) √9
(C) √0
(D) √1
(E) √196
(F) √324
(G) √225
(H) √400
(I) √576
(J) √441
(K) √28 × 32
(L) √52 × 34
2. Evaluate the number by prime factorization
3
(B) √125
3
(C) √216
3
(E) √512
3
(F) √729
(A) √1
(D) √1000
3
3
23
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 2.NUMBER SYSTEMS
CHAPTER 2.1 NATURAL NUMBERS AND RATIONAL NUMBERS
NEGATIVE NUMBERS:
INTEGERS
When we assign a certain meaning of a quantity to be
Integers = {
positive, a value of the quantity that has the opposite
Positive integers: 1, 2, 3, β‹―
0
negative integers: −1, −2, −3, β‹―
meaning is considered as a negative.
+ Profit
−
Debt
Deposit
After
withdrawal Before
integers
REAL LINE: The number line is used to represent
Rational numbers =
integers. This is shown below.
terminal decimals: 0.9, 0.86, β‹―
{repeating decimals: 0.22222 β‹― , β‹―
Numbers are arranged in ascending order from left to
right.
Example 1: Write an integer to represent the following situation
(A) Earnings of 15 dollars
(B) A loss of 20 yards
Solution:
(A) +15 dollars
(B) −20 yards
Example 2: Write an integer to represent the following situation
(A) 10 degrees above zero
(B) a loss of 16 dollars
(A) +10
(B) −16
(C) a gain of 5 points
(D) 8 steps backward
(C) +5
(D) −8
Example 3: What is the opposite number of the following?
(A) 231
Solution:
(A) −231
(B) −1056
(B) 1056
Example 4: The highest elevation in North America is Mt. McKinley, which is 20,320 feet
above sea level. The lowest elevation is Death Valley, which is 282 feet below sea level.
Represent these altitude using positive and negative numbers.
+20,320 ft, −282 ft
24
7 GRADE-I: DR. EUNKYUNG YOU
Example 5: Consider the list of the number
1
− ,
2
+4,
2
+ ,
3
0,
−3.2,
10,
−7,
+6.1
(A) Find all positive integers
(B) Find all rational numbers which are not integers
(C) Arrange the given numbers in ascending order
Solution:
(A) +4, 10
1
2
(B) − 2 , + 3 , −3.2, +6.1
1
2
(C) −7, −3.2, − , 0,
2
,
3
4, +6.1, 10
Example 6: Consider the list of the number
1
+3,
− ,
+0.19,
2
(A) Find all positive integers
6
+ ,
3
3
,
5
0,
(A) +3, +
4
− ,
2
1
6
3
3
(B) − , +0.19, , +4.9
+4.9
2
4
1
2
2
5
3
(C) − , − , 0, +0.19, ,
5
6
+ , +3, +4.9
3
(B) Find all rational numbers which are not integers
(C) Arrange the given numbers in ascending order
𝐴 = −2.5, 𝐡 = −1, 𝐢 = 3.5
Example 7: Using the following real line, find the numbers A,B, and C in the figure.
A
4
3
B
2
C
1
3
2
1
0
4
Example 8: Represent the following numbers on a number line.
(A) −3
(B) 1.5
Solution:
1.5
5
4
3
2
1
0
1
2
3
4
5
Example 9: Represent the following numbers on a number line.
(A) 3
(B) −2.5
5
4
3
(C) −4
2
1
0
1
2
(D)
3
4
5
14
3
25
7 GRADE-I: DR. EUNKYUNG YOU
ABSOLUTE VALUE: The absolute value of a number is
the distance that number is from 0 on the number line.
−π‘₯,
|π‘₯| = {
π‘₯,
π‘₯<0
π‘₯≥0
|πŸ“| = πŸ“
INEQUALITY:
𝒙>𝒂
𝒙<𝒂
π‘₯ is bigger than π‘Ž
π‘₯ is smaller than π‘Ž
𝒙≥𝒂
𝒙≤𝒂
π‘₯ is bigger or equal than π‘Ž
π‘₯ is smaller or equal than π‘Ž
|−πŸ“| = πŸ“
Example 10: Fill in each
(A) +1
0
(D) −1.7
with " < " or " > ".
(B) −12
2.7
(E)
2
3
0
−
(C) +2
5
13
(F) −1.2
−3
4
3
(A)
(B)
(C)
(D)
(E)
(F)
>
<
>
<
>
<
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
2
9
4.1
2.3
−10
4
−5
4.5
−6
(A)
(B)
(C)
(D)
(E)
(F)
>
<
=
>
>
<
Example 11: Evaluate
(A) |3|
(B) | − 7|
(C) −| − 6|
Solution:
(A) |3| = 3
(B) |−7| = −(−7) = 7
(C) −|−6| = −(6) = −6 since |−6| = −(−6) = 6
Example 12: Evaluate
(A) |2|
(B) | − 9|
(C) |4.1|
(D) | − 2.3|
(E) −| − 10|
(F) |4|
(G) −| − 5|
(H) | − 4.5|
(I) −|6|
Example 13: Fill in each
(A) |4|
(D) |−4|
| − 2|
− | − 3|
with <, " = " or " > ".
(B) −| − 5|
(E) | − 9|
0
−9
(C) | − 23|
(F) −| − 10|
|23|
10
26
7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1. Write an integer to represent each quantity
(A) +320
(A) A profit of $320
(B) +13
(C) −18
(B) gaining 13 pound weight
(D) −8
(E) −3250
(C) 18℉ below zero
(D) A loss of 8 feet
(E) A debt of $3250
2. Find the number represented by the points A, B and C on the real line.
A
6
5
B
3
4
2
1
C
0
1
2
4
3
5
6
with <, " = " or " > ".
3. Fill in each
(A) <
(A) 5
13
(B) −5
2
(D) −7
0
(F) 23
− 29
(B) <
(C) >
(C) 14
−8
(D) <
(E) <
(E) −15
−6
(F)
>
4. Simplify
(A) |23|
(B) |−16|
(C) −|−18|
(D) −|5|
(A) 23
(B) 16
(C) −18
(D) −5
(E) |−52|
(F) −|−19|
(E) 52
(F)
5. Fill in each
−19
with <, " = " or " > ".
(A) <
(A) |4|
|5|
(B) |−7|
|9|
(B) <
(C) =
(C) |−23|
|23|
(D) |−5|
− |5|
(E) |−2|
− |−3|
(F) −|9|
|−9|
(D) >
(E) >
1
2
(F)
<
6. Given the numbers 5, −3 , 4, and 0.9,
(A) represent the numbers on the real line
(B) arrange the numbers in ascending order
(A)
1
(B) −3 , 0.9, 4, 5
2
27
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 2.2 ADDITION IN INTEGERS
For any π‘Ž > 0 and 𝑏 > 0
ADDITION OF TWO POSITIVE (OR NEGATIVE)
π‘Ž+𝑏 =π‘Ž+𝑏
(+4) + (+3) = +(4 + 3)
(−π‘Ž) + (−𝑏) = −(π‘Ž + 𝑏)
(−5) + (−3) = −(5 + 3)
π‘Ž + (−𝑏) = +(π‘Ž − 𝑏) if π‘Ž ≥ 𝑏
ADDITION OF ONE POSITIVE AND ONE NEGATIVE
π‘Ž + (−𝑏) = −(𝑏 − π‘Ž) if π‘Ž < 𝑏
(−4) + (+3) = −(4 − 3)
−π‘Ž + 𝑏 = −(π‘Ž − 𝑏) if π‘Ž ≥ 𝑏
(−5) + (+8) = +(8 − 5)
−π‘Ž + 𝑏 = +(𝑏 − π‘Ž) if π‘Ž < 𝑏
Example 1: Use the real line to evaluate the following
(A) (−2) + 5
(C) (−1) + (−3)
(B) 2 + 3
(D) 4 + (−3)
Solution:
(−2) + 5 = 3
2+3=5
+5
5
4
2
3
1
0
1
2
3
+3
+2
Final
4
5
5
4
3
2
1
0
1
3
2
4
5
Final
2
(−1) + (−3) = −4
3
5
4
Final
3
4 + (−3) = 1
+4
1
2
1
0
1
2
3
4
5
5
4
3
2
1
0
1
Final
3
2
4
5
3
Example 2: Use the real line to evaluate the following
(A) 12 + (−7)
(B) (−13) + 8
(A) 5
(B) −5
(C) −13
(D) 2
(C) (−7) + (−6)
(D) (−9) + 11
(E) 10
(F)
(E) 14 + (−4)
(F) (−8) + (−6)
−14
28
7 GRADE-I: DR. EUNKYUNG YOU
Example 3: Evaluate the sum
(A) (−7) + 9
(B) (−8) + (−2)
(A) 2
(B) −10
(C) 1
(D) 4
(C) (−4) + 5
(D) (−8) + 12
(E) −13
(F)
9
(G) 34
(E) (−8) + (−5)
(F) 11 + (−2)
(H) −17
(I)
18
(J)
−13
(K) −79
(G) 49 + (−15)
(H) (−47) + 30
(L) −35
(M) −27
(N) 98
(I) 35 + (−27)
(J) (−27) + 14
(K) −46 + (−33)
(L) (−35) + 0
(M) (−56) + 29
(N) (34) + 64
(O) (−12) + (−33)
(P) 24 + (−30)
(O) −45
(P)
−6
Example 4: Evaluate the sum
(A) (−2) + 11 + 4
(B) 12 + 7 + (−4)
(A) 13
(B) 15
(C) 13
(D) −13
(E) −28
(C) 7 + (−3) + 9
(D) (−11) + 3 + (−5)
(F)
−22
(G) −18
(H) 15
(E) 13 + (−7) + (−34)
(F) (−6) + (−9) + (−7)
(G) −25 + (−8) + 15
(H) −27 + 10 + 32
29
7 GRADE-I: DR. EUNKYUNG YOU
Example 5: Evaluate
(A) 51
(B) |6 + (−9)|
(A) |34 + 17|
(B) 3
(C) −3
(D) 17
(D) |−12 + (−5)|
(C) −|(−5) + 8|
Example 6: Evaluate
(B) |−6| + (−14)
(A) 8 + | − 11|
(A) 19
(B) −8
(C) −37
(D) −18
(C) −|25| + (−12)
(D) −|−13| + (−5)
Example 7: Find the missing number in the equation
(A)
+7=5
(B) 13 +
=7
(A) −2
(B) −6
(C) −15
(D) −14
(C) 6 +
= −9
(D)
+ (−13) = −27
ADDITIVE INVERSE:
−π‘Ž is the additive inverse of π‘Ž if π‘Ž + (−π‘Ž) = 0
 −3 is the additive inverse of 3 and
 3 is the additive inverse of −3
(A) 14
Example 8: Find the additive inverse of the following.
(A) −14
(B) 8
(C) 125
(B) −8
(C) −125
30
7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1. Evaluate
(A) (−2) + (3)
(B) (−15) + (−7)
(C) (−9) + (14)
(D) (5) + (−8)
(E) (−33) + (−15)
(F) (7) + (−1) + 12
(G) (−6) + (−8)
(H) (−4) + (+9)
(I) (+4) + (−5) + (−7)
(J) (−12) + (−4) + (+7)
(K) 0 + (−9)
(L) −46 + (−54) + (+63)
(M) −29 + 45 + (−37)
(N) (−36) + (−21) + (18)
(O) 13 + (−17) + (−38)
(P) (−16) + (−6) + (+22)
(Q) 1 + (−5) + 4 + (−9)
(R) −35 + (60) + (−85) + (120)
2. Evaluate
(A) |37 + (−19)|
(B) |6 + (−14)|
(C) | − 5 + 19|
(D) −|6 − 12|
(E) | − 12 − (−5)|
(F) 7 + | − 11|
(G) |−2| + (−16)
(H) −|−25| + |25|
(I) −|−14| + (−4)
(J) −|−8| + | − 8|
3. Find the missing number in the equation
(A)
+ (−3) = 15
(C) −2 +
= 17
(B)
+ 5 = 18
(D) (−3) +
= −5
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
(J)
(K)
(L)
(M)
(N)
(O)
(P)
1
−22
5
−3
−48
18
−14
5
−8
−9
−9
−37
−21
−39
−42
0
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
(J)
18
8
14
−6
7
18
−14
0
−18
0
(A)
(B)
(C)
(D)
18
13
19
−2
31
7 GRADE-I: DR. EUNKYUNG YOU
4. Fill in each
with <, " = " or " > ".
(A) (−6) + 1
5+1
(C) 49 + (−15)
| − 28|
(B) 10 + (−2)
(−4) + 12
(D) |−16| + (−7)
5. Write a numerical expression for the following and find the sum
(A) Tom save $537 and spend $350
(A)
(B)
(C)
(D)
<
=
>
>
|25 + (−19)|
(A) $187
(B) $12,500
(C) −1800 ft
(B) Kim deposited $5,000 in her bank account and withdraws $3,750
(C) A submarine at 3,500 ft below sea level moves up 1700 ft
6. Find the additive inverse of each following numbers
(A) −167
(B) −19
(C) 234
(D) 435
(A)
(B)
(C)
(D)
167
19
−234
−435
32
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 2.3 SUBTRACTION
SUBTRACTION: To subtract numbers, we changed the
(NO SIGN CASE)
sign of the number being subtracted and add them.
7−9−3
(−3) − (+2)
= (+πŸ•) + (−πŸ—) + (−3)
= (−3) + (−2)
= −(πŸ— − πŸ•) + (−3)
= −(3 + 2)
= −2 + (−3)
= −5
= (−2) + (−3)
= −(2 + 3)
(−3) − (−5)
= −5
= (−3) + (+5)
= +(5 − 3)
= +2
Example 1: Evaluate
(A) 3 − 4
(B) (−3) − (+5)
(A) −1
(B) −8
(C) 8
(D) −7
(E) 7
(C) 5 − (−3)
(D) −4 − 3
(F)
−8
(G) −8
(H) −7
(E) 0 − (−7)
(F) 23 − 31
(I)
15
(J)
33
(K) 31
(L) 24
(G) −15 − (−7)
(H) −12 − (−5)
(I) 10 − (−5)
(J) 15 − (−18)
(K) −33 − (−64)
(L) 6 − (−18)
33
7 GRADE-I: DR. EUNKYUNG YOU
Example 2: Evaluate
(A) 2 − 15 + 4
(B) 16 − 13 − (−5)
(A) −9
(B) 8
(C) −1
(D) 21
(E) −12
(F)
−1
(G) −3
(H) −36
(C) −2 − (−9) − 8
(D) 10 + 3 − (−8)
(I)
−2
(J)
−10
(K) −11
(L) 23
(E) 4 − 3 − 13
(F) 6 − (11 − 4)
(G) 4 − (−5) − 12
(H) −(12 − (−6)) − 18
(I) −17 − (−6) + 9
(J) 7 − 9 − 8
(K) 11 − 5 − 17
(L) 17 − (7 − 13)
34
7 GRADE-I: DR. EUNKYUNG YOU
Example 3: Evaluate
(A) |14 − 7|
(B) |5 − (−6)|
(A) 7
(B) 11
(C) 19
(D) 15
(E) 4
(F)
11
(G) −2
(H) −20
(C) |6 − 25|
(D) | − 6 − 9|
(I)
38
(J)
21
(K) −8
(L) −22
(E) | − 7 − (−11)|
(F) | − 13 − (−24)|
(G) |−5| − 7
(H) −|−9| − |11|
(I) 6 − | − 32|
(J) 12 − (−|−9|)
(K) 5 − |−(−13)|
(L) −|−14| − |8|
35
7 GRADE-I: DR. EUNKYUNG YOU
Example 4: Find the missing number in the equation
(A) 16 −
=7
(B) 7 −
(A) 9
= −9
(B) 16
(C) −25
(D) 14
(C) −8 −
= 17
Example 5: Fill in each
(A) |5 − 8|
(D) −13 −
= −27
with <, " = " or " > ".
|8 − 5|
(B) |8 − 13|
8 − 13
(A) =
(B) >
(C) >
(D) >
(C) −(−4)
− | − 4|
(D) −11 − (−5)
− 11 − | − 5|
Example 6: Tom and Mary droves from same station. Tom droves 40 miles due east and Mary
59 miles
droves 19 miles due west. Find the distance between them.
Example 7: Kim borrows $187 from his mother. Then he pay back $98. Write a numerical
expression of this and find the sum
−$187 + $98 = −$89
36
7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
(A) 3
1. Evaluate
(A) (+5) − (+2)
(B) (−7) − (−9)
(B) 2
(C) 12
(D) −9
(C) (+2) − (−10)
(D) (−6) − (+3)
(E) −8
(F)
(E) −5 − 3
(G) −35 − (−45)
(F) −6 − (−18)
(H) 8 − (−27)
12
(G) 10
(H) 35
(I)
11
(J)
−4
(K) −51
(L) 61
(I) (3) − (−8)
(J) (−8) − (−4)
(M) −27
(N) 19
(K) (−27) − (24)
(L) (38) − (−23)
(O) −39
(P)
−62
(Q) −75
(M) (−8) − (19)
(N) (−16) − (−35)
(O) −58 − (−19)
(P) −79 − (−17)
(Q) −49 − 26
(R) −54 − 29
(A) −3
2. Evaluate
(A) (+4) − (−1) − (+8)
(R) −83
(B) (+5) + (−12) − (−2)
(B) −5
(C) −27
(D) −14
(C) (−16) − 6 + (−5)
(D) (−7) − (−2) − 9
(E) 10
(F)
−4
(G) 6
(E) (−11) − (−14) + 7
(F) 6 + (−7) + (−5) − (−2)
(H) −3
(I)
−9
(J)
−1
(K) −13
(G) 3 − 2 + 5
(H) −2 + 5 − 6
(L) 3
(M) 7
(N) −4
(I) −7 + 3 − 5
(J) 6 − (10 − 3)
(K) −15 − (−7) − 5
(L) −23 − (−17) + 9
(M) 5 − (6 − 8)
(N) 3 − [5 − {2 − (8 − 4)}]
37
7 GRADE-I: DR. EUNKYUNG YOU
(A) 7
3. Evaluate
(B) 47
(B) |34 − (−13)|
(A) |5 − 12|
(C) 22
(D) 4
(D) | − 12 − (−8)|
(C) | − 6 − 14|
(E) −3
(F)
(E) |−9| − 12
(G) −7
(F) 35 − | − 12|
(G) −12 − | − 5|
(H) −|−5| + |4|
(I) |−2| − | − 9|
(J) 4 − | − 3|2
(H) −1
(I)
−7
(J)
−5
(A) =
with <, " = " or " > ".
4. Fill in each
23
(B) >
(A) |11 − 8|
(B) |5 − 9|
|8 − 11|
5−9
(C) =
(D) >
(C) −(−7)
(D) −13 − (−8)
− | − 7|
− 13 − | − 8|
5. Find the missing number in the equation
(A) 7 −
(B) (−5) −
= 11
= −17
(A) −4
(B) 12
(C) 11
(C)
− (−4) = 15
(D)
− (−3) = −23
(D) −26
−50
6. Evaluate
1 − 2 + 3 − 4 + 5 − 6 + β‹― + 99 − 100
7. In a square, each row, column, and diagonal has the same sum. Find the constants π‘Ž, 𝑏
π‘Ž
π‘Ž = 5, 𝑏 = −3
−2
1
4
𝑏
2
8. In a square, each row, column, and diagonal has the same sum. Find the constants π‘Ž, 𝑏
2
π‘Ž = 1, 𝑏 = −1
π‘Ž
0
𝑏
−2
3
9. The temperature of a 7000 ft mountain at the base is 80℉. It decreasing by 4℉ every 1000
ft. What is the temperature at the top of the mountain?
52℉
38
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 2-4 MULTIPLICATION AND DIVISION
MULTIPLICATION OF SAME SIGN NUMBERS
DISTRIBUTION:
If two numbers has the same sign, their product is positive
𝒂 × (𝑏 + 𝑐) = (𝒂 × π‘) + (𝒂 × π‘)
(+5) × (+3) = +(5 × 3) = +15
(𝑏 + 𝑐) × π’‚ = (𝑏 × π’‚) + (𝑐 × π’‚)
(−5) × (−3) = +(5 × 3) = +15
MULTIPLICATION OF DIFFERENT SIGN NUMBERS
EXPONENTIAL EXPRESSION
If two numbers has the opposite sign, their product is
(−3)πŸ’ = +81: even exponent case
negative
(−2)πŸ‘ = −8: odd exponent case
(+5) × (−3) = −(5 × 3) = −15
Example 1: Evaluate the following
(A) (−6) × 5
(B) (−7) × (−8)
Solution:
(A) (−6) × 5 = (−6) × (+5) = −(6 × 5) = −30
(C) 2 × (−3) × (−4)
(B) (−7) × (−8) = +(7 × 8) = +56
(C) 2 × (−3) × (−4) = (−(2 × 3)) × (−4) = (−6) × (−4) = +(6 × 4) = +24
Example 2: Evaluate the following
(A) (+3) × (+6)
(B) (−5) × (−8)
(C) (−9) × (2)
(D) (−3) × (−4)
(E) 5 × (−2) × (−4)
(F) (−4) × (−3) × (−6)
(G) (−12) × 2 × (−7)
(H) (−9) × (−8) × (0)
(I) (−2)5
(J) −33
(K) (−2)2 × (−3)3
(L) (−2) × (−3)3
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
(J)
(K)
(L)
18
40
−18
12
40
−72
168
0
−32
−27
−108
54
39
7 GRADE-I: DR. EUNKYUNG YOU
DIVIONS OF SAME SIGN NUMBERS
INVERSE:
If two numbers has the same sign, their division is positive
The inverse of
(+15) ÷ (+3) = +(15 ÷ 3) = +5
2
3
If two numbers has the opposite sign, their division is
(−15) ÷ (−3) = −(55 ÷ 3) = −5
3
Inverse of 3 = 1 :
MULTIPLICATION OF DIFFERENT SIGN NUMBERS
(+15) ÷ (−3) = −(15 ÷ 3) = −5
𝑏
π‘Ž
in multiplication is where π‘Ž ≠ 0, 𝑏 ≠ 0
Inverse of − 3 : − 2
(−15) ÷ (−3) = +(15 ÷ 3) = +5
negative
π‘Ž
𝑏
1
3
THE ORDER OF OPERATIONS: (from left to right)
 First, simplify the exponent expression
 Simplify the parentheses
 Simplify multiplication and division
 Simplify addition and subtraction
Example 1: Evaluate
(A) 35 ÷ (−5)
Solution
(A) 35 ÷ (−5) = −(35 ÷ 5) = −7
(B) (−16) × (−9) ÷ 6
(C) 23 − 56 ÷ (−4) + (−7)
(B) (−16) × (−9) ÷ 6
= (16 × 9) ÷ 6
= 144 ÷ 6 = 24
(C) 23 − 56 ÷ (−4) + (−7)
= 23 − (−(56 ÷ 4)) + (−7)
= 23 − (−14) + (−7)
= 23 + 14 − 7
= 37 − 7 = 30
Example 2: Evaluate
(A) −8 ÷ 4
(C) (−132) ÷ (−11)
(B) (−24) ÷ (−2)
(D) 6 ÷ (−6)
(A)
(B)
(C)
(D)
−2
12
12
−1
40
7 GRADE-I: DR. EUNKYUNG YOU
Example 3: Evaluate
(A) 32 ÷ (−4) × (−7)
(B) (−8) − (−7) + (−24) ÷ (−6)
(C) 9 × (3 + 3) ÷ 6
(D) (9 + 18 − 3) ÷ 8
(E) (4 − 1 + 8 ÷ 8) × 5
(F) (10 × (−2)) ÷ (1 + 1)
Example 4: Evaluate
(A) (−2)2 × (−3) − [5 − (−1)]2 × 6
(A)
(B)
(C)
(D)
(E)
(F)
56
3
9
3
20
−6
(A)
(B)
(C)
(D)
−18
−30
14
16
(A)
(B)
(C)
(D)
25
−25
−4
9
(B) 6 − ({(−3)3 − 8 ÷ (−2)} + 5)
(C) 7 − ({60 ÷ −5} − (7 − 12))
(D) (−6)3 ÷ (−12) + [−7 − (−5)]3 ÷ 4
Example 5: Evaluate
(A) |(−5)2 |
(C) |5 − (−3)| ÷ (−2)
(B) −| − 5|2
(D) 3 × |−4| ÷ 2 − (−3)
41
7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1. Evaluate.
(A) (+2) × (+3)
(B) (−5) × (−2)
(C) (+3) × (−5)
(D) (−5) × (+6)
(E) (−2)4
(F) −(−2)3
(G) (−6) × (−3) × (2)
(H) (−35) × 0 × (−15)
(I) 24(−16)(−3)
(J) (−9)(−8)(−2)
2. Evaluate
(A) (+10) ÷ (+5)
(B) (−14) ÷ (−7)
(C) (+8) ÷ (−2)
(D) (−6) ÷ (+3)
(E) (−38) ÷ (−2)
(F) (132) ÷ (−11)
(G) (60) ÷ (−2) ÷ (−5)
(H) (−72) ÷ (−3) ÷ (−2)
3. Evaluate
(A) (30 − 3) ÷ 3
(B) (21 − 5) ÷ 8
(C) 1 + (−5)2
(D) 5 × 4 − 8
(E) 8 + 6 × (−9)
(F) 3 + (−17) × (−5)
(G) 15 + 40 ÷ 20
(H) (9 + 18 − 3) ÷ 8
(I) (−6) + (5 + 8) × (−4)
(J) (9 × (−2)) ÷ (2 + 1)
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
(J)
6
10
−15
−30
16
8
36
0
1152
−144
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
2
2
−4
−2
16
−12
6
−12
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
(I)
(J)
9
2
26
12
−46
88
17
3
−58
−6
42
7 GRADE-I: DR. EUNKYUNG YOU
4. Evaluate
(A) 63 ÷ (−9) + (−3) × (−6)
(B) (−32) − 24 ÷ (−3) − 4 × (−3)
(C) 196 ÷ ((−6) − 8) × (−2)
(D) [17 − (−7)] ÷ (−2)2
(E) (−30) × (−6) ÷ (−3)2
(F) (−6)3 ÷ (−3)2 ÷ [−5 − (−3)]2
(G) −4 ÷ | − 4|
(H) −| − 2|2
5. Fill in each
with <, " = " or " > ".
(A) (−9)(−6)
(C) |−9| ÷ |−3|
(6)(−10)
|−9 ÷ (−3)|
(B) −10 ÷ (−2)
25 ÷ (−5)
(D) −|−8| ÷ (2)
− 8 ÷ (−2)
(A)
(B)
(C)
(D)
(E)
(F)
(G)
(H)
11
−12
−7
6
20
−6
−1
−4
(A)
(B)
(C)
(D)
>
>
=
<
0
6. Evaluate
20
(−1)
17
+ (−1)
9
30
− (−1) − (−1)
0
7. Evaluate
(−1) + (−1)2 + (−1)3 + β‹― + (−1)100
8. A theater sold out its evening shows four night in a row. If the theater has 234 seats, how
many people attended the theater in the four nights?
9. Consider the natural numbers from 1 to 500, the sum of even numbers is A and the sum of
odd numbers is B. Find 𝐴 − 𝐡
10. Find all possible integers π‘Ž and 𝑏 such that π‘Ž < 𝑏, |π‘Ž + 𝑏| = 2, and π‘Ž × π‘ = −8
11. Find the largest value of π‘Ž − 𝑏 if |2 × π‘Ž| = 8 and |𝑏 ÷ 3| = 4 where π‘Ž and 𝑏 are integers
936 peoples
250
π‘Ž = −2, 𝑏 = 4 or
π‘Ž = −4, 𝑏 = 2
π‘Ž − 𝑏 = 16
43
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 2.5 RATIONAL NUMBERS
RATIONAL NUMBERS
ADDITION AND SUBTRACTION
 Make sure the bottom numbers (the
The rational numbers are the set of all numbers that can be in
𝑝
denominators) are the same
the form where 𝑝 and π‘ž are integers and π‘ž is not zero.
π‘ž
3÷8=
 Add the top numbers (the numerators), put the
3
8
answer over the denominator
9
(−9) ÷ 7 = −
7
3
3=
1
 Simplify the fraction (if needed)
3 1 3×3 1×2
9
2
11
+ =
+
=
+
=
⏟
4 6 4 × 3 6 × 2 12 12 12
LCM(4,6)=12
LOWEST TERMS (simplest form)
MULTIPLICATION OF RATIONAL NUMBERS
If the numerator and denominator of a fraction has no
 Multiply the top numbers (the numerators).
common factor except 1, we say that it is in the lowest
 Multiply the bottom numbers (the denominators).
terms.
 Simplify the fraction if needed.
8
4×𝟐 πŸ’
=
=
10 5 × πŸ πŸ“
1 4 1×4
4
2×2 2
× =
=
=
=
2 5 2 × 5 10 5 × 2 5
MULTIPLICATION OF RATIONAL NUMBERS
MIXED RATIONAL NUMBERS
 Turn the second fraction (the one you want to
A Mixed Fraction is a whole number and a proper fraction
divide by) upside down (this is now a reciprocal).
combined.
 Multiply the first fraction by that reciprocal
1
1 7
2 is a mixed fraction which is 2 + =
3
3 3
 Simplify the fraction (if needed)
1 4 1 5
5
÷ = × =
3 5 3 4 12
Example 1: Evaluate
1
2
3
8
(A) (− ) + ( )
(B) (−
5
)−
12
1
6
2
5
(+ )
3
4
(C) (+ ) × (− )
Solution
(−1)×4
1
2
3
8
3
+
2×4
8
LCM(2,8)=8
5
)−
12
(+ ) =
(A) (− ) + ( ) = ⏟
(B) (−
1
6
−5
+
12
2
3
2
2
4
2
=
1
−4
8
3
8
(−4)+3
−5
(−1)×2
+ =
(− ) = ⏟ +
6
12
8
6×2
LCM(12,6)=12
3
2×3
6
=−
=
1
8
−5
−2
+
12
12
3×2
=
(−5)+(−2)
12
3
(C) (+ 5) × (− 4) = − 5 × 4 = − 5×4 = − 20 = − 10×2 = − 10
πŸ’
2
πŸ‘
6
1×6
1
(D) (+ 9) ÷ (− 3) = − 9 ÷ (πŸ‘) = − 9 × πŸ’ = − 36 = − 6×6 = − 6
=−
7
12
2
9
4
3
(D) (+ ) ÷ (− )
44
7 GRADE-I: DR. EUNKYUNG YOU
Example 2: Evaluate
1
5
3
7
1
4
(A) ( ) + ( )
1
3
(A)
(B) (− ) + (− )
22
35
(B) −
7
12
(C) −
1
(D) −
8
4
5
(E) −3
(F)
3
1
(G)
2
5
(D) (−2) − (− )
(C) (− 4) + (2)
(J)
1
1
3
4
3
8
5
4
(F) (2 ) − (3
2
3
(G) ( ) − (− ) + (− )
1
4
5
3
5
6
(I) 1 + (− ) − (− )
5
)
12
5
3
(H) −1 − (6) + (− 5)
2
3
(J) (−2 ) − (−
11
)+
4
1
6
(1 )
2
3
23
24
(H) −
(I)
(E) (−2 5) + (−1 3)
−
8
15
5
12
5
4
73
30
=−
53
15
45
7 GRADE-I: DR. EUNKYUNG YOU
Example 3: Evaluate
(A)
−6
−6
(B)
−3
−
(A) 1
2
5
(B)
15
2
(C) 10
(D) −
(E) −
3
−
5
(F)
(C) (−12) ×
5
(− 6)
1
4
(D) ×
(G) −
7
(− 8)
7
32
(H)
(I)
(J)
8
9
1
12
1
4
−
3
10
35
4
(K) 2
(L) −
5
9
(E) (− 12) × (+ 10)
1
3
5
2
2
9
1
2
2
3
(F) (− ) × (+ )
2
(G) (+ 4) × (− 2) × (+ 9)
3
4
(H) (+ ) × (− ) × (− )
(I) (−2 ) × ( )
2
5
3
8
(J) (−6 4) × (− 5)
2 2
3
9
2
(L) (− ) × (−1 )
(K) (− ) × ( )
1
3
4
7
1 2
3
4
3
46
7 GRADE-I: DR. EUNKYUNG YOU
Example 4: Evaluate
1
3
6
5
3
2
(A)
(B) (−0.3) ÷ (+ )
(A) (− ) ÷ (− )
5
18
(B) −
(C)
1
5
6
5
(D) −
55
12
(E) 7
(F)
1
48
(G) −1
(H) −
3
1
11
6
(C) (−1 5) ÷ (−1 3)
1
7
(E) (−2) ÷ (4) × (− 8)
5
6
(I)
2 3
9
1
1
(J)
(F) (− 3) × 16 ÷ (−8)
12
7
5
1
2
1
3
(H)
3
4
1
5
− )
11
22
(J) (− − (− )) ÷ ( − )
(G) (− ) ÷ ( − (− ))
7
8
1
5
(D) 3 × ( ) ÷ (−1 )
(I) ( ÷ (−2 )) ÷ (
× (4 − 9) ÷ (− 4)
1
4
2
3
2
3
3
4
4
21
7
3
−5
47
7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1. Evaluate
(A) −
2
5
(A) +
4
(− 5)
4
(B) 6 −
7
5
6
(B)
(D)
5
(D) 3 − (− 9)
2
3
(E) (− 3) + 8
(F) (−
10
)−
7
1
(− 6)
(G)
(H)
9
5
(G) −
5
8
4
(I)
3
(H) (− 3) − (− 2)
(J)
(K)
2
3
(I) (−2) + (−3 5)
2
(J) (−3 5) − 4 3
(K)
+
1
(−1 7)
(L)
2
15
+
4
(−3 7)
1
(M) 2 3
+
4
(−1 5)
7
8
(N)
1
2
7
(−1 8) +
4
(O) (−1 ) − (−3 )
(P)
1
(−3 4)
1
6
2
3
(C) − + −
8
(− 3)
(B)
2
1
(− 3) − (− 6) +
40
1
6
−
27
−
124
1
5
7
4
3
2
(D) − +
3
3
(− ) − (− ) +
5
4
(C)
9
10
4
9
4
2
(+ )
5
(E)
6
5
(F) − 3 − 1.5 + 6
3
(G) (− 5) + (− 4) − 5 − (− 2)
7
76
35
8
15
41
8
13
8
−
23
10
2
5
(B)
27
− 72
4
10
23
20
11
20
−2
(G) −
11
(H) −
9
4
4
7
(A)
(C)
64
− 80
(D)
98
− 21
4. Find the reciprocal of the following fraction:
(B) −4
3
3
2
5
(B) −
(C) −
(D) −
(A)
15
(H) 3 − (− 4) + (− 6) − 3
3. Express the following in the lowest terms
3
5
5
19
(B) −
1
(− 4)
(F)
(A)
42
47
(A) 0
4
(+ 3) − (+4) −
24
60
53
1
(D)
(E)
−
7
24
(P) (−3 5) − (−1 2)
2. Evaluate
(A)
40
9
(N) −
(O)
67
8
(L) −
(M)
6
37
6
(E) −
(F)
5
31
(C) −
1
(C) (− 5) − 8
2
(A)
(C)
2
19
(D) −2.5
8
4
5
14
3
5
3
(B) −
(C)
3
9
11
(D) −
1
4
2
5
48
7 GRADE-I: DR. EUNKYUNG YOU
5. Evaluate
(A)
(A)
4 7
βˆ™
9 4
(B)
(C) (−2) βˆ™
3
7
(D)
2
5
(− ) βˆ™
3
4
5
9
(− 12) (10)
5
(C) −
6
(D) −
3
(E)
(E)
(F)
1
(−1 ) (9)
4
9
(B) −
(F)
1
3
(−2 5) (−1 4)
7
(G)
5
1
3
5
(G) (−1 7) (−2 2)
2
3
(I) (−2 ) (4
1
)
10
5
7
1
2
(H) (−2 ) (2 )
1
4
8
(K) (+ 21) (− 10) (3)
20
−
7
4
3
2
2
164
15
(K) −
4
(L) −
1
4
9
3
1 3
(L) (− 3) (−3)2
1
2
5
4
(B) −
(B) (− ) ÷ ( )
15
)
8
13
3
(C)
(D)
(C) (− ) ÷ (−
4
7
(A) −
1
5
45
30
9
6. Evaluate
(A) (− ) ÷ ( )
2
−
(J)
(J) (+ 2) (+ 3) (− 8)
7
77
(H) −
(I)
6
1
2
8
7
(D) ( ) ÷ ( )
2
5
4
5
7
16
(E) −
(F)
4
35
4
5
−15
(G) −2
8
(E) (− 5) ÷ (2)
(F) (−
(G) (6.8) ÷ (−3.4)
16
8
) ÷ (15)
3
(H) (−7.2) ÷ (−1.2)
(H) 6
(I)
−
10
(J)
−
2
(K)
(L)
(M)
(I)
5
(3) ÷
(−0.5)
(J) (−1.5) ÷
9
(4)
(P)
(K) (−3.5) ÷ (−14)
3
(M) (−2) ÷ (−3 5)
6
1
(O) (1 7) ÷ (5 5)
5
(L) (−3 9) ÷ (−3)
1
9
1
3
(N) ( ) ÷ (−1 )
7
1
(P) (−3 10) ÷ (2 4)
3
1
4
22
3
5
9
(N) −
(O)
3
5
14
6
5
1
12
49
7 GRADE-I: DR. EUNKYUNG YOU
7. Evaluate
(A)
(A)
2
( )÷
3
5
(− ) ÷
18
8
(− )
5
(B)
1 2
(− 2)
2
3
2
(B) −
1
× (− 3) ÷ (9)
(C)
3
2
4
75
(D) 3
14
2
3
5
(C) ( 5 ) ÷ (−7) ÷ (− 15)
(D) ( ) ÷ (−
1
)÷
10
(E)
(−2)
(F)
(G)
(E)
2
4 2
(− 3) ÷ (− 3)
×
1
(− 2)
(F)
1 5
(− 2)
3 3
(− 2)
3
(− 2) ÷
×
10
)×
3
3
5
(H) (−2)2 × (− ) ÷
(−6)
3 3
(I)
2
18
5
−45
1
12
12
5
8. Evaluate
(A) 3 +
1
72
(J) (2.5) ÷ (−3) ÷ (−10)
(I) (−4) ÷ (−0.3) × (− 2)
(A)
1 3
(− 3)
−
(H) −1
(J)
(G) 2 ÷ (−
3
16
6
(B)
×5
1
4
(− 6) ÷ 9
7
8
−
28
45
(B) −
5
(C) −
4
4
3
(D) −18
(C) 12 ÷ {(−2)3 − 1}
1
1
(D) 9 × (−3)4 − 9 ÷ 3
(E)
(F)
(G)
3
4
(E) +
3
(− 7) ×
3
(− 4)
(F)
1
1
(− 3) — 2
÷
1
4
(G)
×
3
(− ) ÷
10
1
(− )
5
(H) 2 −
1
(−1 + 3) ×
14
−
7
3
3
8
(H) 8
(I)
(J)
1 2
(− )
2
15
9
−
3
4
2
11
(K) −
5
(L) −
10
9
9
(M) −4
(I)
5
(3
−
1 2
)
6
2
3
(J)
÷ −3
1
7
÷ {1 −
2
(7
1
− 14)}
(N) −130
(O) −
(P)
(K) 5 −
1
{(4
−
2
3
) ÷ 5} ×
3
2
3
(−2)
5
(M) 11 ÷ {9 × (9 − 12) − 1}
1
1
1
2
(O) 3 − 2 × (5 ÷ 0.15 − 3 × (0.5)2 )
5
3
(L) 2 − +
5
(6 — 2) 2
21
÷ 13
1
1
(N) (−2)3 ÷ 10 + (−5)2 ÷ (− 2)
1
1 2
3
(P) 1 − {− + 12 × (− ) } ×
5
3
17
4
5
1
4
50
7 GRADE-I: DR. EUNKYUNG YOU
9. In the real line, find the number represented by M which is the midpoint of A and B
M
A
1
4
2
10. In the real line, find the number represented by B if M is the midpoint of A and B
M
5
1
6
3
3
2
B
11. In the real line, find the number represented by Q where Q is the one-third from A to B
A
1
8
B
3
A
−
P
2
9
B
2
2
3
3
12. If a fraction is equal to 5, and the sum of the denominator and the numberator is 24, what si
9
15
the fraction?
2
3
13. Let π‘Ž ⊝ 𝑏 = π‘Ž ÷ 𝑏 + 1 for all rational numbers π‘Ž and 𝑏. Find 13 ⊝ { ⊝ (−5)}
3
3
14. Let π‘Ž ⊞ 𝑏 = 3 ÷ (𝑏 − π‘Ž) for all rational numbers π‘Ž and 𝑏. Find (− 5) ⊞ (− 2)
1
1
1
15. We know that 𝑛×(𝑛+1) = 𝑛 − 𝑛+1 for all natural number 𝑛. Evaluate
16
−
10
3
99
100
1
1
1
1
+
+
+ β‹―+
1×2 2×3 3×4
99 × 100
16. In a square, each row and column has the same sum. Find the constants π‘Ž, 𝑏, and 𝑐
3
2
0.4
−1
π‘Ž
−
6
5
𝑐
2
3
𝑏
1
1
26
π‘Ž = ,𝑏 = ,𝑐 =
5
6
15
51
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 3. INTRODUCTION TO ALGEBRA
CHAPTER 3.1 ALGEBRAIC EXPRESSIONS
DEFINITION:
RULES IN ALGEBRIC EXPRESSIONS:
 A Variable is a symbol for a number we don't
 (number) × (variable) : omit " × " notation
know yet. It is usually a letter like π‘₯ or 𝑦.
2 × π‘₯ = πŸπ’™
 A number on its own is called a Constant
write number in front of
𝑦 × (−5) = −πŸ“π’™
letters
 A Coefficient is a number used to multiply a
 Multiplication with 1 or (−1)
variable
Omit “1”, but not sign
1×π‘Ž =𝒂
 An Operator is a symbol (such as +,×, etc) that
(−1) × π‘ = −𝒃
shows an operation
 An algebraic expression involves numbers and
 Multiplication of same variables: exponent form
letters that are connected with operations symbols
π‘₯ × π‘₯ × π‘₯ = π‘₯3
such as “+, −,×, and ÷ ”
WARNING:
 (0.1) × π‘Ž ≠ 0. π‘Ž (consider π‘Ž =
3
4
3
 π‘Ž ÷ 4 𝑏 ≠ π‘Ž × 3 𝑏 since 4 𝑏 =
1
)
10
3𝑏
4
VALUE
WRITING EXPRESSIONS
When π‘₯ = −4, find 3π‘₯ + 2.
Nine more than π‘Ž
π‘Ž+9
3π‘₯ + 2
4 less than 𝑛
𝑛−4
= 3 × π‘₯ + 2 : put ×
𝑧 times three
3𝑧
π‘₯
3
π‘₯ divided by 3
= 3(−4) + 2 : do not forget ( )
= −10
Example 1: Kim is 16 years old. Find his age after π‘₯ years.
Solution: Right now, his age is 16 years.
Age after π‘₯ years = 16 + π‘₯
Example 2: The length of a rectangle is 6 inches longer than twice of its width. Find the length of the rectangle if its
width is 30 inches.
Solution: Length = 2 × Width +6
Since width is 30 inches, the length is 2(30) + 6 = 66 inches
52
7 GRADE-I: DR. EUNKYUNG YOU
Example 3: Write an expression for the following situation
(A) The area of a square such that the length of its side is π‘Ž cm
(A)
(B)
(C)
(D)
(E)
π‘Ž2 cm2
2(π‘₯ + 𝑦) cm
$ 0.05π‘₯ + 2𝑦
(π‘₯ + 5) cm
3
𝑏
4
(B) The perimeter of a rectangle such that its length is π‘₯ cm and its width is 𝑦 cm
(C) The price of π‘₯ pens and 𝑦 notebooks when the price of a pen is 5 cents and the price
of a notebook is $ 2.
(D) Tom is 5 cm taller than Mary if the height of Mary is π‘₯ cm.
(E) Lee’s test score is three quarters that of Mary if Mary’s test score is 𝑏
Example 4: Express the following without × or ÷ notations
(A) (π‘Ž + 𝑏) × (−5) × π‘
(B) π‘₯ ÷ 𝑏 ÷ 𝑦
Solution
(A) (π‘Ž + 𝑏) × (−5) × π‘ = −5𝑐(π‘Ž + 𝑏)
1
1
π‘₯
(B) π‘₯ ÷ 𝑏 ÷ 𝑦 = π‘₯ × π‘ × π‘¦ = 𝑏𝑦
Example 5: Express the following without × or ÷ notations
(B) π‘Ž × (−1) × π‘Ž × π‘Ž
(A) 2 × π‘Ž × π‘
(A) 2π‘Žπ‘
(B) −π‘Ž3
(C) 2π‘₯ − 5𝑦
(C) π‘₯ × 2 + 𝑦 × (−5)
(D) (π‘₯ − 𝑦) ÷ 2
(D)
(E)
(F)
(E) 3 ÷ (π‘Ž − 𝑏)
(F) π‘₯ ÷ 2 + (𝑦 − 𝑧) ÷ 5
π‘₯−𝑦
2
3
π‘Ž−𝑏
π‘₯
2
+
𝑦−𝑧
5
53
7 GRADE-I: DR. EUNKYUNG YOU
Example 6: When π‘₯ = −4, find π‘₯ 2 − 2π‘₯.
Solution:
π‘₯ 2 − 2π‘₯ = π‘₯ × π‘₯ − 2 × π‘₯ = (−4)(−4) − 2(−4) = 16 − (−8) = 24
Example 7: When π‘₯ = 2, find the following.
2
(A) −π‘₯
(B) π‘₯
(A) −2
(B) 1
(C) 4
(D) −1
(E) 8
(C) π‘₯ 2
(D) −3π‘Ž + 5
(F)
−2
(G) 36
(H) 12
(E) (−π‘₯)2 + 2π‘₯
(F) −π‘₯ 3 + 6
(G) (3π‘₯)2
(H) 3π‘₯ 2
Example 8: Find the following
(A) −19
(A) 3π‘₯ − 5𝑦 when π‘₯ = −3 and 𝑦 = 2
(B) 23
(C) 3
(B) 2π‘₯ 2 − 𝑦 2 when π‘₯ = 4 and 𝑦 = −3
1
1
(C) 9π‘₯ 2 − 24π‘₯𝑦 when π‘₯ = 3 and 𝑦 = − 4
54
7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1. Express the following expression with × or ÷ notations
(A) π‘₯ × 7
(B) π‘₯ × π‘₯ × 6
(C) 4π‘₯ ÷ 𝑦
(D) π‘₯ ÷ 𝑦 × π‘Ž
(E) π‘₯ ÷ 3 × π‘¦
(F) π‘₯ × (π‘Ž + 𝑏) ÷ 5
(G) 5π‘₯ × 4𝑦
(H) 3π‘₯ × 4π‘₯ × 2π‘₯
(I) π‘₯ ÷ 5 + 3 × π‘¦
(J) π‘₯ + 6𝑦 ÷ 2π‘Ž
(K) π‘Ž × π‘Ž − 𝑏 ÷ 3
(L) π‘Ž ÷ (π‘₯ + 𝑦) + 𝑏 × 6
(M) 0.5 × π‘₯ × π‘¦
(N) π‘₯ ÷ 𝑦 ÷ 4
1
(O) π‘Ž ÷ 5 × π‘
(P) π‘Ž ÷ π‘₯ 2 × 2
(Q) π‘₯ + 𝑦 ÷ 2
(R) 3 × π‘Ž + 2 ÷ (π‘₯ + 𝑦)
0
2. When π‘Ž = −2 and 𝑏 = 3, evaluate the following
(A) 3π‘Ž + 2𝑏
−22
(B) 5π‘Ž − 4𝑏
−2
3
−9
𝑏
(C) π‘Žπ‘ + 4
(D) −2π‘Ž − 3
9
(E) 3π‘Ž − 𝑏
(F)
π‘Žπ‘
π‘Ž−𝑏
3. Evaluate the following
(A) π‘₯ 2 −
π‘₯𝑦
2
(B) 9π‘₯ 2 +
+ 2𝑦 when π‘₯ = 2 and 𝑦 = −5
3π‘₯𝑦
5
when π‘₯ = −3 and 𝑦 = −2
1
1
(C) 4π‘₯ 2 − 24π‘₯𝑦 when π‘₯ = − 2 and 𝑦 = 3
π‘₯𝑦
(D) π‘₯ 2 −4𝑦 when π‘₯ = −3 and 𝑦 = 2
(A)
(B)
(C)
(D)
−1
84.6
5
−6
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7 GRADE-I: DR. EUNKYUNG YOU
(A) −93
4. Evaluate the following
(A) π‘₯ 2 𝑦 − (3π‘₯)2 when π‘₯ = 2 and 𝑦 = −3
(B) π‘₯ 2 + 3𝑦 2 − 2𝑀 2 when π‘₯ = 1, 𝑦 = −2 and 𝑀 = −3
(C)
π‘₯ 2 −𝑦 2
𝑀 3 −4𝑦
(B) −5
(C) 0
(D) −96
when π‘₯ = 2, 𝑦 = −2 and 𝑀 = −1
(D) π‘Žπ‘₯ 𝑛+1 when π‘Ž = −3, π‘₯ = 2 and 𝑛 = 4
5. Write an expression for the following situation
(A) The height of Tom if Tom is 15 cm taller than his brother if the height of his brother
is β„Ž cm
(A) (β„Ž + 15) cm
(B) 7π‘₯ days
(C) 𝑑 + 13 pounds
(D) $ (5700 − π‘₯ − 𝑦)
(E) $(1.5π‘₯ + 3.2𝑦)
(B) The number of days in π‘₯ weeks
(F)
60π‘₯ miles
(G) 100 − 5π‘₯
(C) The weight of Kim if Kim is 13 pounds heavier than her mother if the weight of her
mother is 𝑑 pounds
(D) The amount left if Lee has $ 5700 and he spends $π‘₯ on pants and $𝑦 on a shirt
(E) The price of π‘₯ apples and 𝑦 watermelon if the price of apple is $1.5 and the price of
watermelon is $3.2
(F) Driven miles if you drive at the rate of 60 mph for π‘₯ hours
(G) The resulting number if you subtract 5π‘₯ from 100.
6. The minimum pay in a country is $7 per hour. The wage for 𝑑 hour working hours is $7𝑑.
$58
If Tom works for 8 hours, find the wage.
7. The admission fee in museum, $F for one class of π‘₯ girls and 𝑦 boys is given by
𝐹 = 6π‘₯ + 7𝑦
$118
Admission fee for girl is $6
Admission fee for boy is $7
(A) Find the total admission fee for 8 girls and 10 boys.
(B) Find the meaning of 6 and 7 in the formula.
1
3
8. The volume of a cone is πœ‹π‘Ÿ 2 β„Ž where π‘Ÿ is the radius of its base and β„Ž is the height of the
cone.
(A) Find the volume if π‘Ÿ = 6 cm and β„Ž = 4 cm
(B) Find the increase of the volume if the height of the cone increase 2 cm
(A) 48πœ‹ cm3
(B) 24πœ‹ cm3
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7 GRADE-I: DR. EUNKYUNG YOU
9. Mary has $35 in her bank account. If she is saving $15 each week in her bank account,
$(15π‘₯ + 35)
write her account balance after π‘₯ weeks.
10. Tom and Mary collect stamps. Tom has 4 less than twice as many stamps as Mary. If Mary
2π‘₯ − 4
has π‘₯ stamps, write the number of stamps which Tom has.
11. The length of a rectangle is 5 cm bigger than three times of its width. If the length of the
(3𝑦 + 5)cm, 38 cm
width is 𝑦 cm, write the length of the rectangle. Find the length of the rectangle if its width
is 11 cm.
12. The age of John’s father is 2 year younger than three times of John’s age. The age of
John’s grandfather is 27 year younger than twice of the age of John’s father. If John is π‘₯
year old, write the age of his grandfather.
2(3π‘₯ − 2) − 27
57
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 3.2 LIKE TERMS AND UNLIKE TERMS
DEFINITION:
COMBINING LIKE TERMS
 A Term is either a single number or a variable, or
 Add or subtract the coefficient of the like terms.
 Use this as the coefficient of the term’s variable
numbers and variables multiplied together.
 An Expression is a group of terms (the terms are
factor.
separated by + or − signs)
2π‘₯ − 6 + 3π‘₯ + 4
= 2π‘₯ + 3π‘₯ + (−6) + 4
= (2 + 3)π‘₯ + (−6) + 4
= 5π‘₯ − 2
LIKE TERMS: Terms with exactly the same variables
that have the same exponents are like terms:
9π‘₯ 2 , 6π‘₯ 2 ∢ like terms
9π‘₯, 6π‘₯ 2 ∢ NOT like term
Example 1: State the number of terms and the constant term in each expression.
(A) π‘Ž + 3
(B) π‘₯ 2 + 2π‘₯ − 5
Solution
(A) π‘Ž + 3 : two terms and 3 is the constant term
(B) π‘₯ 2 + 2π‘₯ − 5 = π‘₯ 2 + 2π‘₯ + (−5) : three terms and −5 is the constant term
Example 2: State the number of terms and the constant term in each expression.
(A) 2π‘₯ + 3𝑦 + 4
(B) −2π‘₯ 2 + 3𝑦 + 1
(A) 3 terms and 4
(B) 3 terms and 1
(C) 2 terms and 0
(D) 2 terms and 0
(C) π‘₯ 2 − π‘₯
(D) 2(π‘₯ + 𝑦) − 4𝑦
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7 GRADE-I: DR. EUNKYUNG YOU
Example 3: State the coefficients of π‘₯ and 𝑦 in each expression
(B) 2π‘₯ 2 − 3𝑦 + 4π‘₯ + 6
(A) 5π‘₯ − 𝑦 + 4
Solution
Example 4: State the coefficients of π‘₯ and 𝑦 in each expression
π‘₯
(B) π‘₯ 2 + 3π‘₯ − 4𝑦
(A) 4 + 5𝑦 + 6
(A) π‘₯:
1
4
𝑦: 5
(B) π‘₯: 3 𝑦: −4
(C) −3π‘₯ 2 + 5π‘₯ + 8
(D) π‘₯ 2 + 3π‘₯ − 3π‘₯𝑦 − 5𝑦
(C) π‘₯: 5 𝑦: 0
(D) π‘₯: 3 𝑦: −5
Example 5: Simplify the following expression
9
(A) 5π‘₯ × 2
(C) 3π‘₯ − 5 − 6π‘₯ + 9
(B) 6π‘₯ ÷ 2
Solution
(A) 5π‘₯ × 2 = 5 × π‘₯ × 2 = 5 × 2 × π‘₯ = 10π‘₯
9
2
2
9
2
9
4
3
(B) 6π‘₯ ÷ = 6 × π‘₯ × = 6 × × π‘₯ = π‘₯
(C) 3π‘₯ − 5 − 6π‘₯ + 9 = 3π‘₯ − 6π‘₯ − 5 + 9 = (3 − 6)π‘₯ + (−5 + 9) = −3π‘₯ + 4
Example 6: Simplify the following expression
(A) 2𝑦 × 5
(A) 10𝑦
(B) 2π‘Ž × (−5)
(B) −10π‘₯
(C) 18π‘₯
(C) (−3π‘₯) × (−6)
(D) 5π‘₯
(D) 15π‘₯ ÷ 3
(E) −6π‘Ž
(F)
1
(E) 24π‘Ž ÷ (−4)
(F) (−12π‘Ž) ÷ 3
(G) π‘₯ ÷ 𝑦 × 3
(H) 5π‘₯ × 2𝑦
(G)
−36π‘Ž
3π‘₯
𝑦
Example 7: Simplify the following expression
(A) 2𝑦 + 5𝑦
(B) −8π‘₯ + 3π‘₯
(A) 7𝑦
(B) −5π‘₯
(C) 14π‘Ž
(C) 10π‘Ž − 8π‘Ž + 12π‘Ž
(D) −7𝑏 − 11𝑏 + 3𝑏
(D) −14𝑏
(E) 2π‘Ž − 4
(F)
(E) 5π‘Ž + 4 − 3π‘Ž − 8
(F) 7π‘₯ − 8𝑦 + 3π‘₯ + 4𝑦
10π‘₯ − 4𝑦
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7 GRADE-I: DR. EUNKYUNG YOU
Example 8: Simplify the following expression
1
2
(A) 4 − π‘₯ + 6π‘₯ − 9
2
3
2
5
(A)
(B) π‘₯ − 5 + π‘₯ + 9
(B)
(C)
(D)
(E)
(F)
3
1
2
8
3
(C) 4 π‘₯ − 2 𝑦 + 3 π‘₯ + 2𝑦
8
(E) 7π‘₯ − 5 π‘₯ + 4 − 3
1
2
3
9
5
7
4
5
8
11
π‘₯−5
2
16
15
17
12
17
10
27
5
π‘₯+4
3
π‘₯+ 𝑦
2
π‘₯+
π‘₯−
19
5
8
4
Example 9: Find the value of the expression
(A) −11
(B) 15
(C) 7.7
1
3
(C) 4 π‘₯ − 5 𝑦 + π‘₯ − 2𝑦 when π‘₯ = 2 and 𝑦 = −2
12
1
(F) π‘₯ − 𝑦 − 𝑦 − 2π‘₯
(B) 2π‘₯ − 3𝑦 + 5π‘₯ + 9𝑦 when π‘₯ = 3 and 𝑦 = −1
𝑦
23
− π‘₯−
(D) 5 π‘₯ − 3 𝑦 + 2 π‘₯ + 5 𝑦
(A) 6π‘₯ − 8 − 4π‘₯ + 3 when π‘₯ = −3
2
15
𝑦
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7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
(A) 5π‘₯
1. Simplify
(A) 2π‘₯ + 3π‘₯
(B) 9π‘₯
(B) 5π‘₯ + 4π‘₯
(C) 6π‘Ž
(D) −18π‘₯
(C) 11π‘Ž − 5π‘Ž
(E) 11π‘₯
(D) −13π‘₯ − 5π‘₯
(F)
−5π‘₯
(G) −7π‘Ž
(E) 7π‘₯ + 4π‘₯
(H) 3π‘₯
(F) 3π‘₯ − 8π‘₯
(I)
−4π‘Ž
(J)
π‘₯𝑦
(K) −8π‘₯𝑦
(G) −4π‘Ž − 3π‘Ž
(H) π‘₯ + π‘₯ + π‘₯
(L) −3π‘Ž
(M) 0.8𝑦
(N) 2.3π‘Ž
(I) π‘Ž + 4π‘Ž − 9π‘Ž
(J) 6π‘₯𝑦 − 9π‘₯𝑦 + 4π‘₯𝑦
(O)
(P)
(K) 3π‘₯𝑦 − 5π‘₯𝑦 − 6π‘₯𝑦
(L) −4π‘Ž − 3π‘Ž + 5π‘Ž − π‘Ž
(M) 0.5𝑦 + 0.3𝑦
(N) 6π‘Ž − 3.7π‘Ž
1
1
3
4
(O) 2 𝑦 + 3 𝑦
5
6
1
8
𝑦
π‘Ž
2
3
(P) π‘Ž − π‘Ž
(A) 0
2. Simplify
(A) 4π‘₯ + 2π‘₯ − 6π‘₯
(B) −2π‘₯
(B) 3π‘₯ − 7π‘₯ + 2π‘₯
(C) −4π‘₯ + 7
(D) −8π‘Ž − 5
(C) 9π‘₯ − 13π‘₯ + 7
(D) 8π‘Ž − 5 − 16π‘Ž
(E) −11π‘Ž
(F)
(E) −3π‘Ž − 2π‘Ž − 6π‘Ž
(G) −2π‘₯ − 2
(F) 14π‘Ž + 4 − 9π‘Ž − 6
(G) 3π‘₯ + 9 − 5π‘₯ − 11
5π‘Ž − 2
(H) −3π‘Ž + 2
(H) −6 + 3π‘Ž − 6π‘Ž + 8
(I)
−6π‘₯ + 2𝑦
(J)
5π‘₯
(K) −10π‘₯ − 14
(I) 2π‘₯ − 𝑦 + 3𝑦 − 8π‘₯
(J) −4π‘₯ − 2π‘₯𝑦 + 9π‘₯ + 2π‘₯𝑦
(L)
7
20
π‘₯−3
1
(M) − π‘₯ +
6
(K) −3π‘₯ + 3 − 7π‘₯ − 17
2
1
2
3
5
1
1
(N)
2
2
1
(O) 3 π‘₯ − 4 𝑦 − 2 π‘₯ + 2 𝑦
(P)
3
5
6
π‘₯+
(O) −
(N) 2 − 2 π‘₯ + 3 π‘₯ − 5
(M) 3 π‘₯ − 4 − 6 π‘₯ + 2
5
3
(L) −3 + 4 π‘₯ − 5 π‘₯
1
1
(P) 7π‘₯ + 4π‘₯𝑦 − 3 π‘₯ + 4 π‘₯𝑦
16
3
11
6
1
4
7
5
1
π‘₯− 𝑦
π‘₯+
4
17
4
π‘₯𝑦
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7 GRADE-I: DR. EUNKYUNG YOU
(A) −5
3. Find the value of the following expression
(B) 20
(A) −2π‘₯ + 5 − π‘₯ − 4 when π‘₯ = 2
(C) −30
(B) 7π‘Ž − 3𝑏 + 5π‘Ž + 7𝑏 when π‘Ž = 2 and 𝑏 = −1
6
2
4
1
1
1
(C) π‘Ž + 𝑏 − 𝑐 when π‘Ž = − 3 , 𝑏 = 2, and 𝑐 = 4
4. State the number of terms and find the constant term in each expression.
(A) π‘₯ 3 − 3π‘₯𝑦 + 𝑦 2 − 7
5
(B) 3π‘₯𝑦 − 5π‘₯ 2 𝑦 + π‘₯ + 𝑦 5 + 2
(A) 4 terms ; −7
(B) 5 terms; 2
5. Kim throws a ball vertically upward. Find the ball’s height above the ground after 3
45 ft
seconds if the ball’s height above the ground is (30𝑑 − 5𝑑 2 ) ft after 𝑑 seconds.
15π‘₯
6. Find the perimeter of a triangle such that the lengths of the sides of the triangle are 3π‘₯ cm,
5π‘₯ cm, and 7π‘₯ cm.
7. Find the perimeter and area of a rectangle such that its length is 3π‘Ž cm and its width is 2π‘Ž
cm.
8. Find the volume and surface area of a square box whose length of one sides of the square
box is π‘Ž cm.
Perimeter 5π‘Ž
Area 6π‘Ž2
Volume : π‘Ž3 cm3
Surface : 6π‘Ž2 cm2
62
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 3-3 DISTRIBUTION IN ALGEBRAIC EXPRESSIONS
DISTRIBUTION LAW
WARNING
 𝑨(𝐡 + 𝐢) = 𝑨𝐡 + 𝑨𝐢
(2π‘₯ − 5) − 3(3π‘₯ − 2)
 (𝐡 + 𝐢)𝑨 = 𝐡𝑨 + 𝐢𝑨
= 2π‘₯ − 5 − 9π‘₯ + 6 : distribute carefully
1
 (𝐴 + 𝐡) ÷ π‘ͺ = (𝐴 + 𝐡) × πΆ
= 2π‘₯ − 9π‘₯ + (−5) + 6
= (2 − 9)π‘₯ + (−5 + 6)
3(5π‘₯ + 2)
= −7π‘₯ + 1
= 3 × 5π‘₯ + 3 × 2
Wrong
= 15π‘₯ + 6
2
(2π‘₯ − 6) ÷ (− )
3
3
= (2π‘₯ − 6) × (− )
2
3
3
= 2π‘₯ × (− ) + (−6) × (− )
2
2
3π‘₯ π‘₯ − 1
+
4
2
3π‘₯ 2π‘₯ − 1
=
+
∢ wrong!
4
4
5π‘₯ − 1
=
4
Right
3π‘₯ π‘₯ − 1
+
4
2
3π‘₯ 𝟐(𝒙 − 𝟏)
=
+
4
4
5π‘₯ − 1
=
4
= −3π‘₯ + 9
Example 1: Simplify the following expression
(A) 2(π‘₯ + 4)
(B) (−4)(2π‘₯ − 3)
(A) 2π‘₯ + 8
(B) −8π‘₯ + 12
(C) −4π‘₯ + 18
(D) 12π‘₯ − 24
2
(C) 3 (−6π‘₯ + 27)
(E) 3π‘Žπ‘ + π‘Žπ‘
(D) (4π‘₯ − 8) × 3
(F)
−2π‘Žπ‘₯ + 𝑏π‘₯
(G) 12π‘₯ + 20𝑦
(H) −8π‘₯ + 6𝑦
(E) π‘Ž(3𝑏 + 𝑐)
(F) −π‘₯(2π‘Ž − 𝑏)
(G) 4(3π‘₯ + 5𝑦)
(H) −3(5π‘₯ − 2𝑦) + 7π‘₯
(I) π‘₯(2π‘Ž − 3𝑏 + 5𝑐)
(J) (3π‘Ž − 4𝑏 − 5𝑐)(−2π‘₯)
(I)
2π‘Žπ‘₯ − 3𝑏π‘₯ + 5𝑐π‘₯
(J)
−6π‘Žπ‘₯ + 8𝑏π‘₯ +
10𝑐π‘₯
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7 GRADE-I: DR. EUNKYUNG YOU
Example 2: Simplify the following expression
(A) (4π‘₯ − 8) ÷ 2
(B) (25π‘₯ − 15) ÷ (−5)
(A) 2π‘₯ − 4
(B) −5π‘₯ + 3
(C) 48π‘₯ − 16
(D) −3π‘₯ + 9
(E) 6π‘₯ − 9
(C) (24π‘₯ − 8)
1
÷2
1
(E) (2π‘₯ − 3) ÷ 3
(D) (9π‘₯ − 27) ÷ (−3)
(F)
−32π‘₯ + 128
1
(F) (8π‘₯ − 32) ÷ (− 4)
Example 3: Simplify the following expression
(A) (π‘₯ − 4) + (1 − 3π‘₯)
(B) (2π‘Ž + 3) − (−4π‘Ž − 5)
(A) −2π‘₯ − 3
(B) 6π‘Ž + 8
(C) 7π‘₯ − 𝑦
(D) −5π‘₯ + 4
(E) −5π‘₯ + 10
(C) (2π‘₯ + 3𝑦) + (5π‘₯ − 4𝑦)
(D) (3π‘₯ − 5) − (8π‘₯ − 9)
(F)
13π‘₯ − 12
(G) 3π‘₯ − 3
(H) 2π‘₯ + 1
(E) π‘₯ − 2(3π‘₯ − 5)
(F) (3π‘₯ − 6) − 2(3 − 5π‘₯)
(G) (π‘₯ − 2) + (2π‘₯ − 1)
(H) (3π‘₯ − 4) − (π‘₯ − 5)
(I) (−2π‘₯ + 3𝑦 − 4) + (π‘₯ + 5𝑦 + 9)
(J) (5π‘₯ − 3𝑦 + 6) − (2π‘₯ − 5𝑦 − 2)
(I)
−π‘₯ + 8𝑦 + 5
(J)
3π‘₯ + 2𝑦 + 8
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7 GRADE-I: DR. EUNKYUNG YOU
Example 4: Simplify the following expression
(A) 3(π‘₯ − 2) + 4(2π‘₯ − 1)
(B) 2(3π‘₯ − 4) − 3(π‘₯ − 5)
(A) 11π‘₯ − 10
(B) 3π‘₯ − 3
(C) −3π‘₯ − 13
(D) 3π‘₯ + 19
(E) 6π‘₯ − 1
(F)
−11π‘₯ − 38
(G) 8π‘₯ + 3
(C) 2(4 − 3π‘₯) + 3(π‘₯ − 7)
(D) 5(π‘₯ + 1) − 2(π‘₯ − 7)
(E) −2(3π‘₯ − 7) + 3(4π‘₯ − 5)
(F) −3(π‘₯ + 2) − 4(2π‘₯ + 8)
(G) −3(4 − π‘₯) + 5(π‘₯ + 3)
(H) −4(3 − 2π‘₯) − 5(π‘₯ − 9)
(I) 2 − 3(π‘₯ − 3𝑦 + 11)
(J) −2π‘₯ − 3(π‘₯ − 5𝑦 + 2)
(H) 3π‘₯ + 33
(I)
−3π‘₯ + 9𝑦 − 29
(J)
−5π‘₯ + 15𝑦 − 6
Example 5: Subtract π‘Ž − 2𝑏 from 3π‘Ž − 𝑏 and simplify
Solution:
(3π‘Ž − 𝑏) − (π‘Ž − 2𝑏) = 3π‘Ž − 𝑏 − π‘Ž + 2𝑏 = 2π‘Ž + 𝑏
Example 6: Express the statement and simplify
(A) Add (2π‘₯ − 4) to (3π‘₯ − 5𝑦 + 8)
(A) 5π‘₯ − 5𝑦 + 4
(B) −8π‘₯ + 7𝑦
(B) Subtract 3π‘₯ − 4𝑏 from −5π‘₯ + 3𝑦
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7 GRADE-I: DR. EUNKYUNG YOU
PROPERTIES OF EXPONENTS: Let π‘š, 𝑛 be positive
WARNING:
integers.
𝑛
π‘Ž ×π‘Ž
(π‘Žπ‘)𝑛
π‘š
=π‘Ž
𝑛+π‘š
(π‘Žπ‘› )π‘š
𝑛 𝑛
𝑛
=π‘Ž
2πŸ‘ ≠ 2 × 3 = 6
π‘›π‘š
( 23 = 2 × 2 × 2 = 8)
𝑛
π‘Ž
π‘Ž
( ) = 𝑛
𝑏
𝑏
=π‘Ž 𝑏
00 = 0
π‘₯0 ≠ 0
π‘₯ 3 × π‘₯ 2 ≠ π‘₯ 3×2 = π‘₯ 6
(π‘₯ 3 × π‘₯ 2 = π‘₯ 5 )
(2π‘₯)3 ≠ 2π‘₯ 3
π‘₯ 2 π‘₯2
( ) ≠
3
3
π‘Ž0 = 1
Example 7: Simplify the expression by writing with positive exponent. Assume that all variable is not zero.
3π‘₯ 2
(B) (π‘₯ 2 )4
(C) (2π‘₯ 2 )3
(A) π‘Ž2 × π‘Ž5
(D) ( )
𝑦
Solution:
(A) π‘Ž2 × π‘Ž5 = π‘Ž2+5 = π‘Ž7
(B) (π‘₯ 2 )4 = π‘₯ 2×4 = π‘₯ 8
(C) (2π‘₯ 2 )3 = 23 (π‘₯ 2 )3 = 8π‘₯ 6
3π‘₯ 2
(D) ( 𝑦 ) =
32 π‘₯ 2
𝑦2
=
9π‘₯ 2
𝑦2
Example 8: Simplify the expression by writing with positive exponent. Assume that all variable is not zero.
(A) 3𝑝 × π‘ × 5𝑝
(B) 4π‘Ž × 4𝑏 × π‘Ž
(A) 15𝑝3
(B) 16π‘Ž2 𝑏
(C) 14π‘₯𝑦 3
(C) 7π‘₯ × 2𝑦 × π‘¦ × π‘¦
2
(D) (7π‘₯)
(D) 49π‘₯ 2
(E) π‘₯ 7
(F)
π‘₯4
(G) π‘₯ 24
(E) π‘₯ 3 × π‘₯ 4
(F) (−π‘₯)4
(H) π‘₯ 5 𝑦10
(I)
(J)
(G) (π‘₯ 6 )2 × (π‘₯ 3 )4
(H) (π‘₯𝑦 2 )5
(K)
(L)
π‘Ž3
𝑏4 𝑐2
27π‘Ž6
π‘₯4
𝑦6
8
π‘Ž6
(M) 1
(I) π‘Ž3 ÷ 𝑏 4 ÷ 𝑐 2
π‘₯2
2
(J) (3π‘Ž2 )3
2 3
(K) (− 𝑦3 )
(L) (π‘Ž2 )
(M) 60
(N) (2π‘₯ 3 )0
(N) 1
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7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
(A) 6π‘₯
1. Simplify
(B) −10π‘₯
(B) 2π‘₯ × (−5)
(A) 3π‘₯ × 2
(C) −24π‘₯
(D) 6π‘₯
(C) (−4π‘₯) × 6
(E) 10π‘Ž ÷ 2
5
(G) 25π‘₯ ÷ 3
(D) (−9π‘₯) ×
2
(− 3)
(F)
−2𝑏
(G) 15π‘Ž
(F) (−8𝑏) ÷ 4
(H) −4π‘Ž
7
(H) 14π‘Ž ÷ (− 2)
(A) 8π‘₯ − 4
2. Simplify
(A) 4 × (2π‘₯ − 1)
(E) 5π‘Ž
(B) (−3) × (3π‘₯ − 2)
(B) −9π‘₯ + 6
(C) −20π‘₯ + 5
(D) −12π‘₯ − 3
5
4
(C) × (4 − 16π‘₯)
(E) (12π‘₯ − 15) ÷ (−3)
(D) (36π‘₯ + 9)
1
× (− 3)
(F) (−10 + 4π‘₯) ÷ (−2)
1
(G) (−1) × (3π‘₯ + 2)
(H) (2π‘₯ − 3) ÷ 4
(I) −(−7π‘₯ + 5)
(J) −(8π‘₯ − 𝑦 − 7)
(F)
−2π‘₯ + 5
(G) −3π‘₯ − 2
(H) 8π‘₯ − 12
(I)
7π‘₯ − 5
(J)
−8π‘₯ + 𝑦 + 7
(A) π‘₯ + 8
3. Simplify
(A) (2π‘₯ + 3) + (−π‘₯ + 5)
(E) −4π‘₯ + 5
(B) (2π‘₯ + 4) − (−3π‘₯ + 5)
(B) 5π‘₯ − 1
(C) 8π‘₯ + 11
(D) π‘₯ − 13
(C) (3π‘₯ + 2) + (5π‘₯ + 9)
(D) (4π‘₯ − 7) − (3π‘₯ + 6)
(E) −10π‘₯ − 5
(F)
−11π‘₯ + 9𝑦
(G) 13π‘₯ + 𝑦
(E) (−4π‘₯ − 9) + (−6π‘₯ + 4)
(F) (−9π‘₯ + 6𝑦) − (2π‘₯ − 3𝑦)
(H) 11π‘₯ + 16π‘Ž
(I)
3π‘₯ − 1
(J)
2π‘₯ + 6𝑦
(K) 2π‘₯
(G) (8π‘₯ + 3𝑦) + (5π‘₯ − 2𝑦)
(H) (5π‘₯ + 9π‘Ž) − (−6π‘₯ − 7π‘Ž)
(L) −9π‘₯ + 9𝑦
(M) −7π‘₯ − 2𝑏
(N) −5π‘₯ − 1
(I) (2π‘₯ + 3) + (π‘₯ − 4)
(J) (3π‘₯ + 3𝑦) − (π‘₯ − 3𝑦)
(K) (π‘₯ + 2𝑦) + (π‘₯ − 2𝑦)
(L) (−5π‘₯ + 6𝑦) − (4π‘₯ − 3𝑦)
(M) (−4π‘Ž − 5𝑏) + (−3π‘Ž + 3𝑏)
(N) −3 − (5π‘₯ − 2)
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7 GRADE-I: DR. EUNKYUNG YOU
(A) −4π‘₯ + 15
4. Simplify
(A) 2(π‘₯ + 3) − 3(2π‘₯ − 3)
(B) 10π‘₯ − 6
(B) 6(π‘₯ + 1) + 4(π‘₯ − 3)
(C) 10π‘₯ + 16
(D) −4π‘₯ + 7𝑦
(C) 4(π‘₯ + 5) + 2(3π‘₯ − 2)
(D) 5(π‘₯ + 2𝑦) + 3(−3π‘₯ − 𝑦)
(E) 12π‘₯ + 3
(F)
(E) 6(3π‘₯ − 4) − 3(2π‘₯ − 9)
(F) −4(π‘₯ + 2) + 3(2π‘₯ − 5)
(H) (π‘₯ − 3𝑦)(−2) + (π‘₯ − 𝑦)(6)
(G) −7(2π‘₯ − 1) − 4(3π‘₯ + 2)
2π‘₯ − 23
(G) −26π‘₯ − 1
(H) 4π‘₯
(I)
−2π‘₯ − 5
(J)
4π‘₯ + 5
(K) π‘₯ − 9
(L) 5π‘₯ − 8
(I) 3(−2π‘₯ + 1) − 4(−π‘₯ + 2)
(J) −(3 + 2π‘₯) + 2(4 + 3π‘₯)
(M) 11π‘₯ − 8
(N) π‘₯ +
(K) 3(π‘₯ − 1) −
1
1
(4π‘₯
2
+ 6)
(L)
2
1
(3π‘₯
3
1
4
(M) 2 (6π‘₯ − 4) − 3 (−12π‘₯ + 9)
− 6)
2
− (−10π‘₯
5
1
3
15
4
+ 15)
1
4
(N) 12 ( π‘₯ − ) − (8π‘₯ − 1)
7π‘₯ − 10𝑦 − 5
5. Add 4π‘₯ − 5𝑦 + 2 to 3π‘₯ − 5𝑦 − 7
−6π‘₯ + 8𝑦 − 13
6. Subtract 4π‘₯ − 5𝑦 + 7 from −2π‘₯ + 3𝑦 − 6
7. Find an algebraic expression; if we add this expression to −2π‘₯ + 1, then its result is 4π‘₯ +
6π‘₯ + 4
5
8. In a square, each row and column has the same sum. Find the expression 𝐴
−π‘₯ − 3
𝐴
5π‘₯ + 3
2π‘₯
3π‘₯ + 1
𝐴 = −2
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7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 3-4 MORE SIMPLICATION OF ALGEBRAIC EXPRESSIONS
ORDER OF OPERATIONS
(1) Apply all exponents
2π‘₯ − 3{π‘₯ + 2(π‘₯ − 𝑦)}
(2) Simplify parentheses from inside one
= 2π‘₯ − 3{π‘₯ + 2π‘₯ − 2𝑦} : apply distributive law
(3) Do any multiplication or division
= 2π‘₯ − 3{3π‘₯ − 2𝑦} : simplify inside parentheses
(4) Do addition or subtraction from left to right
= 2π‘₯ − 9π‘₯ + 6𝑦 : apply distributive law
= −7π‘₯ + 6𝑦 : simplify like terms
Example 1: Simplify 3{π‘₯ − 2(4 − π‘₯)}.
Solution
3{π‘₯ − 2(4 − π‘₯)}
= 3{π‘₯ − 8 + 2π‘₯)} : apply distributive law in inside parentheses
= 3{3π‘₯ − 8)} : simplify inside parentheses
= 39π‘₯ − 24 : apply distributive law in the parentheses
Example 2: Simplify the following expression
(A) 7π‘₯ − {5π‘₯ + (6π‘₯ − 4)}
(B) 3π‘Ž − {7π‘Ž − (2π‘Ž − 3)}
(A) −4π‘₯ + 4
(B) −2π‘Ž − 3
(C) −3π‘₯ + 41𝑦
(D) 13π‘₯ − 3𝑦 − 60
(E) −11π‘₯
(F)
(C) 3{2π‘₯ − 3(π‘₯ − 4𝑦)} + 5𝑦
3
(E) −2{4π‘₯ − 3(π‘₯ − 𝑦)} − 4 (12π‘₯ − 8𝑦)
(D) 3(π‘₯ − 𝑦) − 5{π‘₯ + 3(4 − π‘₯)}
(F) 2.5(π‘₯ − 𝑦) − 1.3(2π‘₯ − 𝑦)
0.1π‘₯ − 1.2𝑦
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7 GRADE-I: DR. EUNKYUNG YOU
Example 3: Simplify
(A)
3π‘₯−4
2π‘₯−5
+
2
3
(B)
3π‘₯+1
4π‘₯−2
−
5
3
Solution:
3π‘₯−4
2π‘₯−5
+ 3 :
2
Use the LCM(2,3) = 6
=
3(3π‘₯−4)
2(2π‘₯−5)
+ 6 :
6
=
3(3π‘₯−4)+2(2π‘₯−5)
6
=
9π‘₯−12+4π‘₯−10
6
=
13π‘₯−22
6
Make those as fractions which have same denominator (DO NOT FORGET parentheses)
: Write as one fraction
: apply the distributive law
: simplify
3π‘₯+1
4π‘₯−2
− 3 :
5
Use the LCM(5,3) = 15
=
3(3π‘₯+1)
5(4π‘₯−2)
−
:
15
15
=
3(3π‘₯+1)−5(4π‘₯−2)
15
=
9π‘₯+3−20π‘₯+10
15
=
−11π‘₯+13
15
Make these as fractions which have same denominator (DO NOT FORGET parentheses)
: Write as one fraction
: apply the distributive law
: simplify
Example 4: Simplify the following expression
(A) π‘₯ +
π‘₯−2
3
(B)
3π‘₯+5
π‘₯−3
+
4
2
(A)
(B)
(C)
(D)
(C)
5π‘₯−4
2π‘₯−1
− 3
6
(D)
3π‘₯+2
2π‘₯−1
2π‘₯−3
− 3 + 6
5
4
3
5
4
1
6
π‘₯−
2
π‘₯−
1
π‘₯−
1
4
15
3
4
3
π‘₯+
7
30
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7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1. Simplify
(A)
(C)
π‘₯+3
2
(A)
+
2π‘₯−4
5
4π‘₯−5
−
3
(B)
2π‘₯
(D)
π‘₯+5
2
−
π‘₯−1
5
(B)
9
10
3
10
π‘₯+
π‘₯+
(E) −2 + π‘₯ +
3π‘₯−5
2π‘₯
− 3
6
5
3
3
1
5
(D) − π‘₯ −
6
(F)
(G)
(G)
−π‘₯−1
2π‘₯−5
5π‘₯−3
− 3 + 4
2
(H)
π‘₯+2
3
−
2π‘₯−1
2
10
(C) − π‘₯ −
2π‘₯−1
π‘₯+2
− 2
3
(F)
27
2
(E) 2π‘₯ −
4π‘₯+7
4
7
10
+
3π‘₯−5
6
1
6
π‘₯−
1
12
6
1
4
4
3
π‘₯+
5
12
1
1
6
3
(H) − π‘₯ +
(A) −4π‘Ž + 5
2. Simplify
(B) 4π‘Ž − 3
(A) π‘Ž − {2π‘Ž + (3π‘Ž − 5)}
(C) −3π‘₯ + 3
(D) −3π‘₯ + 6
(B) 3π‘Ž − {π‘Ž − (2π‘Ž − 3)}
(E) 7π‘₯ − 4𝑦
(F)
(C) 4π‘₯ − {5π‘₯ − (3 − 2π‘₯)}
4π‘₯ − 19𝑦
(G) −2π‘₯ + 20
(H) −π‘₯ + 8
(D) 8π‘₯ − {9π‘₯ + 2(π‘₯ − 3)}
(E) 5{π‘₯ − (2𝑦 − π‘₯)} + 3(−π‘₯ + 2𝑦)
(F) 3(π‘₯ − 3𝑦) − {6π‘₯ − π‘₯ − 2(3π‘₯ − 5𝑦)}
(G) 5π‘₯ + 8 − {3π‘₯ − (5 − 4π‘₯) − 7}
(H) −2π‘₯ − {−(4 − 5π‘₯) − 2(3π‘₯ + 2)}
65 2
π‘₯ cm2
2
3. Find the area of the shaded region in the figure
5x cm
10x cm
5x cm
8x cm
4. In a square, each row, column, and diagonal has the same sum. Find 𝐴, 𝐡
𝐴 = 4π‘₯ − 2 𝐡 = −2π‘₯
𝐡
5π‘₯ + 1
π‘₯−1
−4
5. Find the blank : −3π‘₯ + 2 +
−3π‘₯ − 3
𝐴
= −2π‘₯ + 4
π‘₯+2
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7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 4. LINEAR EQUATIONS
CHAPTER 4.1 SIMPLE LINEAR EQUATIONS IN ONE VARIABLE.
LINEAR EQUATIONS: A linear equation in one
variable is an equation that can be written in the form,
π‘Žπ‘₯ + 𝑏 = 𝑐, where π‘Ž, 𝑏, and 𝑐 are numbers.
πŸπ’™ + πŸ“ = 𝟏𝟏
x
SOLUTION FOR AN EQUATION: A value of a variable
2π‘₯ + 5 − 5 = 11 − 5
Subtract 5 from both sides
that makes an equation true.
2π‘₯ = 4
Divide both sides by 2
PROPERTIES OF EQUATIONS: If π‘Ž = 𝑏, then
2π‘₯ 4
=
2
2
𝒙=𝟐
Solution for 2π‘₯ + 5 = 11
 π‘Ž+𝑐 =𝑏+𝑐
 π‘Ž−𝑐 =𝑏−𝑐
 π‘Žπ‘ = 𝑏𝑐

π‘Ž
𝑐
𝑏
𝑐
= , 𝑐≠0
Example 1: Solve the equation.
(A) π‘₯ − 3 = 7
(B) π‘₯ + 2 = 5
(C) −4π‘₯ = 12
2
3
(D) π‘₯ = −2
Solution
π‘₯−3=7
π‘₯−3+3=7+3
π‘₯+2 = 5
π‘₯+2−2 = 5−2
Add 3 to both sides
π‘₯ = 10
π‘₯=3
−4π‘₯ = 12
1
π‘₯ = −2
3
3 2
3
( π‘₯) = (−2)
2 3
2
−4π‘₯ 12
=
−4
−4
π‘₯ = −3
Divide both sides by −4
Subtract 2 from both sides
3
Multiply both side by 2
π‘₯ = −3
Example 2: Solve the equation.
(A) π‘₯ + 8 = −3
(B) π‘₯ − 2 = 11
(A) −11
(B) 13
(C) 24
(C) π‘₯ + 3 = 27
(D) π‘₯ − 9 = 5
(D) 14
(E) −12
(F)
(E) π‘₯ + 27 = 15
(F) π‘₯ − 23 = 18
41
72
7 GRADE-I: DR. EUNKYUNG YOU
Example 3: Solve the equation.
(A) 3π‘₯ = 12
(B) 4π‘₯ = −32
(A) 4
(B) −8
(C) −7
(D) 4
(E) 10
(C) −6π‘₯ = 42
(D) −7π‘₯ = −28
(F)
−4
(G) −4
(H) 12
(E)
π‘₯
5
3
=2
(F) − 4 π‘₯ = 3
1
2
(G) −π‘₯ + 3 = 7
(H) − π‘₯ = −6
Example 4: Solve the equation 2π‘₯ − 3 = 5π‘₯ + 4
Solution
πŸπ’™ − πŸ‘ = πŸ“π’™ + πŸ’
2π‘₯ − 3 + 3 = 5π‘₯ + 4 − 5
Separate variable terms(RHS) and constant terms(LHS)
Add 5 from both sides
2π‘₯ = 5π‘₯ − 1
2π‘₯ − 5π‘₯ = 5π‘₯ − 5π‘₯ − 1
Subtract 5π‘₯ from both sides
−3π‘₯ = −1
−3π‘₯ −1
=
−3
−3
𝟏
𝒙=πŸ‘
Divide both sides by −3
Solution for 2π‘₯ − 3 = 5π‘₯ + 4
Example 5: Solve the equation.
(A) 4π‘₯ = 3π‘₯ + 7
(B) π‘₯ = 6 − 2π‘₯
(A) π‘₯ = 7
(B) π‘₯ = 2
(C) π‘₯ = 2
(D) π‘₯ = 1
(C) 4 − π‘₯ = 5π‘₯ − 8
(D) 7 − 10π‘₯ = 2π‘₯ − 5
(E) π‘₯ = 4
(F)
(E) 2π‘₯ − 3 = 7π‘₯ − 23
(F) 19 − 6π‘₯ = 2π‘₯ + 3
π‘₯=3
73
7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
(A) 13
1. Simplify
(A) π‘₯ − 5 = 8
(B) 17
(B) π‘₯ + 6 = 23
(C) 3
(D) 29
(C) π‘₯ − 11 = −8
(D) π‘₯ + 23 = 52
(E) −24
(F)
(E) 3π‘₯ = −72
(G) 18
(F) −2π‘₯ = 46
2
−23
(H) 10
8
(G) 3 π‘₯ = 12
(H) − 5 π‘₯ = −16
(A) 4
2. Simplify
(A) 3π‘₯ − 5 = 7
(B) 5π‘₯ = −π‘₯ + 4
(B)
2
3
(C) 4
(D) 5
(C) 4π‘₯ + 1 = 2π‘₯ + 9
(D) 6π‘₯ − 8 = 2π‘₯ + 12
(E) −
(F)
(E) 8 − 3π‘₯ = 7π‘₯ + 12
(F) 2π‘₯ − 3 = 5π‘₯ + 9
4
10
−4
(G) 7
(H) −2
(G) −7π‘₯ + 23 = −3π‘₯ − 5
(H) −4 − 6π‘₯ = 12 + 2π‘₯
(I)
5
(J)
−3
(K) −3
(I) 4π‘₯ + 3 = 18 − π‘₯
(J) 0.2π‘₯ + 1.2 = 0.6
(L) 2
(M) −6
(N) 6
(K) 0.3π‘₯ + 1.5 = 0.6
(L) 0.2π‘₯ − 0.8 = 1.3π‘₯ − 3
(O) 12
(P)
(M) 0.3π‘₯ + 0.9 = 0.4π‘₯ + 1.5
(N) 0.08π‘₯ − 0.3 = 0.12π‘₯ − 0.54
5
(Q) 2
(R) 6
(S)
3
4
7
5
(O) π‘₯ − 1 = 8
(P) π‘₯ + 2 = 9
3
1
π‘₯
3
−6 = +4
(Q) 2 π‘₯ + 1 = − 2 π‘₯ + 5
(S)
π‘₯
2
5
(U) 5 + 3 π‘₯ = π‘₯ − 4
−60
(T) 35
(U) −6
3
1
(R) 4 π‘₯ + 1 = − 4 π‘₯ + 7
(V) −4
5
(T) 6 + 7 π‘₯ = π‘₯ − 4
2
5
(V) 2 − 3 π‘₯ = 6 π‘₯ + 8
3. Lisa is cooking muffins. The recipe calls for 7 cups of sugar. She has already put in 2 cups.
How many more cups does she need to put in?
5 cups
4. At a restaurant, Mike and his three friends decided to divide the bill evenly. If each person
paid $13 then what was the total bill?
$52
74
7 GRADE-I: DR. EUNKYUNG YOU
5. How many packages of diapers can you buy with $40 if one package costs $8?
5 package
6. Last Friday Trevon had $29. Over the weekend he received some money for cleaning the
attic. He now has $41. How much money did he receive?
$12
7. Last week John ran 30 miles more than Tom. John ran 47 miles. How many miles did Tom
run?
17 miles
8. How many boxes of envelopes can you buy with $12 if one box costs $3?
4 boxes
9. Amanda and her best friend found some money buried in a field. They split the money
evenly, each getting $24.28. How much money did they find?
$48.56
10. Jenny wants to buy an MP3 player that costs $30.98. How much change does she receive if
she gives the cashier $40?
$9.02
11. After paying $5.12 for a salad, Tom has $27.10. How much money did he have before
buying the salad?
$32.22
1
4
1
9
36
3
$22.2
12. A recipe for cookies calls for 3 cups of sugar. Amy has already put in 3 cups. How
5
cups
many more cups does she need to put in?
13. Your mother gave you $13.32 with which to buy a present. This covered 5 of the cost. How
much did the present cost?
75
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 4.2 MORE LINEAR EQUATIONS
Example 1: Solve the equation
(A) 2(π‘₯ − 3) = 3(2π‘₯ − 5)
(B)
3π‘₯−2
5
=
3π‘₯−5
6
Solution
𝟐(𝒙 − πŸ‘) = πŸ‘(πŸπ’™ − πŸ“)
2π‘₯ − 6 = 6π‘₯ − 15
Apply distributive law
2π‘₯ − 6 + 6 = 6π‘₯ − 15 + 6
Add 6 from both sides
2π‘₯ = 6π‘₯ − 9
2π‘₯ − 6π‘₯ = 6π‘₯ − 6π‘₯ − 9
Subtract 6π‘₯ from both sides
−4π‘₯ = −9
−4π‘₯
−4
= −4
−9
π‘₯=
9
4
πŸ‘π’™−𝟐
πŸ“
30 ×
=
Divide both sides by −4
πŸ‘π’™−πŸ“
πŸ”
(3π‘₯−2)
5
= 30 ×
(3π‘₯−5)
Multiply both sides by LCM(5,6) = 30
6
6(3π‘₯ − 2) = 5(3π‘₯ − 5)
Simplify
18π‘₯ − 12 = 15π‘₯ − 25
Apply distributive law
18π‘₯ − 15π‘₯ − 12 = 15π‘₯ − 15π‘₯ − 25
Subtract 15π‘₯ from both sides
3π‘₯ − 12 = −25
3π‘₯ − 12 + 12 = −25 + 12
Add 12 from both sides
3π‘₯ = −13
3π‘₯
3
=
−13
3
π‘₯=−
Divide both sides by 3
13
3
Example 2: Solve the equation
(A) 5π‘₯ − 3(π‘₯ − 1) = 9
(B) −2(π‘₯ + 1) = 3π‘₯ + 8
(A) 3
(B) −2
(C) 3
(D) 9
(C) 4(π‘₯ + 3) = 27 − π‘₯
1
(D) π‘₯ − 2 (π‘₯ − 1) = 5
76
7 GRADE-I: DR. EUNKYUNG YOU
Example 3: Solve the equation
(A) 7(2 − π‘₯) = −5π‘₯ + 4
(B) −4π‘₯ + 6 = −2(π‘₯ + 5)
(A) 5
(B) 8
(C)
18
5
(D) −9
(E)
(F)
(G)
(C) 3(π‘₯ − 4) = 2(−π‘₯ + 3)
(D) 3π‘₯ + 9 = 2(2π‘₯ + 9)
(H)
5
11
−
31
5
5
4
9
4
(I)
11
(J)
3
(K) −19
(L) −
(E) 2(5π‘₯ − 3) = 4(1 − 3π‘₯)
(F) 4(3π‘₯ − 1) = 7(π‘₯ − 5)
(G) −2(2π‘₯ − 7) = 3(4π‘₯ − 2)
(H) 3(7 − 2π‘₯) − 3π‘₯ = 3(π‘₯ − 2)
(I)
2π‘₯−7
5
=3
(K)
5π‘₯+3
2
=
7π‘₯−5
3
(J)
5π‘₯−9
3
=2
(L)
2π‘₯−7
3
=
11π‘₯−3
5
23
26
77
7 GRADE-I: DR. EUNKYUNG YOU
π‘₯
3
Example 4: Solve the equation : +
π‘₯−5
2
=2
Solution
π‘₯
3
+
π‘₯−5
2
=2
π‘₯
6 × (3 +
π‘₯−5
)
2
π‘₯
=6×2
π‘₯−5
)
2
6 (3 ) + 6 (
= 12
Multiply both sides by LCM(3,2) = 6
Apply distributive law
2π‘₯ + 3(π‘₯ − 5) = 12
2π‘₯ + 3π‘₯ − 15 = 12
Apply distributive law
5π‘₯ − 15 = 12
Simplify
5π‘₯ − 15 + 15 = 12 + 15
Add 15 from both sides
5π‘₯ = 27
5π‘₯
5
=
π‘₯=
27
5
Divide both sides by 5
27
5
Example 5: Solve the equation
π‘₯
(A) 3 +
(B)
π‘₯−5
4
=2
3π‘₯−2
π‘₯−2
− 5
4
=3
39
7
62
11
78
7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
(A) −5.2
1. Simplify
(A) 3(π‘₯ − 2) = 4(2π‘₯ + 5)
(B)
(B) 2(4π‘₯ − 1) = 3(5π‘₯ − 2)
(C)
(C) 5(π‘₯ − 1) = −2(3π‘₯ − 1)
(D) 5 − 2(3π‘₯ + 1) = 3(5 − π‘₯)
(F) 0.02(π‘₯ − 8) = 0.3π‘₯ + 0.2
7
7
11
(D) −4
(E)
(E) 0.2(3π‘₯ + 2) = 0.4(8 − π‘₯)
4
(F)
14
5
−
7
9
(G) 12
(H) 2
(G) 2(π‘₯ + 5) + 2 = 4(π‘₯ − 3)
(H) 0.3(2π‘₯ − 3) = 0.2(π‘₯ + 2) − 0.5
(I)
(J)
(I) 4(π‘₯ − 5) = −(2π‘₯ − 16)
(J) 2(π‘₯ − 3) = 3(π‘₯ − 8) + 7π‘₯
6
9
4
(K) −1
(L) 3
1
1
1
(K) 0.6π‘₯ − 5 = 3 (10 π‘₯ − 6)
(L) 0.6π‘₯ − 1 = 4(0.3π‘₯ − 0.7)
2. Simplify
(A)
(A) −
2π‘₯+1
− 4
= 3(π‘₯ + 1) + 1
1
1
4
(B)
3π‘₯+1
8
1
(C) 6 (π‘₯ − 2) = 4 π‘₯ + 3
(D) 2 −
π‘₯+7
6
=
π‘₯−1
3
+
π‘₯+3
2
(C) −20
(D) 0
2−π‘₯
4
= 0.5π‘₯
1
=1
14
(B) 5
(E) −
(F)
(E)
17
(F) 3 π‘₯ − 0.2π‘₯ =
2π‘₯−7
5
1
5
21
4
(G) 3
(H) 16
2π‘₯+1
3
=
(I)
9π‘₯−5
2
(K)
2π‘₯
3
(M)
π‘₯+2
2
(O)
3π‘₯+4
2
(G)
5π‘₯−1
6
(H) 2 π‘₯ =
1
π‘₯+5
+
3
=
π‘₯−7
3
(J)
5π‘₯−9
+
4
3 = 2π‘₯
5π‘₯
4
π‘₯
6
(L)
𝑦+11
2
𝑦−3
π‘₯
−
4
3
1
(I)
−
(J)
1
(K) −
4
15
32
7
(L) 15
(M) −10
+
−
= −8
2π‘₯−1
3
=
−π‘₯+3
4
4
3
3
4
5
2
(N) (π‘₯ + ) = −
= 1.05(π‘₯ − 2) −
2
5
1
4
(N)
17
4
1−5π‘₯
4
(P) 0.3(π‘₯ − 2) + = 0.1π‘₯ +
1
2
(A) 4
3. Simplify
(A) 5 − {2 − (3π‘₯ − 6)} = π‘₯ + 5
1
3
=
1
2
1
6
π‘₯
2
(B) 2 ( π‘₯ + ) − 3 { − ( − 1)} = 0.5π‘₯ + 1
(B)
21
10
79
7 GRADE-I: DR. EUNKYUNG YOU
4. 331 students went on a field trip. Six buses were filled and 7 students traveled in their own
cars. How many students were in each bus?
54
5. Aliyah had $24 to spend on seven pencils. After buying them she had $10. How much did
each pencil cost?
$2
6. The sum of three consecutive numbers is 72. What are the smallest of these numbers?
23
7. The sum of three consecutive even numbers is 48. What are the smallest of these numbers?
14
8. Maria bought seven boxes. A week later half of all her boxes were destroyed in a fire.
There are now only 22 boxes left. With how many did she start?
37 boxes
9. Mary spent half of her weekly allowance playing mini-golf. To earn more money her
parents let her wash the car for $4. What is her weekly allowance if she ended with $12?
$16
10. Mary had some candy to give to her four children. She first took ten pieces for herself and
then evenly divided the rest among her children. Each child received five pieces. With how
many pieces did she start?
30 pieces
11. For a field trip 4 students rode in their own cars and the rest filled nine buses. How many
students were in each bus if 472 students were on the trip?
52
80
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 4-4 SIMPLE FRATION EQUATIONS
FRACTION EQUSTIONS: When the variable of an
2π‘₯
=3
π‘₯−3
2π‘₯
3
=
π‘₯−3 1
equation is in the denominator of a term, this equation is
called a fraction equation.
2π‘₯(1) = 3(π‘₯ − 3)
2π‘₯ = 3π‘₯ − 9
Apply distributive law
2π‘₯ − 3π‘₯ = 3π‘₯ − 3π‘₯ − 9
Subtract 3π‘₯ from both
HOW TO SOLVE A FRACTION EQUATION?
(1) Make both sides as one fraction
(2) Put Cross products same
sides
(3) Solve the equation
−π‘₯ = −9
(4) Check your answer
π‘₯=9
Example 1: Solve the equation
(A)
3
π‘₯−2
=2
(B)
3π‘₯−2
π‘₯−1
=2
Solution
3
π‘₯−2
=2
3
π‘₯−2
=
2
1
3(1) = 2(π‘₯ − 2)
Cross products are same
3 = 2π‘₯ − 4
Apply distributive law
3 + 4 = 2π‘₯ − 4 + 4
Add 4 from both sides
7 = 2π‘₯
Divided both side by 2
7
2
Check :
=π‘₯
3π‘₯−2
π‘₯−1
=2
3π‘₯−2
π‘₯−1
=1
Cross products are same
3
7
2
( )−2
=
3
3
2
=2
2
(3π‘₯ − 2)(1) = 2(π‘₯ − 1)
Cross products are same
3π‘₯ − 2 = 2π‘₯ − 2
Apply distributive law
3π‘₯ − 2π‘₯ − 2 = 2π‘₯ − 2π‘₯ − 2
Subtract 2π‘₯ from both sides
π‘₯ − 2 = −2
π‘₯ − 2 + 2 = −2 + 2
Add 2 from both side
π‘₯=0
Check :
3(0)−2
0−1
−2
= −1 = 2
Divided both side by −1
2(9)
Check : 9−3 =
18
6
=3
81
7 GRADE-I: DR. EUNKYUNG YOU
Example 2: Solve the equation
(A)
2
π‘₯−3
=4
(B)
2
2π‘₯−1
= −3
(A)
(B)
(C)
7
2
1
6
17
15
20
(D)
11
(E) No solution
21
(F)
2
3
(C) 4−3π‘₯ = 5
(E)
2π‘₯
3−π‘₯
= −2
(D)
π‘₯
3π‘₯−5
=4
(F)
4π‘₯−2
2π‘₯+3
=3
5
82
7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1. Simplify
4
(A) π‘₯ = 2
(B)
8
π‘₯
= −3
3
π‘₯
15
− 4
(A) 2
8
(B) −
(C)
(D)
2
π‘₯
(C) − 5 = 0
(E)
3
π‘₯
(D)
−2 = 5
(F)
5
π‘₯−3
=0
(E)
(F)
(G)
=3
(H)
(I)
(G)
3
2π‘₯−1
=4
(I)
7
2π‘₯−5
=3
2
2. Simplify
4π‘₯
(A) π‘₯−2 = 1
(C)
2π‘₯
3π‘₯−5
(H)
5
3−4π‘₯
= −2
(J)
4
3π‘₯−2
= −4
(B)
π‘₯
π‘₯−3
=5
π‘₯+1
2
(J)
2
5
3
7
14
3
7
8
11
4
31
4
1
3
(A) −
(B)
(H)
(E)
3π‘₯+2
2π‘₯−7
=4
(F)
1
π‘₯+3
(G)
π‘₯−2
4−3π‘₯
=
5
3
(H)
5π‘₯+3
π‘₯−7
2
3
4
5
9
1
11
2
= 3π‘₯
=−
1
2
16
3. Find the value of the expression
π‘Ž−𝑏
(A)
when π‘Ž = 4, 𝑏 = −32, and 𝑐 = 9
(A)
9
(B) 6
(C) 25
2𝑐
π‘Ž
15
(C)
2
(D) −11
(E) 6
(F) 6
13
(G)
(D) π‘₯−4 = 3
=2
3
5
4
1
(B) 1−π‘Ÿ when π‘Ž = 4 and π‘Ÿ = 3
1
π‘Ž
2
𝑏
3
𝑐
1
4
1
3
(C) − + when π‘Ž = , 𝑏 = − , and 𝑐 =
π‘˜π‘Ž
1
5
1
4. Find the value of 𝑏 if 𝑇 = π‘Ž+𝑏 where 𝑇 = 2 2 , π‘Ž = 5, and π‘˜ = 4
1
1
1
1
1
5. Find the value of π‘Ž if π‘Ž + 𝑏 = 𝑐 where 𝑏 = − 5, and 𝑐 = 3
3π‘Ž
6. Find the value of π‘Ÿ if 𝐴 = 1−π‘Ÿ where 𝐴 = −6 and π‘Ž = 2
𝑏=3
π‘Ž=
1
8
π‘Ÿ=2
83
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 4.5 PERCENTAGE APPLICATION
PERCENTAGE : A percentage is a fraction with 100 as
PERCENT AND EQUATION:
the denominator
When 𝑦 is π‘Ž % of π‘₯,
100
=1
100
 Percentage to decimal:
𝑦 = π‘Ž × 0.01 × π‘₯
100% =
π‘₯%=π‘₯×
1
or π‘₯ × 0.01
100
 Decimal to percentage: 𝑦 = 𝑦 × 100 %
Example 1: Write each expression as a percentage
(A) 0.32
2
(B) 0.257
(C) 2 3
Solution:
(A) 0.32 = 0.32 × 100 % = 32%
(B) 0.257 = 0.257 × 100 % = 25.7 %
2
5
(C) 2 =
12
5
=
12
×
5
100 % = 240 %
(A) 75%
650
(B)
%
3
(C) 98.7%
(D) 120%
Example 2: Write each expression as a percentage
3
1
(A) 4
(B) 2 6
(C) 0.987
(D) 1.2
Example 3: Write each expression as decimals
(A) 45%
(B) 2.5%
(C) 265%
Solution:
(A) 45 % = 45 × 0.01 = 0.45
(B) 2.5 % = 2.5 × 0.01 = 0.025
(C) 256 % = 256 × 0.01 = 2.56
Example 3: Write each expression as decimals
(A) 3%
(B) 46 %
(C) 2.35 %
(D) 673 %
(A)
(B)
(C)
(D)
0.03
0.46
0.0235
6.73
84
7 GRADE-I: DR. EUNKYUNG YOU
Example 4: Solve the each problems.
(A) What is 270% of 60?
(B) What percent of 125 is 45?
(C) 81 is 25 % of what?
Solution:
(A) π‘Ž is 270 % of 60
decimal
π‘Ž=⏞
270 × 0.01 × 60
⇒ π‘Ž = 162
(B) 45 is π‘₯ % of 125
decimal
⏞
45 = 0.01π‘₯
× 125 ⇒ 45 = 1.25 π‘₯ ⇒ π‘₯ =
45
= 36 %
1.25
(C) 81 is 56 % of 𝑦
decimal
81 = ⏞
25 × 0.01 × π‘¦ ⇒ 81 = 0.25𝑦
⇒ 𝑦=
81
= 324
0.25
(A) 24
(B) 155
Example 5: Solve the each problems
(A) What is 30% of 80?
(B) What is 25% of 620?
(C)
100
7
%
(D) 9%
(E) 800
(C) What percent of 17500 is 2500?
(D) What percent of 25000 is 2250?
(E) 200 is 25 % of what?
(F) 80 is 60 % of what?
(F)
400
3
Example 6: The monthly budget for the front of the house is $5,000. You spent 10% of the budget on fresh flowers.
How much did you spend on fresh flowers?
Solution: $ π‘Ž is 10 % of $5000
π‘Ž = 10 × 0.01 × 5000 ⇒ π‘Ž = 500
Therefore, you spend $500 on fresh flowers
Example 7: You have a large container of olive oil. You have used 25 quarts of oil. Thirty
7.5 quarts
percent of the olive oil remains. How many quarts of olive oil remain?
Example 8: Out of 3500 students of a school, only 36% passed. Find how many students failed.
2240 students
85
7 GRADE-I: DR. EUNKYUNG YOU
Example 9: Emily just hired a new employee to work in your bakeshop. In one hour the employee burned 750
chocolate chip cookies. If this represented 15% of the day’s production, how many cookies did you plan on producing
that day?
Solution: 750 is 15 % of π‘₯ production
750 = 15 × 0.01 × π‘₯ ⇒ 750 = 0.15π‘₯ ⇒ π‘₯ =
750
= 5000
0.15
Therefore, they produce 5000 cookies
Example10: In a survey, 1023 people, which represent 93 %, answered yes. How many people
1100 people
were surveyed?
Example 11: 45% of Mary’s class are girls. There are 72 girls. How many student are in
160 students
Mary’s class?
Example 12: Your food costs are $1700. Your total food sales are $8500. What percent of your food sales do the food
costs represent?
Solution: $1700 is 𝑦 % of $8500
1700 = 𝑦 × 0.01 × 8500 ⇒ 1700 = 85𝑦
⇒𝑦=
1700
= 20
85
Therefore, the cost of your food is 20 % of the total food sales.
Example 13: A serving of ice cream contains 1200 calories. One hundred forty-four calories
12%
come from fat. What percent of the total calories come from fat?
Example 14: 3082 students out of 6700 students in ABAC passed. What percent of student
passed?
46%
86
7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1
A baseball pitcher won 80% of the games he pitched. If he pitched 35 ballgames, how
28 games
many games did he win?
2
A student earned a grade of 80% on a math test that had 20 problems. How many
16 problems
problems on this test did the student answer correctly?
3
Jerry, an electrician, worked 7 months out of the year. What percent of the year did he
1
58 %
3
work?
4
Manuel found a wrecked Trans-Am that he could fix. He bought the car for 65% of the
$4680
original price of $7200. What did he pay for the car?
5
Ben earns $12,800 a year. About 15% is taken out for taxes. How much is taken out for
$1920
taxes?
6
Alma brought balls of green and blue color. Thirty five percent of balls are blue. If she
130
brought total of 70 blue balls, how many green balls she had?
7
There are 28 students in a class. Sixteen of those students are men. What percent of the
6
42 %
7
class are women?
8
There are 36 carpenters in a crew. On a certain day, 29 were present. What percent
5
80 %
9
showed up for work?
9
Donovan took a math test and got 35 correct and 10 incorrect answers. What was the
7
693 %
9
percentage of correct answers?
10
There are 32 students in a class. Nine of those students are women. What percent are
71.875 %
men?
11
The Royals softball team played 75 games and won 55 of them. What percent of the
1
73 %
3
games did they lose?
12
A metal bar weighs 8 ounces. 93% of the bar is silver. How many ounces of silver are in
7.44 ounces
the bar?
13
A woman put $580 into a savings account for one year. The rate of interest on the
$ 34.8
account was 6%. How much was the interest for the year in dollars and cents?
14
A student answered 147 problems on a test correctly and received a grade 98%. How
150 problems
many problems were on the test, if all the problems were worth the same number of
points?
15
Pamela bought an electric drill at 85% of the regular price. She paid $32 for the drill.
$40
What was the regular price?
16
At a sale, shirts were sold for $15 each. This price was 80% of their original price. What
was the original price?
$18.75
87
7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 4.6 MIXTURE PROBLEMS
MIXTURE WORD PROBLEMS:
INTEREST
interest = interest rate(decimal) × amount money
amont of material
= concentation × amount of solution
Example 1: How many liters of water should be added to 200 liters of a 8% salt solution to make a 5% solution?
Solution: Let π‘₯ be a volume of water in liters which is added to a 8% solution: its concentration is 0% (no salt)
Liters of solution Percent salt Total amount of salt
8%
200
0.08
16
0%
π‘₯
0
0
5%
200 + π‘₯
0.05
0.05(200 + π‘₯)
Then
16 = 0.05(200 + π‘₯)
16 = 10 + 0.05π‘₯
6 = 0.05π‘₯
π‘₯=
6
= 120
0.05
Therefore, 120 g of water should be added.
Example 2: How much water should be added to 200 liters of a 10% salt solution to get a 2%
800 liters
salt solution?
Example 3: How much water must be evaporated from 500 liters of a 2% salt solution to get a
100 liters
10% salt solution?
Example 4: How much water must be evaporated from 100 g of a 3% salt solution to make a
60 g
5% salt solution?
Example 5: When 320 liters of a salt solution is added to 80 liters of water to make a 8% salt
10%
solution, find the percent concentration of the salty solution.
Example 6: How many grams of salt should be added to 300 g of a 20% salt solution to make a 40% solution?
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7 GRADE-I: DR. EUNKYUNG YOU
Solution: Let π‘₯ be the a mass of salt in grams which is add to a 20% solution : concentration is 100%
grams of solution Percent salt Total amount of salt
20%
300
0.2
60
100%
π‘₯
1
π‘₯
40%
300 + π‘₯
0.4
0.4(300 + π‘₯)
Then
60 + π‘₯ = 0.4(300 + π‘₯)
60 + π‘₯ = 120 + 0.4π‘₯
0.6π‘₯ = 60
π‘₯=
60
= 100
0.6
Therefore, 120 g of salt should be added.
Example 7: 127 g of water is added to 250 g of a 10% solution. If salt is added this solution to
23 g
make a 12% solution, how many grams of salt should be added?
Example 8: After salt is added to 300 g of a 8% salt solution, equal mass of water evaporates
30 g
to leave a 18% salt solution. How much of the water evaporated?
Example 9: How much salt should be added to 1000 liters of a 2% salt solution to get a 4% salt
solution?
37.5 g
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7 GRADE-I: DR. EUNKYUNG YOU
Example 10: How many grams of a 5% sugar solution should be added to 200 g of a 10% sugar solution to make a 7%
solution?
Solution: Let π‘₯ be a mass of a 5% solution in gram.
Grams of solution Percent salt Total amount of salt
10%
200
0.1
20
5%
π‘₯
0.05
0.05π‘₯
7%
200 + π‘₯
0.07
0.07(200 + π‘₯)
Then
20 + 0.05π‘₯ = 0.07(200 + π‘₯)
20 + 0.05π‘₯ = 14 + 0.07π‘₯
6 = 0.02π‘₯
6
= 300
0.02
Therefore, 130 g of a 5% salt solution should be added.
π‘₯=
Example 11: 320 g of a 7% sugar solution was mixed with 80 g of a sugar solution to make a
12%
8% solution. Find the concentration of the second sugar solution.
Example 12: How many grams of the 11% solution should be added to 800 g of a 7% solution
2400 g
to make a 10% solution?
Example 13: How much of the 10% acid solution should be added to 1000 liters of a 2% acid
solution to get a 5% acid solution?
600 g
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7 GRADE-I: DR. EUNKYUNG YOU
Example 14: A bank loaned out $50,000. Part of the money earned 10 % per year, and the rest of it earned 5 % per
year. If the total interest received for one year was $3,500, how much was loaned at 10%?
Solution: Let π‘₯ be amount of money in 10 % account.
Grams of solution Interest rate Amount of interest
π‘₯
0.1
0.1π‘₯
50000 − π‘₯
0.05
0.05(50000 − π‘₯)
10 %
5%
3500
Then
0.1π‘₯ + 0.05(50000 − π‘₯) = 3500
0.1π‘₯ + 2500 − 0.05π‘₯ = 3500
0.04π‘₯ = 1000
π‘₯=
1000
= 25000
0.04
Therefore, $25,000 is loaned at 10%.
Example 15: A bank loaned out $10,000. Part of the money earned 7 % per year, and the rest
$3000
of it earned 5 % per year. If the total interest received for one year was $560, how much was
loaned at 7%?
Example 16: Tom has $20,000 to invest, some at 4% and some at 10%. If the annual interest is
$1130, how much is invested at 4%?
$14,500
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7 GRADE-I: DR. EUNKYUNG YOU
BASIC PROBLEMS
1
How many liters of water should be added to 300 liters of a 20% salt solution to make a
100 liters
15% solution?
2
How many liters of water should be added to 240 liters of a 25% acid solution to make a
60 liters
20% solution?
3
How much water must be evaporated from 450 liters of a 6% salt solution to get a 9 %
150 liters
salt solution?
4
How much water must be evaporated from 60 ml of a 2% salt solution to get a 3% salt
40 ml
solution?
5
When 160 g of a salt solution is added to 40 g of water to make a 8% salt solution, find
10 %
the percent concentration of the first solution.
6
How much salt should be added to 200 g of a 10% salt solution to get a 20% salt
25 g
solution?
7
When 40 g of salt is added to 300 g of a salt solution to make a twice saltier solution,
12.5%
find the concentration of the first solution.
8
A chemist mixes some 70% solution with some 40% solution to obtain 120 gallons of
40 g
50% solution. Find the number of gallons of the 70% solution.
9
A 25% acid solution must be added to a 40% solution to get 240 liters of 30% acid
160 g
solution. How many of the 25% solution must be used?
10
How much of the 8% acid solution should be added to a 14% acid solution to get 300 g
200g
of a 10 % acid solution?
11
400 g of a 6% sugar solution was mixed with 600 g of a sugar solution to make a 9%
solution. Find the concentration of the 500 g sugar solution.
11%
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7 GRADE-I: DR. EUNKYUNG YOU
12
100 g of a 6% acid solution was mixed with 300g of a 10% acid solution. How much
100 g
water must be evaporated from the resulting solution to get a 12 % acid solution?
13
A certain metal is 40% tin. A 40% tin metal must be mixed with a 70% tin metal to get
90 kg
150 kilograms of metal that is 52% tin. How much of the 40% tin metal must be used?
14
A person had $10,000 to invest; some was invested at 6% and some at 8 %. If the total
$3500
annual interest was $730, find the amount invested at 6% interest rate.
15
Jenn invested some money at 5% and $6000 more than this at 7 %. If the total annual
$110,000
interest is $1740, find the amount invested at 5% rate.
16
Gloria invested some money at 18%, and $3000 less than this at 20%. The total annual
$7,000
interest is $3200. How much is invested at 20% rate?
17
Joe invested some money at 8 %, and $3000 more than twice as much at 10%. The total
$8,000
annual interest was $2540. How much was invested at each rate?
18
Marty inherited a sum of money from a relative. He deposits some of the money at 16%,
$12,000
and $4000 more than this at 12%. He earns $3840 in interest per year. How much is
invested at 16%?
19
Evelyn invested some money at 10%, and $5000 more than this at 14%. Her total annual
$10,000
interest was $3100. How much was invested at 10% rate?
20
A company manufactures some spigots at $25 each and others at $32 each. If the
38
company sold 50 spigots and made $1334, find the number of $25- type spigot sold.
21
A merchant wishes to mix candy worth $5 per pound with candy worth $2 per pound to
30 pounds
get 60 pounds of a mixture that can be sold for $3 per pound. How many pounds of $5type of candy should be used?
22
A grocer mixed $1.80 a pound peanuts with $3.20 a pound chocolate to obtain 50 pounds
10 pounds
of a mixture worth $2.92 a pound. Find the number of pounds of peanuts used.
23
Helen has 2 more dimes than nickels. Altogether she has $1.70. How many dimes does
she have?
12
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7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 4.7 DISTANCE-SPEED APPLICATION
DISTANCE, SPEED, AND TIME:
WARNING: Check the units
distance = speed × time
When Tom drives at 50 mph for 35 minutes,
distance
,
time
distance
time =
speed
50 mph : 50 miles per hour
speed =
35 minutes =
35
60
hour
Example 1: Solve the following problems
(A) Calculate the speed for a car that went a distance of 125 kilometers in 2 hours.
(B) How much time does it take for a bird flying at a speed of 45 mph to travel a distance of 1,800 miles?
(C) A comet is cruising through the solar system at a speed of 50,000 mph for 4 hours. What is the total distance
traveled by the comet during this time?
Solution:
(A) Distance : 125 km, Time : 2 hours Speed : π‘₯ km/h
125 = π‘₯ × 2 ⇒ π‘₯ = 62.5 km/h
(B) Distance : 1800 km, Time : π‘₯ hours Speed : 45 mph
1800 = 45 × π‘₯ ⇒ π‘₯ =
1800
= 40 hours
45
(C) Distance : π‘₯ miles, Time : 5 hours Speed : 50,000 km/h
π‘₯ = 50000 × 5 ⇒ π‘₯ = 250,000 miles
Example 2: An airplane travels 1120 miles in seven hours. Find the rate of the plane in still air.
160 mph
Example 3: Tom plan a 425 mile trip to a museum at 50 mph. How long will it take?
8.5 hours
Example 4: Mary drove to library at 67 mph for 5 hours. Find the driven miles.
335 miles
Example 5: A driver traveled 1325 miles for 25 hours. Find the speed.
53 hours
Example 6: John took 40 minutes to drive from A to B at 60 mph. Find the distance between A
40 miles
and B
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7 GRADE-I: DR. EUNKYUNG YOU
Example 7: John took a drive to town at an average rate of 40 mph. In the evening, he drove back at 20 mph. If he spent
a total of 3 hours traveling, what is the distance traveled by John?
Solution: Let π‘₯ miles be the distance between his house to town.
(driving time to town) + (returning time) = total time
π‘₯
π‘₯
+
=3
40 20
π‘₯ + 2π‘₯ = 120
π‘₯ = 40 miles
The traveled distance is 2π‘₯ = 80 miles
Example 8: Mary took a drive to town at an average rate of 60 mph. In the evening, he drove
360 miles
back at 90 mph. If he spent a total of 5 hours traveling, what is the distance traveled by John?
Example 9: Tom run at an average rate of 2 mph from his house to the library. He returns along
9 miles
the same route at an average rate of 3 mph. If the round running took 7 and a half hours, find
the distance between his house and the library?
Example 10: Susan drove from city A to city B at 50 mph. She returned along the same routes
at 75 mph. She spend 40 minutes more in returning trip. Find the distance between cities A and
B.
20miles
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7 GRADE-I: DR. EUNKYUNG YOU
Example 11: James leaves his home town traveling at 70 mph. At the same time Paul leaves
4 hours
home traveling at 75 mph. The two live 580 miles apart and are traveling to meet each other for
a lunch. How long will it take the two to meet each other?
Example 12: A biker covered half the distance between two towns in 2 hr 30 min. After that he
28 km and 11.2 km/hr
increased his speed by 2 km/hr. He covered the second half of the distance in 2 hr 20min. Find
the distance between the two towns and the initial speed of the biker.
Example 13: Two cyclists start at the same time from opposite ends of a course that is 45 miles
long. One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after
they begin will they meet?
1.5 hours
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7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 5 FUNCTIONS
CHAPTER 5.1 BASIC CONCEPTS OF FUNCTIONS
DEFINITION: In the rectangular coordinate system,
 The horizontal and vertical number lines are
called the π‘₯-axis and the 𝑦-axis respectively.
 The two lines intersect at right angle at the point
(called the origin)
 The position of any point, say P, on the plane can
be expressed by an ordered pair (π‘Ž, 𝑏) where π‘Ž is
the π‘₯-value and 𝑏 is the 𝑦-value. We say that 𝑃
has coordinate (π‘Ž, 𝑏) and refer the point 𝑃 as
𝑃(π‘Ž, 𝑏).
Example 1: Given the figure,
4
(A) State the coordinate of the points 𝐴, 𝐡, 𝐢, 𝐷 and 𝐸.
B
(B) State the quadrants in which those points lie
A
2
(C) Plot the points 𝑃(3,4), 𝑄(−1,2), 𝑅(3, −3),
and 𝑆(−2, −3)
4
2
2
C
2
4
D
E
4
Solution:
4
𝐴
𝐡
𝐢
𝐷
𝐸
Ordered pair
(2,2)
(−1,3)
(−2, −1)
(0, −2)
(3, −3)
Quadrants
𝐼
𝐼𝐼
𝐼𝐼
P
2
Q
𝐼𝑉
5
4
2
4
2
2
R
S
4
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7 GRADE-I: DR. EUNKYUNG YOU
Example 2: Given the figure,
(A) State the coordinate of the points
4
𝐴, 𝐡, 𝐢, 𝐷 and 𝐸.
B
2
(B) State the quadrants in which those
A
points lie
4
2
D
2
4
E
2
C
4
Example 3: Given the figure,
(A) State the coordinate of the points
4
𝐴, 𝐡, 𝐢, 𝐷 and 𝐸.
C
A
(B) State the quadrants in which those
2
points lie
B
5
4
2
2
4
2
D
E
4
Example 4: Plot the points 𝐴(1, −3), 𝐡(4, −3), 𝐢(4,1) and 𝐷(1,1). Then find the area of a
rectangle ABCD.
4
2
5
4
2
2
2
4
4
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7 GRADE-I: DR. EUNKYUNG YOU
DEFINITION: A function relates each element (input,
NOTATION:
usually π‘₯ values) of a set with exactly one element
(output, usually 𝑦 values) of another set.
y = f (x)
y = 3x + 1
f (x) = 3x + 1
same function with
different expressions
Example 5: The price of an apple is $5. Let 𝑦 be the price of π‘₯ apples.
(A) Fill the blank in the table
π‘₯ apples
1
$𝑦
$5
2
3
4
5
β‹―
β‹―
(B) Determine whether 𝑦 is a function of π‘₯ or not
(C) Find the relation between 𝑦 and π‘₯ and write it as an equation.
Solution:
(A)
π‘₯ apples
1
2
3
4
5
$𝑦
$5
$10
$15
$20
$25
(B) Since there is only one price for π‘₯ apples, it is a function.
(C) Since the price is increasing $5 for one more apple, 𝑦 = 5π‘₯
Example 6: The price of a pen is $2. Let 𝑦 be the price of π‘₯ pens.
(A) Fill the blank in the table
π‘₯ apples
1
$𝑦
$2
2
3
4
(B) Determine whether 𝑦 is a function of π‘₯ or not
(C) Find the relation between 𝑦 and π‘₯ and write it as an equation.
5
β‹―
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7 GRADE-I: DR. EUNKYUNG YOU
Example 7: Consider a function 𝑓(π‘₯) = 4π‘₯. Find the value of function;
(A) 𝑓(2)
(B) 𝑓(−3)
(C) 𝑓(−1)
(B) 𝑓(−3) = −12
(C) 𝑓(−1) = −4
Solution:
(A) 𝑓(2) = 8
𝑓(−3) = 4 × (−πŸ‘)
= −𝟏𝟐
⏟
𝑓(2) = 4 × (𝟐) = πŸ–
replace π‘₯ with − 3
replace π‘₯ with 2
Example 8: Evaluate 𝑓(−2) if 𝑓(π‘₯) = 3π‘₯
Example 9: Evaluate 𝑓(−3) if 𝑓(π‘₯) = 2π‘₯ + 3
1
Example 10: Evaluate 𝑓(6) if 𝑓(π‘₯) = 2 π‘₯ − 4
Example 11: Evaluate 𝑓(4) if 𝑓(π‘₯) = 5 − 3π‘₯
4
Example 12: Evaluate 𝑓(−2) if 𝑓(π‘₯) = π‘₯ + 5
Example 13: Find the value of constant π‘Ž if 𝑓(π‘₯) = 3π‘₯ + π‘Ž and 𝑓(2) = 0
𝑓(−1) = 4 × (−𝟏)
= −πŸ’
⏟
replace π‘₯ with − 1
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7 GRADE-I: DR. EUNKYUNG YOU
CHAPTER 5.2 THE GRAPH OF A LINEAR RELATION.
 If a function 𝑦 = 𝑓(π‘₯) contains a point (π‘Ž, 𝑏),
HOW TO DRAW THE GRAPH
 Plot some points on the function
then it satisfies
 Connect with those points to draw the graph
𝑓(π‘Ž) = 𝑏
 A linear function is of the form 𝑦 = π‘šπ‘₯ + 𝑏,
where π‘š, 𝑏 are constants.
Example 1: Consider 𝑦 = −π‘₯ + 2
(A) Fill the blank in the table
π‘₯ value
1
2
3
4
5
β‹―
𝑦 value
(B) Using (A), find the points on the graph of 𝑦 = −π‘₯ + 2 as the coordinate ordered pairs.
(C) Plot those points in the plane.
(D) Draw the graph of this function.
(E) Does the points 𝐴(3, −1) and 𝐡(4,3) lie on the graph?
Solution:
(A)
π‘₯ value
1
2
3
4
5
𝑦 value
1
0
−1
−2
−3
β‹―
(B) (1,1), (2,0), (3, −1), (4, −2), (5, −3)
(C) And (D)
2
2
5
5
2
2
4
4
(E) 𝐴(3, −1) is on the graph since −1 = −(3) + 2. But
𝐡(4,3) is not on the graph since 3 ≠ −(4) + 2
y= x+2
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7 GRADE-I: DR. EUNKYUNG YOU
Example 2: Consider 𝑦 = 3π‘₯ − 2
(A) Fill the blank in the table
π‘₯ value
1
2
3
4
5
β‹―
𝑦 value
(B) Using (A), find the points on the graph of 𝑦 = 3π‘₯ − 2 as the coordinate ordered pairs.
(C) Plot those points in the plane.
(D) Draw the graph of this function.
(E) Does the points 𝐴(1, −1) and 𝐡(0, −2) lie on the graph?
Example 3: Consider 𝑦 = −2π‘₯ + 1
(A) Fill the blank in the table
π‘₯ value
1
2
3
4
5
β‹―
𝑦 value
(B) Using (A), find the points on the graph of 𝑦 = −2π‘₯ + 1 as the coordinate ordered
pairs.
(C) Plot those points in the plane.
(D) Draw the graph of this function.
(E) Does the points 𝐴(2, −3) and 𝐡(0,3) lie on the graph?
1
Example 4: Consider 𝑦 = 2 π‘₯ + 3
(A) Fill the blank in the table
π‘₯ value
−2
0
2
4
6
β‹―
𝑦 value
1
(B) Using (A), find the points on the graph of 𝑦 = 2 π‘₯ + 3 as the coordinate ordered pairs.
(C) Plot those points in the plane.
(D) Draw the graph of this function.
(E) Does the points 𝐴(2,4) and 𝐡(−4,2) lie on the graph?
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7 GRADE-I: DR. EUNKYUNG YOU
Example 5: Consider 𝑦 = 3
(A) Fill the blank in the table
π‘₯ value
1
2
3
4
5
β‹―
𝑦 value
(B) Using (A), find the points on the graph of 𝑦 = 3 as the coordinate ordered pairs.
(C) Plot those points in the plane.
(D) Draw the graph of this function.
(E) Does the points 𝐴(3, −1) and 𝐡(4,3) lie on the graph?
VERTICAL LINES: The graph of π‘₯ = 𝑐 is a vertical
WARNING: It is not a function!
straight line which is parallel to the 𝑦-axis.
Example 6: Draw the graph of a line π‘₯ = 2
Solution: Every points in line π‘₯ = 2 has π‘₯-value 2 for any 𝑦-value.
π‘₯ value
2
2
2
2
2
𝑦 value
1
2
3
4
5
4
β‹―
2
2
4
Example 7: Draw the graph of a line π‘₯ = −1
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