GEC 4 Mathematics in the Modern World 1 PDF

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Warning: TT: undefined function: 32 Warning: TT: undefined function: 32MODULEINGEC 4MATHEMATICS IN THE MODERNWORLDPrepared by:Leonardo G. GoreroReneboy H. CalcañaEden T. PatnongonKorina D. VillanuevaCHAPTER 5PROBLEM SOLVINGRegardless of what we do for a living or where we live, most of us spend our ...

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MODULE IN GEC 4 MATHEMATICS IN THE MODERN WORLD

Prepared by: Leonardo G. Gorero Reneboy H. Calcaña Eden T. Patnongon Korina D. Villanueva

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CHAPTER 5 PROBLEM SOLVING Regardless of what we do for a living or where we live, most of us spend our waking hours, at school, at work or at home, solving problems. Most problems we face are small, some are large and complex, but they all need to be solved in a suitable way. The ability to solve problems is a basic life skill and is essential to our-to-day lives. One of the leading mathematicians to make a study on problem solving was George Polya (1887-1985). He was born in Hungary and moved to the United States in 1940. George Polya after a brief stay at Brown University moved to Stanford University in 1942 and taught there until his retirement. While at Stanford, he published 10 books and a number of articles for mathematics journals. Of the book Polya published, How to solve it (1945) is one of his best known. In this book, Polya outlines a strategy for solving problems from virtually any discipline. “A great discovery solves a great problem but there is a grain of discovery in the solution of any problem. Your problem may be modest; but if it challenges your curiosity and brings into play your inventive faculties, and if you solve it by your own means, you may experience the tension and enjoy the triumph of discovery.” The basic problem solving strategy that Polya advocated consisted of the following four steps (Jacobs, Martin, Ambrose, Philipp, 2014) 1. Understanding the problem. Read the problem slowly, jotting down the key ideas 2. Devising a plan. Draw a diagram, find a formula, look for patterns 3. Carrying out the plan. Create an equation and solve the problem 4. Looking back. Check your answer. Does your answer make sense? Did you solve for the requested unknown? Lesson 1 - Translating Words into Mathematical Symbols  Starting Accurately The knowledge on how to translate key words from English language into mathematical symbols is very important in problem solving. The first step in any problem solving situation in mathematics is always to read the problem, then it follows by translating the words into mathematical symbols. Learning Objectives: At the end of the lesson, the students are expected to: 1. translate mathematical sentence to mathematical symbols. 2. translate mathematical symbols to mathematical sentence.  Stimulating Learning In Problem solving, words such as gain, more, sum, total, increase, plus all mean to add. Words such as difference between, minus, decrease, less, fewer, and loss all mean to subtract. Words such as the product of, double (2x), twice (2x), triple (3x), a fraction of, a percent of, or times all mean to multiply. And finally, words such as the quotient of, divided equally, and per mean to divide. Experience and practice with problem solving will help better acquaint you with the key words that translate into these operations.  Inculcating Concepts

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Let's translate the following mathematical sentence/statements into mathematical symbols: Addition Mathematical statement 1.) 2.) 3.) 4.) 5.) 6.) 7.)

Mathematical symbol

1.) The product of 14 and x 2.) 4 times b 3.) twice x 4.)

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𝑜𝑓 𝑝

5.) 9 multiplied by n

1.) the difference of 25 and y 2.) 560 minus h 3.) w less than 32 4.) 9 decreased by m 5.) p reduced by x 6.) 13 subtracted from j 7.) 500 less d

𝑎 + 18 14 + 𝑐 𝑛 + 15 𝑓 + 22 𝑡+𝑚 𝑝 + 13 𝑞 + 33

The sum of a and 18 14 plus c 15 added to n 22 more than f t increased by m 13 greater than p exceeds q by 33

Multiplication Mathematical statement

Subtraction Mathematical statement

Division Mathematical statement

Mathematical symbol 14𝑥 4𝑏 2𝑥 3 𝑝 4 9𝑛

1.) the quotient of s and 17 2.) x divided by y 3.) the ratio of c and m 4.) the price p per gallon g

Mixed operations Mathematical statement 1.) Five more than one-fourth of k is 55 2.) 3.) 4.) 5.)

Eight less than twice a number n is greater than 90 Nine increased by the product of two numbers x and y A number t plus seven times p is less than or equal to 20 The quotient of two numbers m and n subtracted from 30.

6.) The product of five and three less than the number y 7.) Twice the sum of a number x and nine

Mathematical symbol 25 − 𝑦 560 − ℎ 32 − 𝑤 9−𝑚 𝑝−𝑥 𝑗 − 13 500 − 𝑑

Mathematical symbol 𝑠 17 𝑥 𝑦 𝑐 𝑚 𝑝 𝑔

Mathematical symbol 1 𝑘 + 5 = 55 4 2𝑛 − 8 > 90 9 + (𝑥𝑦) 𝑡 + 7𝑝) ≤ 20 𝑚 30 − 𝑛 5(𝑦 − 3) 2(𝑥 + 9)

 Using/Applying Knowledge The following problems serve as your practice exercises. Be honest in answering them and compare your answers with the indicated answer key. Exercise 1A. Translate the following mathematical statements into mathematical symbols or equations. Use x and y if no specific variable is indicated. 1. Six less than twice a number is less than or equal to forty-five. 2. A number minus seven yields ten 3. A total of six and a number m 4. Twelve added to a number 5. Eight times a number is forty-eight 6. The produce of fourteen and a number 7. Twice a number minus eight 8. The quotient of a number and seven is two 9. Three-fourths of a number h 10. The product of a number and ten is eighty 2

11. Eight less than a number is greater than five 12. How many times does five go into twenty? Exercise 1B. Translate the following mathematical symbols into mathematical statements 1.) x + 12 = 8 2.) 3x = 15 3.) w/15 4.) 10/m 5.) x – 6 6.) 5(x + 4) ≤ 20 7.) 2(x-3) =12 8.) 7/x 9.) (6 –x)/9 10.) 4(12+y) Answers for Exercise 1A 1.) 2x - 6 ≤ 45 2.) x – 7 = 10 2x – 8 8.) x/7 = 2 9.) ¾ h

3.) 6 + m

4.) x + 12

10.) 10x = 80 11.) x – 8 > 5

5.) 8x = 48

6.) 14x

7.)

12.) 20/5 = x

Answers for Exercise 1B 1. Twelve added to a number is the same as eight 2. Three times a number equals fifteen 3. The ratio of a number w to fifteen 4. Ten divided by a number m 5. Six subtracted from a number 6. Five times the sum of a number and four is less than or equal to 20 7. Twice the difference of a number and three totals twelve 8. The quotient of a seven and a number 9. The difference between six and a number divided by nine 10. Four times the sum of twelve and y  Evaluating Understanding A. Translate the following mathematical statements into mathematical symbols or equations. Use x and y if no specific variable is indicated. 1. The sum of m and thirteen is forty-five. 2. Eight is subtracted from twice the number is greater than fifteen. 3. Twelve is five more than k 4. One-half of a number is equal to fifty-six. 5. Six times a number decreased by eight is two hundred 6. The product of fifteen and thrice a number p 7. The quotient of a number and eleven is less than or equal to seventy-five 8. Four-fifths of a number is sixteen 9. The number reduced by twenty-four is five times a number 10. The product of eight and six less than a number is greater than four hundred five B. Translate the following mathematical symbols into mathematical statements 1.) 2x + 4 = 38 2.) m - 3 = 65 3.) 2(3x + 3) 4.) 5/2x 5.) 3x – 5 > 32 6.) 4(x - 8) ≤ 72 3 7.) 𝑡 + 12 8 8.) 4p - 13 9.) (2x – 8) /9 10.) 6(2q + 4)

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Lesson 2A - Number Problems  Starting Accurately A number problem is asking us to imagine a scenario by looking at the numbers and figuring out what the problem is asking us to do. At the end of the lesson, students are expected to solve number related problems.  Stimulating Learning Whether we like it or not, whatever our nature of work, number problems are everywhere. Some people may think that we either can do it or we can't. Contrary to that belief, it can be learned. Even the best athletes and musicians had some coaching along the way and lots of practice. That's what it also takes to be good at problem solving.  Inculcating Concepts Example 1. The sum of two numbers is 60. The larger number is four times the smaller. Find the smaller number. Understanding the problem There are two numbers whose sum is 60. One number is larger than the other. The larger number is four times the smaller. We are going to find the smaller number. Devising a plan 𝐿𝑒𝑡 𝑥 = 𝑡ℎ𝑒 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑛𝑢𝑚𝑏𝑒𝑟 4𝑥 = 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑟 𝑛𝑢𝑚𝑏𝑒𝑟

Looking back: 𝑥 + 4𝑥 = 60

12 + 4(12) = 60

The sum of two numbers is 60: 𝑥 + 4𝑥 = 60

12 + 48

= 60

 60 = 60

Carrying out the plan 𝑥 + 4𝑥 = 60 5𝑥 = 60 60 𝑥= 5

𝑥 = 12 → 𝑡ℎ𝑒 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑛𝑢𝑚𝑏𝑒𝑟

Answer: The smaller number is 12. Example 2. The sum of two numbers is 50. The second number is 8 more than the first number. Find the numbers. Understanding the problem There are two numbers whose sum is 50. The second number is larger than the first number. The second number is greater than the first number by 8. We are going to find the two numbers.

Devising a plan Looking back:

𝐿𝑒𝑡 𝑥 = 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑥 + 8 = 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑛𝑢𝑚𝑏𝑒𝑟

𝑥 + 𝑥 + 8 = 50 21 + 21 + 8 = 50

The sum of two numbers is 50: 𝑥 + 𝑥 + 8 = 50

 50 = 50

Carrying out the plan 4

𝑥 + 𝑥 + 8 = 50 2𝑥 + 8 = 50 2𝑥 = 50 − 8 2𝑥 = 42 42 𝑥= 2 𝑥 = 21 → 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑥 + 8 = 21 + 8 = 29 → 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑛𝑢𝑚𝑏𝑒𝑟. Answer: The two numbers are 21 and 29. Example 3. Find three consecutive even integers whose sum is 138. Understanding the problem Solving consecutive even integers and solving consecutive odd integers use the same formula. Numbers such as 8, 10, 12 are consecutive even numbers. Numbers such as 5, 7, 9 are consecutive odd numbers. It is understood that we add 2 to 8, to get 10 and add 2 to 10, to get 12 for consecutive even numbers. The same process will be used for consecutive odd numbers, we add 2 to 5, to get 7 and add 2 to 7, to get 9. Here, we are going to find 3 successive even numbers whose sum is 138. Devising a plan

Looking back:

𝐿𝑒𝑡 𝑥 = 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑒𝑣𝑒𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑥 + 2 = 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑒𝑣𝑒𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑥 + 4 = 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑒𝑣𝑒𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟

𝑥 + 𝑥 + 2 + 𝑥 + 4 = 138 44 + 44 + 2 + 44 + 4 = 138  138

Three consecutive even integers whose sum is 138: 𝑥 + 𝑥 + 2 + 𝑥 + 4 = 138

=

Carrying out the plan 𝑥 + 𝑥 + 2 + 𝑥 + 4 = 138 3𝑥 + 6 = 138 3𝑥 = 138 − 6 3𝑥 = 132 132 𝑥= 3 𝑥 = 44 → 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑒𝑣𝑒𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑥 + 2 = 44 + 2 = 46 → 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑒𝑣𝑒𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑥 + 4 = 44 + 4 = 48 → 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑒𝑣𝑒𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 Answer: The three consecutive even integers are 44, 46, and 48. Example 4. Three times a number increased by 8, equals 41. What is a number? Understanding the problem We look for a number multiplied by 3 and added by 8 should be equal to 41. Devising a plan

Looking back:

𝐿𝑒𝑡 𝑛 = 𝑎 𝑛𝑢𝑚𝑏𝑒𝑟

3𝑛 + 8 = 41

Three times a number increased by 8, equals 41: 3𝑛 + 8 = 41 Carrying out the plan 3𝑛 + 8 = 41 3𝑛 = 41 − 8 3𝑛 = 33 5

3(11) + 8 = 41

33 + 8 = 41  41 = 41

138

33 𝑛 = 3 𝑛 = 11 → 𝑖𝑠 𝑎 𝑛𝑢𝑚𝑏𝑒𝑟 Answer: The number is 11. Example 5. The sum of three numbers is 63. The second number is three greater than five times the first number and the third number is four less than twice the first number. What are the numbers? Understanding the problem There are three numbers whose sum is 63. These are the first, second and the third numbers. The second number is 5 times the first number plus 3. The third number is two times the first number minus 4. We are to find the three numbers. Devising a plan Looking back:

𝐿𝑒𝑡 𝑛 = 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 5𝑛 + 3 = 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 2𝑛 − 4 = 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑛𝑢𝑚𝑏𝑒𝑟

𝑛 + 5𝑛 + 3 + 2𝑛 − 4 = 63 8 + 5(8) + 3 + 2(8) − 4 = 63 8 + 40 + 3 + 16 − 4 = 63  63 = 63

The sum of these three numbers is 63: 𝑛 + 5𝑛 + 3 + 2𝑛 − 4 = 63 Carrying out the plan

𝑛 + 5𝑛 + 3 + 2𝑛 − 4 = 63 8𝑛 − 1 = 63 8𝑛 = 63 + 1 8𝑛 = 64 64 𝑛 = 8 𝑛 = 8 → 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 5𝑛 + 3 = 5 (8) + 3 = 40 + 3 = 43 → 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 2𝑛 − 4 = 2 (8) − 4 = 16 − 4 = 12 → 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 Answer: The three numbers are 8, 43, and 12. Example 6. A string is cut into 3 pieces, A, B, and C in the ratio 1: 2: 4. If C is longer than B by 24 feet, find the length of the string. Understanding the problem A string is to be cut into 3 pieces, A, B, and C in the ratio 1: 2: 4, respectively. It is understood that the length of A times 2 is equal to the length of B. The length of C is equal to the length of A times 4. Furthermore, the length of C is longer than B by 24 feet. Then, we are to find the length of the string. Devising a plan 𝐿𝑒𝑡 𝑛 = 𝑎 𝑐𝑜𝑚𝑚𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑡ℎ𝑎𝑡 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑎 𝑝𝑖𝑒𝑐𝑒 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 𝐴 𝑤𝑖𝑡ℎ 1 𝑎𝑠 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜 2𝑛 = 𝑎 𝑝𝑖𝑒𝑐𝑒 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝐵 𝑤𝑖𝑡ℎ 2 𝑎𝑠 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜 4𝑛 = 𝑎 𝑝𝑖𝑒𝑐𝑒 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝐶 𝑤𝑖𝑡ℎ 4 𝑎𝑠 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜 The length of C, which is 4n is longer than the length of B, which is 2n by 24 feet: 4𝑛 = 2𝑛 + 24 Looking back:

Carrying out the plan 4𝑛 = 2𝑛 + 24 4𝑛 − 2𝑛 = 24 2𝑛 = 24

4𝑛 = 2𝑛 + 24 4(12) = 2(12) + 24 48 = 24 + 24  48 = 48 6

24 𝑛= 2 𝑛 = 12 → 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 𝐴 2𝑛 = 2(12) = 24 → 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 𝐵 4𝑛 = 4(12) = 48 → 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 𝐶 Answer: The total length of a string is 84 feet. This string is being cut into three pieces 12 ft., 24 ft., and 48 ft., in the ratio of 1:2:4. Example 7. The total of two numbers is 23, double the larger number exceeds three times the smaller number by 16. Find the numbers. Understanding the problem There are two numbers, the larger number and the smaller number whose sum is 23. When the larger number is doubled, then it is equal to 3 times the smaller number plus 16. Then we are to find the two numbers. Devising a plan 𝐿𝑒𝑡 𝑥 = 𝑡ℎ𝑒 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑛𝑢𝑚𝑏𝑒𝑟 system of two equations in two unknowns

𝑦 = 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑟 𝑛𝑢𝑚𝑏𝑒𝑟 𝑥 + 𝑦 = 23 2𝑦 = 3𝑥 + 16 We can solve for the values of x and y using substitution or elimination by addition or elimination by subtraction. Carrying out the plan We will use elimination by addition: 𝑥 + 𝑦 = 23 → 𝑥 + 𝑦 = 23 2𝑦 = 3𝑥 + 16 → −3𝑥 + 2𝑦 = 16 To eliminate x, multiply 3 to (x + y = 23): 3𝑥 + 3𝑦 = 69 Addition −3𝑥 + 2𝑦 = 16 5𝑦 = 85 85 𝑦= 5

Looking back: 𝑥 + 𝑦 = 23 6 + 17 = 23  23 = 23

2𝑦 = 3𝑥 + 16 2(17) = 3(6) + 16 34 = 18 + 16  34 = 34

𝑦 = 17 → 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑟 𝑛𝑢𝑚𝑏𝑒𝑟 𝑥 + 𝑦 = 23 → 𝑥 + 17 = 23 → 𝑥 = 23 − 17 = 6 → 𝑡ℎ𝑒 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑛𝑢𝑚𝑏𝑒𝑟 4𝑛 = 4(12) = 48 → 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 𝐶 Answer: The two numbers are 17 and 6.  Using/Applying Knowledge The following problems serve as your practice exercises. Be honest in answering them and compare your answers with the indicated answer key. Exercise 2: Solve the following number problems. 1. Two-fifths a number equals 18. Find the number. 2. A number added to one-fourth of itself equals 45. Find the number. 7

3. When eight times a number is diminished by 20, the result is 28. Find the number. 4. The sum of three integers is 27. The second integer is 2 less than twice the first integer, and the third integer is 5 greater than three times the first. Find the numbers. 5. Two numbers are in the ratio 3:8 and their sum is 77. What are the numbers?

Answers for Exercise 2: 1.) 18

2.) 36

3.) 6

4.) 4, 6, 17

5.) 21, 56

 Evaluating Understanding Evaluation 2: Solve the following number problems. 1. The sum of two numbers is 30. The second number is 6 more than the first number. Find the numbers. 2. Find three consecutive even integers whose sum is 168. 3. Four times a number increased by 5 is 37. Find the number. 4. Forty-five is equal to twelve more than thrice a number. Find the number. 5. The difference between eight times a number and 16 is 40. Find the number. 6. Thrice a number, decreased by 4, is the same as twice the number, increased by 11. Find the number. 7. Four times a number, increased by 17, is 9 less than six times the number. Find the number. 8. The total of two integers is 43. Double the bigger integer exceeds three times the smaller integer by 16. What are the integers? 9. The difference of two integers is 9. Five times the smaller is 9 more than three times the larger. Find the integers. 10. Find three consecutive odd integers such that the sum of the first, twice the second, and three times the third is 70.

 Upgrading Competence and Expanding Insights 1.) The sum of two numbers is 15. Three times one of the numbers is 19 less than five times the other. Find the numbers. 2.) The difference of two integers is 9. Five times the smaller is 9 more than three times the larger. Find the integers. 3.) A piece of string is 80 cm long. It is cut into three pieces. The longest piece is 3 times as long as the middle sized piece and the shortest piece is 46 cm shorter than the longest piece. Find the length of the shortest piece.

Lesson 2B - DIGIT WORD PROBLEMS  Starting Accurately Digit word problems are problems that involve individual digits in integers and how digits are related according to the question. At the end of the lesson, students are expected to solve digit word problems.  Stimulating Learning

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To solve digit word problems, we need a knowledge on solving a system of linear equations.

 Inculcating Concepts Example 1. The sum of the digits of a two-digit number is 11. If we interchange the digits, then the new number formed is 45 less than the original. Find the original number. Understanding the problem The sum of the digits of a two-digit number is 11. Interchanging the digits, the new number formed is equal to the original number minus 45. Then, we are to find the original two-digit number. Devising a plan 𝐿𝑒𝑡 𝑥 = 𝑜𝑛𝑒’𝑠 𝑑𝑖𝑔𝑖𝑡, 𝑡 = 𝑡𝑒𝑛’𝑠 𝑑𝑖𝑔𝑖𝑡 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑔𝑖𝑡𝑠 𝑖𝑠 11, 𝑠𝑜 𝑥 + 𝑡 = 11 → 𝑥 = 11 − 𝑡 ′ 𝑆𝑖𝑛𝑐𝑒 𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑡𝑒𝑛 𝑠 𝑑𝑖𝑔𝑖𝑡, 𝑠𝑜 10𝑡 + 𝑥 = 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝐼𝑛𝑡𝑒𝑟𝑐ℎ𝑎𝑛𝑔𝑖𝑛𝑔 𝑡ℎ𝑒 𝑑𝑖𝑔𝑖𝑡𝑠, 𝑥 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑡𝑒𝑛′ 𝑠 𝑑𝑖𝑔𝑖𝑡 𝑎𝑛𝑑 𝑡 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑜𝑛𝑒 ′ 𝑠 𝑑𝑖𝑔𝑖𝑡 𝑆𝑜, 10𝑥 + 𝑡 = 𝑖𝑛𝑡𝑒𝑟𝑐ℎ𝑎𝑛𝑔𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑑𝑖𝑔𝑖𝑡𝑠 𝑇ℎ𝑒 𝑛𝑒𝑤 𝑛𝑢𝑚𝑏𝑒𝑟 𝑓𝑜𝑟𝑚𝑒𝑑 𝑖𝑠 45 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑆𝑜, 10𝑥 + 𝑡 = 10𝑡 + 𝑥 − 45 Carrying out the plan 10𝑥 + 𝑡 = 10𝑡 + 𝑥 − 45 10𝑥 − 𝑥 = 10𝑡 − 𝑡 − 45 9𝑥 = 9𝑡 − 45 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑥 𝑏𝑦 11 − 𝑡 ∶ 9𝑥 = 9𝑡 − 45 9(11 − 𝑡) = 9𝑡 − 45 99 − 9𝑡 = 9𝑡 − 45 −9𝑡 − 9𝑡 = −45 − 99 −18𝑡 = −144 −144 𝑡 = −18

Looking back: 10𝑥 + 𝑡 = 10𝑡 + 𝑥 − 45 10(3) + 8 = 10(8) + 3 − 45 30 + 8 = 80 + 3 − 45 38 = 83 − 45  38 = 38

𝑡 = 8 → 𝑡𝑒𝑛′ 𝑠 𝑑𝑖𝑔𝑖𝑡 𝑥 = 11 − 𝑡 = 11 − 8 = 3 → 𝑜𝑛𝑒 ′ 𝑠 𝑑𝑖𝑔𝑖𝑡 Answer: The original number is 83. Example 2. The ten’s digit of a number is twice the unit’s digit. If the digits are reversed, the new number is 27 less than the original. Find the original number. Understanding the problem The ten’s digit of a number is equal to the unit’s digit times 2. If the digits are reversed, the new number is equal to the original number minus 27. Then, we are to find the original number. Devising a plan 𝐿𝑒𝑡 𝑥 = 𝑢𝑛𝑖𝑡’𝑠 𝑑𝑖𝑔𝑖𝑡, 2𝑥 = 𝑡𝑒𝑛’𝑠 𝑑𝑖𝑔𝑖𝑡 ′ 𝑆𝑖𝑛𝑐𝑒 2𝑥 𝑖𝑠 𝑡ℎ𝑒 𝑡𝑒𝑛 𝑠 𝑑𝑖𝑔𝑖𝑡, 𝑠𝑜 10(2𝑥) + 𝑥 = 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑅𝑒𝑣𝑒𝑟𝑠𝑒𝑑 𝑡ℎ𝑒 𝑑𝑖𝑔𝑖𝑡𝑠, 𝑥 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑡𝑒𝑛′ 𝑠 𝑑𝑖𝑔𝑖𝑡 𝑎𝑛𝑑 2𝑥 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑢𝑛𝑖𝑡′𝑠 𝑑𝑖𝑔𝑖𝑡 𝑆𝑜, 10𝑥 + 2𝑥 = 𝑟𝑒𝑣𝑒𝑟𝑠𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 𝑑𝑖𝑔𝑖𝑡𝑠 𝑇ℎ𝑒 𝑛𝑒𝑤 𝑛𝑢𝑚𝑏𝑒𝑟 𝑓𝑜𝑟𝑚𝑒𝑑 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 27 𝑆𝑜, 10𝑥 + 2𝑥 = 10(2𝑥) + 𝑥 − 27

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Carrying out the plan 10𝑥 + 2𝑥 = 10(2𝑥) + 𝑥 − 27 10𝑥 + 2𝑥 = 20𝑥 + 𝑥 − 27 10𝑥 + 2𝑥 − 20𝑥 − 𝑥 = −27 −9𝑥 = −27 −27 𝑥 = −9 𝑥 = 3 → 𝑢𝑛𝑖𝑡′𝑠 𝑑𝑖𝑔𝑖𝑡 2𝑥 = 2(3) = 6 → 𝑡𝑒𝑛′ 𝑠 𝑑𝑖𝑔𝑖𝑡

Looking back: 10𝑥 + 2𝑥 = 10(2𝑥) + 𝑥 − 27 10(3) + 2(3) = 10{2(3)} + 3 − 27 30 + 6 = 60 + 3 − 27 36 = 63 − 27  36 = 36

Answer: The original number is 63

Example 3. The sum of the digits of a three-digit number is 6. The hundred’s digit is twice the unit’s digit, and the ten’s digit equals the sum of the other two. Find the number. Understanding the problem The sum of the digits of a three-digit number is 6. The hundred’s digit is equal to the unit’s digit times 2, and the ten’s digit equals the sum of the hundred’s and unit’s digits. Then we are to find the number. Devising a plan 𝐿𝑒𝑡 𝑥 = 𝑢𝑛𝑖𝑡’𝑠 𝑑𝑖𝑔𝑖𝑡, 2𝑥 = 𝑡𝑒𝑛’𝑠 𝑑𝑖𝑔𝑖𝑡, 𝑥 + 2𝑥 = 𝑡𝑒𝑛′ 𝑠 𝑑𝑖𝑔𝑖𝑡 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑔𝑖𝑡𝑠 𝑜𝑓 𝑎 𝑡ℎ𝑟𝑒𝑒 − 𝑑𝑖𝑔𝑖𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠 6: 𝑆𝑜, 𝑥 + 2...