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 = 3π β two 3 3 × 3 × 3 × 3 = 3π β four 3 2 × 2 × 2 × π × π = 23 × 7π 1 1 = 3, 2×2×2 2 BASE AND EXPONENT WARNING: 2 + 2 + 2 ≠ 23 , 2 βΆππ±π©π¨π§ππ§π β 3 1 1 = 2 2 × 2 × 3 × 3 2 × 32 23 ≠ 6 πππ¬π π0 = 1 (π ≠ 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 π2 × π3 (G) π2 × π4 (H) π₯ 4 × π¦ 2 (I) (J) (E) 5 × 5 × 7 × 7 × 7 × 11 (F) π × π × π × 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 × π × π × π × π × 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) π5 × π7 (P) π₯ 4 × π₯ 9 (I) ×3×2×2 1 2 3 1 5 × × × 1 5 (O) π12 (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 × π × π × π × π = π2 × π 3 (D) π + π + π + π + π = π5 1 1 1 3 (E) 4 × 4 × 4 = 4 9. Sierpinski Triangle: We will create Sierpinski Triangle in the following way: ο¨ Start with a filled-in equilateral triangle. ο¨ Connect the midpoints of each side, dividing the triangle into four smaller equilateral triangles. ο¨ Remove the middle triangle. ο¨ Repeat on all filled-in triangles infinitely. Find the number of the shaded equilateral triangles in π9 π1 π2 π3 First four stages of the Sierpinski triangle. π4 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 without a remainder is called a multiple of the number π ππ = π × π : 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 ππ = π × ππ = π × π = π × π 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 4 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 5 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 6 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 7 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, π be two natural numbers. When π is divided by π, its quotient and remainder are 10 and 8, respectively. Find the remainder when π 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 8 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 π 90 = 2 × 45 45 β° =2×5×9 β± π =2×5×3×3 9 β° β± = 2 × 5 × 32 π π DIVISION METHOD WARNING: 90 = 2 × 5 × 32 since π) 90 30 = 2 × ππ : Not prime factorization π) 45 30 = 2 × 3 × 5: Prime factorization π) 9 π 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 9 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 10 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 such that 200 × π is a perfect square Solution: Step1: Find a prime factorization of 200 200 = 2 × 100 = 22 × 50 = 23 × 25 = 2π × 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 = 2 Example 8: Find the smallest natural number π such that 26 × 5 × π is a perfect square 5 11 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 12 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 if 28 × π is a perfect square. 9. Find the smallest natural number π if 360 π is a perfect square. 3, 5, 7, 15,21, β― π=7 π = 10 10. Find a natural number π₯ such that 60 × π₯ become a perfect square. π₯ = 15 11. Find the smallest natural number π if 27 × π is a perfect square and a multiple of 4 π = 12 12. Find the number of all natural numbers π if 180 π is a natural number. 13. Find the smallest natural number π if 23 × π 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 and π if 2π = 64 and 33 = π 18 6 π₯=2 1,2,3,6,7,14,21,42 π = 6, π = 27 13 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 and π, usually denoted by LCM(π, π), is the integer that divides the numbers without a remainder. smallest positive integer that is divisible by both π and π. If either π or π is 0, LCM(π, π) is defined to be zero. Factors of 16 : 1, 2, π , 8, 16 Multiples of 2 : 2, 4, π , 8, 10, 12 , 14, β― Factors of 20 : 1, 2, π , 5, 10, 20 Multiples of 3 : 3, π , 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 = π GCF OF TWO NUMBERS 2 ) 54 72 ×π×π×3 72 = π × 2 × 2 × π × π GCF = π × ×π×π = 18 36 3 ) 9 12 3 4 GCF(54,72) = 2 × 3 × 3 = 18 GCF OF THREE NUMBERS GCF OF THREE NUMBERS 54 = π 3 ) 27 ×π×π×3 72 = π × 2 × 2 × π × π 90 = π × ×π×π GCF = π × ×π×π ×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 = ππ × ππ : 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 17 7 GRADE-I: DR. EUNKYUNG YOU LCM OF TWO NUMBERS 54 = π LCM OF TWO NUMBERS 2 ) 54 72 ×π×π×π 72 = π × π × π × π × π LCM = π × π × π × π × π × π = 216 3 ) 27 36 3 ) 9 12 3 4 LCM (54,72) = 2 × 3 × 3 × 4 = 216 LCM OF THREE NUMBERS 54 = π LCM OF THREE NUMBERS ×π×π×π 72 = π × π × π × π × π 90 = π × ×π×π× ×π LCM= π × π × π × π × π × π × π = 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 = ππ × ππ × 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= β 2 ×β GCF(6,8) = 2 πππ LCM (6,8) = 2 × 3 × 4 = 24 πππ 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₯, π¦ are relatively prime. 12 × π₯ × 12 × π¦ = 12 × 180 ⇒ π₯π¦ = 15 So π₯ = 1, π¦ = 15 or π₯ = 3, π¦ = 5. Therefore the two numbers are 12, 180 or 36, 60 Example 14: The GCF of a natural number π΄ 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 19 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 20 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, π and π 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 π + 28 π where π, 70 28 , π π 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 βΆ 3 is a square root of 9 or √9 = 3 23 = 8 βΆ 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 π΄ = −2.5, π΅ = −1, πΆ = 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| = π INEQUALITY: π>π π<π π₯ is bigger than π π₯ is smaller than π π≥π π≤π π₯ is bigger or equal than π π₯ is smaller or equal than π |−π| = π 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) 18β 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 > 0 and π > 0 ADDITION OF TWO POSITIVE (OR NEGATIVE) π+π =π+π (+4) + (+3) = +(4 + 3) (−π) + (−π) = −(π + π) (−5) + (−3) = −(5 + 3) π + (−π) = +(π − π) if π ≥ π ADDITION OF ONE POSITIVE AND ONE NEGATIVE π + (−π) = −(π − π) if π < π (−4) + (+3) = −(4 − 3) −π + π = −(π − π) if π ≥ π (−5) + (+8) = +(8 − 5) −π + π = +(π − π) if π < π 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 is the additive inverse of π if π + (−π) = 0 ο¨ −3 is the additive inverse of 3 and ο¨ 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) + (−π) + (−3) = (−3) + (−2) = −(π − π) + (−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, π π π = 5, π = −3 −2 1 4 π 2 8. In a square, each row, column, and diagonal has the same sum. Find the constants π, π 2 π = 1, π = −1 π 0 π −2 3 9. The temperature of a 7000 ft mountain at the base is 80β. It decreasing by 4β every 1000 ft. What is the temperature at the top of the mountain? 52β 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 π × (π + π) = (π × π) + (π × π) (+5) × (+3) = +(5 × 3) = +15 (π + π) × π = (π × π) + (π × 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 = +81: even exponent case negative (−2)π = −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 π π in multiplication is where π ≠ 0, π ≠ 0 Inverse of − 3 : − 2 (−15) ÷ (−3) = +(15 ÷ 3) = +5 negative π π 1 3 THE ORDER OF OPERATIONS: (from left to right) ο¨ First, simplify the exponent expression ο¨ Simplify the parentheses ο¨ Simplify multiplication and division ο¨ 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 π΄ − π΅ 10. Find all possible integers π and π such that π < π, |π + π| = 2, and π × π = −8 11. Find the largest value of π − π if |2 × π| = 8 and |π ÷ 3| = 4 where π and π are integers 936 peoples 250 π = −2, π = 4 or π = −4, π = 2 π − π = 16 43 7 GRADE-I: DR. EUNKYUNG YOU CHAPTER 2.5 RATIONAL NUMBERS RATIONAL NUMBERS ADDITION AND SUBTRACTION ο¨ Make sure the bottom numbers (the The rational numbers are the set of all numbers that can be in π denominators) are the same the form where π and π are integers and π is not zero. π 3÷8= ο¨ Add the top numbers (the numerators), put the 3 8 answer over the denominator 9 (−9) ÷ 7 = − 7 3 3= 1 ο¨ Simplify the fraction (if needed) 3 1 3×3 1×2 9 2 11 + = + = + = β 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 ο¨ Multiply the top numbers (the numerators). common factor except 1, we say that it is in the lowest ο¨ Multiply the bottom numbers (the denominators). terms. ο¨ Simplify the fraction if needed. 8 4×π π = = 10 5 × π π 1 4 1×4 4 2×2 2 × = = = = 2 5 2 × 5 10 5 × 2 5 MULTIPLICATION OF RATIONAL NUMBERS MIXED RATIONAL NUMBERS ο¨ 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. ο¨ Multiply the first fraction by that reciprocal 1 1 7 2 is a mixed fraction which is 2 + = 3 3 3 ο¨ 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) (− ) + ( ) = β (B) (− 1 6 −5 + 12 2 3 2 2 4 2 = 1 −4 8 3 8 (−4)+3 −5 (−1)×2 + = (− ) = β + 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 2 π 6 1×6 1 (D) (+ 9) ÷ (− 3) = − 9 ÷ (π) = − 9 × π = − 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 β 9 4 (B) (C) (−2) β 3 7 (D) 2 5 (− ) β 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 β π = π ÷ π + 1 for all rational numbers π and π. Find 13 β { β (−5)} 3 3 14. Let π β π = 3 ÷ (π − π) for all rational numbers π and π. Find (− 5) β (− 2) 1 1 1 15. We know that π×(π+1) = π − π+1 for all natural number π. 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, π, and π 3 2 0.4 −1 π − 6 5 π 2 3 π 1 1 26 π = ,π = ,π = 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: ο¨ A Variable is a symbol for a number we don't ο¨ (number) × (variable) : omit " × " notation know yet. It is usually a letter like π₯ or π¦. 2 × π₯ = ππ ο¨ A number on its own is called a Constant write number in front of π¦ × (−5) = −ππ letters ο¨ A Coefficient is a number used to multiply a ο¨ Multiplication with 1 or (−1) variable Omit “1”, but not sign 1×π =π ο¨ An Operator is a symbol (such as +,×, etc) that (−1) × π = −π shows an operation ο¨ An algebraic expression involves numbers and ο¨ Multiplication of same variables: exponent form letters that are connected with operations symbols π₯ × π₯ × π₯ = π₯3 such as “+, −,×, and ÷ ” WARNING: ο¨ (0.1) × π ≠ 0. π (consider π = 3 4 3 ο¨ π ÷ 4 π ≠ π × 3 π since 4 π = 1 ) 10 3π 4 VALUE WRITING EXPRESSIONS When π₯ = −4, find 3π₯ + 2. Nine more than π π+9 3π₯ + 2 4 less than π π−4 = 3 × π₯ + 2 : put × π§ times three 3π§ 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 cm (A) (B) (C) (D) (E) π2 cm2 2(π₯ + π¦) cm $ 0.05π₯ + 2π¦ (π₯ + 5) cm 3 π 4 (B) The perimeter of a rectangle such that its length is π₯ cm and its width is π¦ cm (C) The price of π₯ pens and π¦ 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 π Example 4: Express the following without × or ÷ notations (A) (π + π) × (−5) × π (B) π₯ ÷ π ÷ π¦ Solution (A) (π + π) × (−5) × π = −5π(π + π) 1 1 π₯ (B) π₯ ÷ π ÷ π¦ = π₯ × π × π¦ = ππ¦ Example 5: Express the following without × or ÷ notations (B) π × (−1) × π × π (A) 2 × π × π (A) 2ππ (B) −π3 (C) 2π₯ − 5π¦ (C) π₯ × 2 + π¦ × (−5) (D) (π₯ − π¦) ÷ 2 (D) (E) (F) (E) 3 ÷ (π − π) (F) π₯ ÷ 2 + (π¦ − π§) ÷ 5 π₯−π¦ 2 3 π−π π₯ 2 + π¦−π§ 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 + 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₯ − 5π¦ when π₯ = −3 and π¦ = 2 (B) 23 (C) 3 (B) 2π₯ 2 − π¦ 2 when π₯ = 4 and π¦ = −3 1 1 (C) 9π₯ 2 − 24π₯π¦ when π₯ = 3 and π¦ = − 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₯ ÷ π¦ (D) π₯ ÷ π¦ × π (E) π₯ ÷ 3 × π¦ (F) π₯ × (π + π) ÷ 5 (G) 5π₯ × 4π¦ (H) 3π₯ × 4π₯ × 2π₯ (I) π₯ ÷ 5 + 3 × π¦ (J) π₯ + 6π¦ ÷ 2π (K) π × π − π ÷ 3 (L) π ÷ (π₯ + π¦) + π × 6 (M) 0.5 × π₯ × π¦ (N) π₯ ÷ π¦ ÷ 4 1 (O) π ÷ 5 × π (P) π ÷ π₯ 2 × 2 (Q) π₯ + π¦ ÷ 2 (R) 3 × π + 2 ÷ (π₯ + π¦) 0 2. When π = −2 and π = 3, evaluate the following (A) 3π + 2π −22 (B) 5π − 4π −2 3 −9 π (C) ππ + 4 (D) −2π − 3 9 (E) 3π − π (F) ππ π−π 3. Evaluate the following (A) π₯ 2 − π₯π¦ 2 (B) 9π₯ 2 + + 2π¦ when π₯ = 2 and π¦ = −5 3π₯π¦ 5 when π₯ = −3 and π¦ = −2 1 1 (C) 4π₯ 2 − 24π₯π¦ when π₯ = − 2 and π¦ = 3 π₯π¦ (D) π₯ 2 −4π¦ when π₯ = −3 and π¦ = 2 (A) (B) (C) (D) −1 84.6 5 −6 55 7 GRADE-I: DR. EUNKYUNG YOU (A) −93 4. Evaluate the following (A) π₯ 2 π¦ − (3π₯)2 when π₯ = 2 and π¦ = −3 (B) π₯ 2 + 3π¦ 2 − 2π€ 2 when π₯ = 1, π¦ = −2 and π€ = −3 (C) π₯ 2 −π¦ 2 π€ 3 −4π¦ (B) −5 (C) 0 (D) −96 when π₯ = 2, π¦ = −2 and π€ = −1 (D) ππ₯ π+1 when π = −3, π₯ = 2 and π = 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 cm (A) (β + 15) cm (B) 7π₯ days (C) π + 13 pounds (D) $ (5700 − π₯ − π¦) (E) $(1.5π₯ + 3.2π¦) (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 π pounds (D) The amount left if Lee has $ 5700 and he spends $π₯ on pants and $π¦ on a shirt (E) The price of π₯ apples and π¦ 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 π‘ hour working hours is $7π‘. $58 If Tom works for 8 hours, find the wage. 7. The admission fee in museum, $F for one class of π₯ girls and π¦ boys is given by πΉ = 6π₯ + 7π¦ $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 ππ 2 β where π is the radius of its base and β is the height of the cone. (A) Find the volume if π = 6 cm and β = 4 cm (B) Find the increase of the volume if the height of the cone increase 2 cm (A) 48π cm3 (B) 24π cm3 56 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 (3π¦ + 5)cm, 38 cm width is π¦ 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 ο¨ A Term is either a single number or a variable, or ο¨ Add or subtract the coefficient of the like terms. ο¨ Use this as the coefficient of the term’s variable numbers and variables multiplied together. ο¨ 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 βΆ like terms 9π₯, 6π₯ 2 βΆ NOT like term Example 1: State the number of terms and the constant term in each expression. (A) π + 3 (B) π₯ 2 + 2π₯ − 5 Solution (A) π + 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₯ + 3π¦ + 4 (B) −2π₯ 2 + 3π¦ + 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₯ + π¦) − 4π¦ 58 7 GRADE-I: DR. EUNKYUNG YOU Example 3: State the coefficients of π₯ and π¦ in each expression (B) 2π₯ 2 − 3π¦ + 4π₯ + 6 (A) 5π₯ − π¦ + 4 Solution Example 4: State the coefficients of π₯ and π¦ in each expression π₯ (B) π₯ 2 + 3π₯ − 4π¦ (A) 4 + 5π¦ + 6 (A) π₯: 1 4 π¦: 5 (B) π₯: 3 π¦: −4 (C) −3π₯ 2 + 5π₯ + 8 (D) π₯ 2 + 3π₯ − 3π₯π¦ − 5π¦ (C) π₯: 5 π¦: 0 (D) π₯: 3 π¦: −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) 2π¦ × 5 (A) 10π¦ (B) 2π × (−5) (B) −10π₯ (C) 18π₯ (C) (−3π₯) × (−6) (D) 5π₯ (D) 15π₯ ÷ 3 (E) −6π (F) 1 (E) 24π ÷ (−4) (F) (−12π) ÷ 3 (G) π₯ ÷ π¦ × 3 (H) 5π₯ × 2π¦ (G) −36π 3π₯ π¦ Example 7: Simplify the following expression (A) 2π¦ + 5π¦ (B) −8π₯ + 3π₯ (A) 7π¦ (B) −5π₯ (C) 14π (C) 10π − 8π + 12π (D) −7π − 11π + 3π (D) −14π (E) 2π − 4 (F) (E) 5π + 4 − 3π − 8 (F) 7π₯ − 8π¦ + 3π₯ + 4π¦ 10π₯ − 4π¦ 59 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 π¦ + 3 π₯ + 2π¦ 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₯+ π¦ 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 π¦ + π₯ − 2π¦ when π₯ = 2 and π¦ = −2 12 1 (F) π₯ − π¦ − π¦ − 2π₯ (B) 2π₯ − 3π¦ + 5π₯ + 9π¦ when π₯ = 3 and π¦ = −1 π¦ 23 − π₯− (D) 5 π₯ − 3 π¦ + 2 π₯ + 5 π¦ (A) 6π₯ − 8 − 4π₯ + 3 when π₯ = −3 2 15 π¦ 60 7 GRADE-I: DR. EUNKYUNG YOU BASIC PROBLEMS (A) 5π₯ 1. Simplify (A) 2π₯ + 3π₯ (B) 9π₯ (B) 5π₯ + 4π₯ (C) 6π (D) −18π₯ (C) 11π − 5π (E) 11π₯ (D) −13π₯ − 5π₯ (F) −5π₯ (G) −7π (E) 7π₯ + 4π₯ (H) 3π₯ (F) 3π₯ − 8π₯ (I) −4π (J) π₯π¦ (K) −8π₯π¦ (G) −4π − 3π (H) π₯ + π₯ + π₯ (L) −3π (M) 0.8π¦ (N) 2.3π (I) π + 4π − 9π (J) 6π₯π¦ − 9π₯π¦ + 4π₯π¦ (O) (P) (K) 3π₯π¦ − 5π₯π¦ − 6π₯π¦ (L) −4π − 3π + 5π − π (M) 0.5π¦ + 0.3π¦ (N) 6π − 3.7π 1 1 3 4 (O) 2 π¦ + 3 π¦ 5 6 1 8 π¦ π 2 3 (P) π − π (A) 0 2. Simplify (A) 4π₯ + 2π₯ − 6π₯ (B) −2π₯ (B) 3π₯ − 7π₯ + 2π₯ (C) −4π₯ + 7 (D) −8π − 5 (C) 9π₯ − 13π₯ + 7 (D) 8π − 5 − 16π (E) −11π (F) (E) −3π − 2π − 6π (G) −2π₯ − 2 (F) 14π + 4 − 9π − 6 (G) 3π₯ + 9 − 5π₯ − 11 5π − 2 (H) −3π + 2 (H) −6 + 3π − 6π + 8 (I) −6π₯ + 2π¦ (J) 5π₯ (K) −10π₯ − 14 (I) 2π₯ − π¦ + 3π¦ − 8π₯ (J) −4π₯ − 2π₯π¦ + 9π₯ + 2π₯π¦ (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 π¦ − 2 π₯ + 2 π¦ (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₯π¦ − 3 π₯ + 4 π₯π¦ 16 3 11 6 1 4 7 5 1 π₯− π¦ π₯+ 4 17 4 π₯π¦ 61 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 − 3π + 5π + 7π when π = 2 and π = −1 6 2 4 1 1 1 (C) π + π − π when π = − 3 , π = 2, and π = 4 4. State the number of terms and find the constant term in each expression. (A) π₯ 3 − 3π₯π¦ + π¦ 2 − 7 5 (B) 3π₯π¦ − 5π₯ 2 π¦ + π₯ + π¦ 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 (30π‘ − 5π‘ 2 ) ft after π‘ 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 cm and its width is 2π cm. 8. Find the volume and surface area of a square box whose length of one sides of the square box is π cm. Perimeter 5π Area 6π2 Volume : π3 cm3 Surface : 6π2 cm2 62 7 GRADE-I: DR. EUNKYUNG YOU CHAPTER 3-3 DISTRIBUTION IN ALGEBRAIC EXPRESSIONS DISTRIBUTION LAW WARNING ο¨ π¨(π΅ + πΆ) = π¨π΅ + π¨πΆ (2π₯ − 5) − 3(3π₯ − 2) ο¨ (π΅ + πΆ)π¨ = π΅π¨ + πΆπ¨ = 2π₯ − 5 − 9π₯ + 6 : distribute carefully 1 ο¨ (π΄ + π΅) ÷ πͺ = (π΄ + π΅) × πΆ = 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 = + βΆ wrong! 4 4 5π₯ − 1 = 4 Right 3π₯ π₯ − 1 + 4 2 3π₯ π(π − π) = + 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) 3ππ + ππ (D) (4π₯ − 8) × 3 (F) −2ππ₯ + ππ₯ (G) 12π₯ + 20π¦ (H) −8π₯ + 6π¦ (E) π(3π + π) (F) −π₯(2π − π) (G) 4(3π₯ + 5π¦) (H) −3(5π₯ − 2π¦) + 7π₯ (I) π₯(2π − 3π + 5π) (J) (3π − 4π − 5π)(−2π₯) (I) 2ππ₯ − 3ππ₯ + 5ππ₯ (J) −6ππ₯ + 8ππ₯ + 10ππ₯ 63 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 + 3) − (−4π − 5) (A) −2π₯ − 3 (B) 6π + 8 (C) 7π₯ − π¦ (D) −5π₯ + 4 (E) −5π₯ + 10 (C) (2π₯ + 3π¦) + (5π₯ − 4π¦) (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₯ + 3π¦ − 4) + (π₯ + 5π¦ + 9) (J) (5π₯ − 3π¦ + 6) − (2π₯ − 5π¦ − 2) (I) −π₯ + 8π¦ + 5 (J) 3π₯ + 2π¦ + 8 64 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₯ − 3π¦ + 11) (J) −2π₯ − 3(π₯ − 5π¦ + 2) (H) 3π₯ + 33 (I) −3π₯ + 9π¦ − 29 (J) −5π₯ + 15π¦ − 6 Example 5: Subtract π − 2π from 3π − π and simplify Solution: (3π − π) − (π − 2π) = 3π − π − π + 2π = 2π + π Example 6: Express the statement and simplify (A) Add (2π₯ − 4) to (3π₯ − 5π¦ + 8) (A) 5π₯ − 5π¦ + 4 (B) −8π₯ + 7π¦ (B) Subtract 3π₯ − 4π from −5π₯ + 3π¦ 65 7 GRADE-I: DR. EUNKYUNG YOU PROPERTIES OF EXPONENTS: Let π, π be positive WARNING: integers. π π ×π (ππ)π π =π π+π (ππ )π π π π =π 2π ≠ 2 × 3 = 6 ππ ( 23 = 2 × 2 × 2 = 8) π π π ( ) = π π π =π π 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 π0 = 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) π2 × π5 (D) ( ) π¦ Solution: (A) π2 × π5 = π2+5 = π7 (B) (π₯ 2 )4 = π₯ 2×4 = π₯ 8 (C) (2π₯ 2 )3 = 23 (π₯ 2 )3 = 8π₯ 6 3π₯ 2 (D) ( π¦ ) = 32 π₯ 2 π¦2 = 9π₯ 2 π¦2 Example 8: Simplify the expression by writing with positive exponent. Assume that all variable is not zero. (A) 3π × π × 5π (B) 4π × 4π × π (A) 15π3 (B) 16π2 π (C) 14π₯π¦ 3 (C) 7π₯ × 2π¦ × 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 π¦10 (I) (J) (G) (π₯ 6 )2 × (π₯ 3 )4 (H) (π₯π¦ 2 )5 (K) (L) π3 π4 π2 27π6 π₯4 π¦6 8 π6 (M) 1 (I) π3 ÷ π 4 ÷ π 2 π₯2 2 (J) (3π2 )3 2 3 (K) (− π¦3 ) (L) (π2 ) (M) 60 (N) (2π₯ 3 )0 (N) 1 66 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 ÷ 2 5 (G) 25π₯ ÷ 3 (D) (−9π₯) × 2 (− 3) (F) −2π (G) 15π (F) (−8π) ÷ 4 (H) −4π 7 (H) 14π ÷ (− 2) (A) 8π₯ − 4 2. Simplify (A) 4 × (2π₯ − 1) (E) 5π (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₯ − π¦ − 7) (F) −2π₯ + 5 (G) −3π₯ − 2 (H) 8π₯ − 12 (I) 7π₯ − 5 (J) −8π₯ + π¦ + 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₯ + 9π¦ (G) 13π₯ + π¦ (E) (−4π₯ − 9) + (−6π₯ + 4) (F) (−9π₯ + 6π¦) − (2π₯ − 3π¦) (H) 11π₯ + 16π (I) 3π₯ − 1 (J) 2π₯ + 6π¦ (K) 2π₯ (G) (8π₯ + 3π¦) + (5π₯ − 2π¦) (H) (5π₯ + 9π) − (−6π₯ − 7π) (L) −9π₯ + 9π¦ (M) −7π₯ − 2π (N) −5π₯ − 1 (I) (2π₯ + 3) + (π₯ − 4) (J) (3π₯ + 3π¦) − (π₯ − 3π¦) (K) (π₯ + 2π¦) + (π₯ − 2π¦) (L) (−5π₯ + 6π¦) − (4π₯ − 3π¦) (M) (−4π − 5π) + (−3π + 3π) (N) −3 − (5π₯ − 2) 67 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₯ + 7π¦ (C) 4(π₯ + 5) + 2(3π₯ − 2) (D) 5(π₯ + 2π¦) + 3(−3π₯ − π¦) (E) 12π₯ + 3 (F) (E) 6(3π₯ − 4) − 3(2π₯ − 9) (F) −4(π₯ + 2) + 3(2π₯ − 5) (H) (π₯ − 3π¦)(−2) + (π₯ − π¦)(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₯ − 10π¦ − 5 5. Add 4π₯ − 5π¦ + 2 to 3π₯ − 5π¦ − 7 −6π₯ + 8π¦ − 13 6. Subtract 4π₯ − 5π¦ + 7 from −2π₯ + 3π¦ − 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 π΄ −π₯ − 3 π΄ 5π₯ + 3 2π₯ 3π₯ + 1 π΄ = −2 68 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₯ − π¦)} (2) Simplify parentheses from inside one = 2π₯ − 3{π₯ + 2π₯ − 2π¦} : apply distributive law (3) Do any multiplication or division = 2π₯ − 3{3π₯ − 2π¦} : simplify inside parentheses (4) Do addition or subtraction from left to right = 2π₯ − 9π₯ + 6π¦ : apply distributive law = −7π₯ + 6π¦ : 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 − {7π − (2π − 3)} (A) −4π₯ + 4 (B) −2π − 3 (C) −3π₯ + 41π¦ (D) 13π₯ − 3π¦ − 60 (E) −11π₯ (F) (C) 3{2π₯ − 3(π₯ − 4π¦)} + 5π¦ 3 (E) −2{4π₯ − 3(π₯ − π¦)} − 4 (12π₯ − 8π¦) (D) 3(π₯ − π¦) − 5{π₯ + 3(4 − π₯)} (F) 2.5(π₯ − π¦) − 1.3(2π₯ − π¦) 0.1π₯ − 1.2π¦ 69 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 70 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 + 5 2. Simplify (B) 4π − 3 (A) π − {2π + (3π − 5)} (C) −3π₯ + 3 (D) −3π₯ + 6 (B) 3π − {π − (2π − 3)} (E) 7π₯ − 4π¦ (F) (C) 4π₯ − {5π₯ − (3 − 2π₯)} 4π₯ − 19π¦ (G) −2π₯ + 20 (H) −π₯ + 8 (D) 8π₯ − {9π₯ + 2(π₯ − 3)} (E) 5{π₯ − (2π¦ − π₯)} + 3(−π₯ + 2π¦) (F) 3(π₯ − 3π¦) − {6π₯ − π₯ − 2(3π₯ − 5π¦)} (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 π΄, π΅ π΄ = 4π₯ − 2 π΅ = −2π₯ π΅ 5π₯ + 1 π₯−1 −4 5. Find the blank : −3π₯ + 2 + −3π₯ − 3 π΄ = −2π₯ + 4 π₯+2 71 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, ππ₯ + π = π, where π, π, and π are numbers. ππ + π = ππ 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 = π, then 2π₯ 4 = 2 2 π=π Solution for 2π₯ + 5 = 11 ο¨ π+π =π+π ο¨ π−π =π−π ο¨ ππ = ππ ο¨ π π π π = , π≠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 ππ − π = ππ + π 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 π π=π 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 π(π − π) = π(ππ − π) 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 ππ−π π 30 × = Divide both sides by −4 ππ−π π (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) π¦+11 2 π¦−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−π (A) when π = 4, π = −32, and π = 9 (A) 9 (B) 6 (C) 25 2π π 15 (C) 2 (D) −11 (E) 6 (F) 6 13 (G) (D) π₯−4 = 3 =2 3 5 4 1 (B) 1−π when π = 4 and π = 3 1 π 2 π 3 π 1 4 1 3 (C) − + when π = , π = − , and π = ππ 1 5 1 4. Find the value of π if π = π+π where π = 2 2 , π = 5, and π = 4 1 1 1 1 1 5. Find the value of π if π + π = π where π = − 5, and π = 3 3π 6. Find the value of π if π΄ = 1−π where π΄ = −6 and π = 2 π=3 π= 1 8 π=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 π¦ is π % of π₯, 100 =1 100 ο¨ Percentage to decimal: π¦ = π × 0.01 × π₯ 100% = π₯%=π₯× 1 or π₯ × 0.01 100 ο¨ Decimal to percentage: π¦ = π¦ × 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 is 270 % of 60 decimal π=β 270 × 0.01 × 60 ⇒ π = 162 (B) 45 is π₯ % of 125 decimal β 45 = 0.01π₯ × 125 ⇒ 45 = 1.25 π₯ ⇒ π₯ = 45 = 36 % 1.25 (C) 81 is 56 % of π¦ decimal 81 = β 25 × 0.01 × π¦ ⇒ 81 = 0.25π¦ ⇒ π¦= 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 is 10 % of $5000 π = 10 × 0.01 × 5000 ⇒ π = 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 π¦ % of $8500 1700 = π¦ × 0.01 × 8500 ⇒ 1700 = 85π¦ ⇒π¦= 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? 88 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 89 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 90 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 91 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% 92 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 93 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 94 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 95 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 96 7 GRADE-I: DR. EUNKYUNG YOU CHAPTER 5 FUNCTIONS CHAPTER 5.1 BASIC CONCEPTS OF FUNCTIONS DEFINITION: In the rectangular coordinate system, ο¨ The horizontal and vertical number lines are called the π₯-axis and the π¦-axis respectively. ο¨ The two lines intersect at right angle at the point (called the origin) ο¨ The position of any point, say P, on the plane can be expressed by an ordered pair (π, π) where π is the π₯-value and π is the π¦-value. We say that π has coordinate (π, π) and refer the point π as π(π, π). Example 1: Given the figure, 4 (A) State the coordinate of the points π΄, π΅, πΆ, π· and πΈ. B (B) State the quadrants in which those points lie A 2 (C) Plot the points π(3,4), π(−1,2), π (3, −3), and π(−2, −3) 4 2 2 C 2 4 D E 4 Solution: 4 π΄ π΅ πΆ π· πΈ Ordered pair (2,2) (−1,3) (−2, −1) (0, −2) (3, −3) Quadrants πΌ πΌπΌ πΌπΌ P 2 Q πΌπ 5 4 2 4 2 2 R S 4 97 7 GRADE-I: DR. EUNKYUNG YOU Example 2: Given the figure, (A) State the coordinate of the points 4 π΄, π΅, πΆ, π· and πΈ. 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 π΄, π΅, πΆ, π· and πΈ. 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 π΄(1, −3), π΅(4, −3), πΆ(4,1) and π·(1,1). Then find the area of a rectangle ABCD. 4 2 5 4 2 2 2 4 4 98 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 π¦ 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 π¦ be the price of π₯ apples. (A) Fill the blank in the table π₯ apples 1 $π¦ $5 2 3 4 5 β― β― (B) Determine whether π¦ is a function of π₯ or not (C) Find the relation between π¦ and π₯ and write it as an equation. Solution: (A) π₯ apples 1 2 3 4 5 $π¦ $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, π¦ = 5π₯ Example 6: The price of a pen is $2. Let π¦ be the price of π₯ pens. (A) Fill the blank in the table π₯ apples 1 $π¦ $2 2 3 4 (B) Determine whether π¦ is a function of π₯ or not (C) Find the relation between π¦ and π₯ and write it as an equation. 5 β― 99 7 GRADE-I: DR. EUNKYUNG YOU Example 7: Consider a function π(π₯) = 4π₯. Find the value of function; (A) π(2) (B) π(−3) (C) π(−1) (B) π(−3) = −12 (C) π(−1) = −4 Solution: (A) π(2) = 8 π(−3) = 4 × (−π) = −ππ β π(2) = 4 × (π) = π replace π₯ with − 3 replace π₯ with 2 Example 8: Evaluate π(−2) if π(π₯) = 3π₯ Example 9: Evaluate π(−3) if π(π₯) = 2π₯ + 3 1 Example 10: Evaluate π(6) if π(π₯) = 2 π₯ − 4 Example 11: Evaluate π(4) if π(π₯) = 5 − 3π₯ 4 Example 12: Evaluate π(−2) if π(π₯) = π₯ + 5 Example 13: Find the value of constant π if π(π₯) = 3π₯ + π and π(2) = 0 π(−1) = 4 × (−π) = −π β replace π₯ with − 1 100 7 GRADE-I: DR. EUNKYUNG YOU CHAPTER 5.2 THE GRAPH OF A LINEAR RELATION. ο¨ If a function π¦ = π(π₯) contains a point (π, π), HOW TO DRAW THE GRAPH ο¨ Plot some points on the function then it satisfies ο¨ Connect with those points to draw the graph π(π) = π ο¨ A linear function is of the form π¦ = ππ₯ + π, where π, π are constants. Example 1: Consider π¦ = −π₯ + 2 (A) Fill the blank in the table π₯ value 1 2 3 4 5 β― π¦ value (B) Using (A), find the points on the graph of π¦ = −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 π΄(3, −1) and π΅(4,3) lie on the graph? Solution: (A) π₯ value 1 2 3 4 5 π¦ 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) π΄(3, −1) is on the graph since −1 = −(3) + 2. But π΅(4,3) is not on the graph since 3 ≠ −(4) + 2 y= x+2 101 7 GRADE-I: DR. EUNKYUNG YOU Example 2: Consider π¦ = 3π₯ − 2 (A) Fill the blank in the table π₯ value 1 2 3 4 5 β― π¦ value (B) Using (A), find the points on the graph of π¦ = 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 π΄(1, −1) and π΅(0, −2) lie on the graph? Example 3: Consider π¦ = −2π₯ + 1 (A) Fill the blank in the table π₯ value 1 2 3 4 5 β― π¦ value (B) Using (A), find the points on the graph of π¦ = −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 π΄(2, −3) and π΅(0,3) lie on the graph? 1 Example 4: Consider π¦ = 2 π₯ + 3 (A) Fill the blank in the table π₯ value −2 0 2 4 6 β― π¦ value 1 (B) Using (A), find the points on the graph of π¦ = 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 π΄(2,4) and π΅(−4,2) lie on the graph? 102 7 GRADE-I: DR. EUNKYUNG YOU Example 5: Consider π¦ = 3 (A) Fill the blank in the table π₯ value 1 2 3 4 5 β― π¦ value (B) Using (A), find the points on the graph of 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 π΄(3, −1) and π΅(4,3) lie on the graph? VERTICAL LINES: The graph of π₯ = π is a vertical WARNING: It is not a function! straight line which is parallel to the π¦-axis. Example 6: Draw the graph of a line π₯ = 2 Solution: Every points in line π₯ = 2 has π₯-value 2 for any π¦-value. π₯ value 2 2 2 2 2 π¦ value 1 2 3 4 5 4 β― 2 2 4 Example 7: Draw the graph of a line π₯ = −1 103