MATH 1

4.3 value of P10 coins 4.4 first degree mathematical sentence in one variable for “The value of all coins is at most P110.”5. Let d be the distance ran by Henry. Diego ran a third as far as Henry. Write the first-degree equation that models the statement “The distance ran by Henry is four times the distance ran by Diego.”. Answer Key on page 22What to do after (Posttest)Multiple Choice. Choose the letter of the correct answer.1. Which of the following is a mathematical sentence?a. x2 + 6x + 9 c. x - 9b. y + 1/y < 5 d. x2 + 4x + 42. The following are inequalities EXCEPTa. 2x – 5 > x + 2 c. 3x + 5 < 2x – 7b. 3n2 + 5n –2 < 3n + 4 d. x2 – x = 23. The following are open sentences EXCEPT?a. She is the first woman president in the Philippines.b. x is a counting number divisible by five.c. The Philippines is composed of 10 regions.d. x2 + 5x + 4 = 04. Which of the following is an equation?a. 5 + 7 c. x > yb. 2x – y d. 6 = a5. What is an equivalent verbal statement for “2 (x + y) – 5 ≥ 12”. 18

a. Twice the sum of x and y less five is grater than or equal to twelve.b. Twice x plus y minus five is greater than twelve.c. Twice the sum of x and y less than five is greater than twelve.d. Twice x plus y minus five is greater than or equal to twelve.6. Which of the following is a verbal sentence for the inequality 3x ≥ 12?a. Three times a number is greater than twelve.b. Thrice a number multiplied by three is greater than or equal to twelve.c. Thrice a number is at least twelve.d. Three times a number is at most twelve.7. What is the mathematical sentence for “A number diminished by eight is thirteen.”?a. x – 8 = 13 c. x + 8 = 13b. 8 – x = 13 d. x – 8 > 138. Which of the following is the mathematical sentence for the verbal sentence “The product of a number n and sixteen added to another number m is 50.”?a. n + 16 + m = 50 c. n + 16m = 50b. 16n + m = 50 d. 16nm = 509. Which of the following is used as a symbol to represent a first-degree equation in one variable?a. = b. < c. ≤ d. >10. The following are first-degree inequalities in one variable EXCEPTa. x2 – 4x > 2x – 5 c. 3 > -5x + 2b. 3x – 2 < x + 7 d. 7x – 10 ≥ 5x – 111. If a is any nonzero real number and b is any real number, which of the following represents a first-degree equation in x?a. ax + b = 0 c. ax2 + b = 0b. ax + by = 0 d. ax + b > 012. Which of the following is a first-degree equation in one variable? i. v – 3 = 15 ii. v2 = 100 iii. 6 = 12 - va. i only b. ii only c. i and ii d. i and iii 19

13. What equation represents the distance d traveled by a bus at the rate of 80 kph in 2 hours?a. d = 80 (2) c. 2 = 80db. 80 = 2d d. d = 80 + 214. The number of P100-bills is 4 more than thrice the number of P50-bills. If there are x-P50 bills, what first-degree equation in one variable will represent the sentence “The total number of bills is 28”.?a. x + 4 + 3x = 28 c. x = 28b. 28 = 4 + 3x d. x + 4 + x = 2815. If Celia’s age at present is represented by c, what first-degree equation in one variable will represent “Celia’s age 15 years from now is 23 less than twice her age at present.”?a. c = 2c – 23 c. c + 15 = 2c – 23b. –15 = 2c – 23 d. 15 + c = 23 – 2c Answer Key on page 22 20

Answer KeyPretest page 31. d 6. d 11. a2. c 7. c 12. a3. c 8. d 13. d4. c 9. a 14. b5. a 10. b 15. dLesson 1 Self-check 1 page 9A. 1. mathematical phrase 6. mathematical sentence 2. mathematical sentence 7. mathematical phrase 3. mathematical sentence 8. mathematical phrase 4. mathematical phrase 9. mathematical sentence 5. mathematical phrase 10. mathematical sentenceB. 1. true 2. open 3. true 4. open 5. falseC. 1. = 2. = 3. > 4. = 5. <Lesson 2 Self-check 3 page 13A. 1. n = 6 = 2 4. 7(8 + a) ≥ 10 2. y – 16 = 38 5. 4 – 6y = 8 3. 4a – 7 < 6B. 1. The sum of twice a number and five is nine. 2. The sum of nine and thrice a number is eighteen. 3. Sixteen less than twice a number is less than or equal to four. 4. Twice the sum of a number and one is eight. 5. Four times a number subtracted by three is at least 16.Lesson 3 Self-check 3 page 151. first-degree equation in one variable; there is only one variable raised to exponent 12. not a first-degree equation in one variable; the exponent of the variable is not 1 21

3. not a first-degree equation in one variable; the exponent of the variable is not 14. not a first-degree equation in one variable; there are 2 variables5. first-degree equation in one variable; there is only one variable raised to exponent 1Lesson 4 Self-check 4 page 171. w – 2 + w = 2102. 6 = 2(1 + w) + w3. 5(c + c + 20) = 8(c + 20)4. 4.1 5f 4.2 17 – f 4.3 10(17-f) 4.4 5f + 10(17 – f) ≤ 1405. 4d ≤ 2( 1/3) dPosttest page 181. b 6. c 11. a2. d 7. a 12. d3. c 8. b 13. a4. d 9. a 14. a5. d 10. a 15. c END OF MODULE BIBLIOGRAPHYFuller, Gordon. (1977). College algebra. 4th ed. New York, NY: Van Nostrand Company.Leithold, Louis. (1989). College algebra. USA: Addison-Wesley Publishing Company, Inc.Mckeague, C.P. (1990). Intermediate algebra: A text/workbook. 3rd ed. USA: Harcourt Brace Jovanovich, Inc. 22

Module 10 Guess, Try and CheckWhat this module is all about This module discusses the definition of the solution set of a first degreeequation or inequality in one variable. It also deals with the representation of thesolutions of equations and inequalities on a number line. Equations andinequalities are useful in industry and in other fields like science. Thus, it isimportant to know the different methods of solving equations and inequalities.Some of these methods are discussed in this module.There are five lessons in this module.Lesson 1 Solving Linear Equations in One VariableLesson 2 Solving Linear Inequalities in One variableLesson 3 Solving Linear Equations and Inequalities on a Number LineLesson 4 Solutions of Linear Equations and Inequalities from Replacement SetLesson 5 Methods of Solving Linear Equations and Inequalities in One Variable What you are expected to learn After working on this module, you are expected to:• define the solution set of a first degree equation or inequality in one variable• illustrate the solution set of equations and inequalities in one variable on a number line• find the solution set of simple equations and inequalities in one variable from a given replacement set• find the solution set of simple equations and inequalities in one variable by inspection 1

What to do before (Pretest)Multiple Choice. Choose the letter of the correct answer.1. Which of the following statements is false?a. 3x = 3, if x = 1b. 4 + x = -6, if x = -10c. 2x = 10, if x = 8d. x > -3, if x = 32. Which equation is true if b = 2?a. 3b = 6 c. 1 – 2b = 3b. 4b – 1 = 0 d. b = 3b – 13. What equation has {0} as solution set?a. y + 1 = 0 c. y > 0b. y - 1 = -1 d. y < 14. What is the solution set of 3y + 1 = 10?a. {2} b. {3} c. {4} d. {5}5. Determine the open sentence with –1 as one of its solutions.a. x > 1 c. x < -2b. x < 1 d. x > 06. Which of the following equations represents the graph below? • -2a. d – 1 = 3 c. d – 1 > 3b. d + 1 = -1 d. d + 1 < - 1 2

7. Illustrate the graph of v = ½. a. • -2 -1 b. • -1 0 c. • 01 d. • 128. Which of the following graphs illustrates the solution set of a > 5? a. ° 5 b. • 5 c. ° 5 d. • 5 3

9. Which of the graphs drawn below represents the solution set of the inequality s - 2 < 0? a. ° 2 b. • 2c. ° 2d. • 210. Determine which sentence has 18 as the only solution.a. b + 2 = -16 c. r ≥ 18b. h/2 = 9 d. w ≤ 1811. What is the solution set of 2z + 1 = 1 taken from the replacement set {-2, -1, 0, 1}?a. { -2 } b. { - 1 } c. { 0 } d. { 1 }12. What sentence is satisfied by all elements of the set {-1, 0}?a. 6m – 1 = 7 c. 5x + 3 = 4x + 3b. –2 > t – 3 d. b – 5 < -413. Given the replacement set {1, 2, 3, 4}, determine the solution set of the inequality 2r + 3 < 8.a. { } b. { 1 } c. { 1, 2 } d. { 1, 2, 3 } 4

14. Find the solution set of the inequality 3a + 1 ≥ 4a. { a/a ≥ -1 } c. { a/a ≥ -5/3}b. { a/a ≥ 1 } d. { a/a ≥ 5/3}15. Taken from the replacement set { 2, 4, 6 }, the solution set of the inequality 3 – 2y ≤ -5 isa. { } b. {2, 4} c. {4} d. {4, 6} What you will doLesson 1 In the previous modules, you learned about open sentences, first degreeequations in one variable and first degree inequalities in one variable. Recall thefollowing. An open sentence is a sentence which becomes true or false when the variable is replaced by a given value. An open sentence of the form ax + c = 0 is called a first degree equation in x. An open sentence of the form ax + b > 0, ax + b ≥ 0, ax + b < 0, or ax + b ≤ 0 is called a first degree inequality in x. Let us consider the following first-degree equations. 5

Example 1. 5 + x = 12Suppose 2 is replaced in terms of x. Let us see what will happen. 5 + x = 12If x = 2, 5 + 2 = 12This is false since 7 ≠ 12We say that 2 is not a solution of the given equation. What value are you going to replace in terms of the variable x to make theequation true? ____ If your answer is 7, then you are right. Since 5 + 7 = 12, it means that 7must be replaced for the variable x to make the equation 5 + x = 12 true. Is that the only value which will make the equation true? Yes, 7 is the onlyvalue that will make the given equation true. We call 7 the solution of the first-degree equation 5 + x = 12 and {7}, the solution set.Example 2. 2y – 7 = 3. What value are you going to replace for the variable y to make thesentence true? How will you check if your answer is correct? An example isdone for you. 2y – 7 = 3If y = 6, then 2(6 ) – 7 = 3This is false since 5 ≠3We say that 6 is not a solution of the given equation. Can you guess what will make the equation true? Write your guess andchecking below. 2y – 7 = 3If y = ___, then 2(6 ) – 7 = 3 ___ = 3 If your answer is 5, then you are correct. The variable y must be replacedby 5 to make 2y – 7 = 3 a true statement. Is 5 the only value that will make thatequation true? Yes, 5 is the only value that will make the equation true. Thatmeans, 5 is the solution of the first degree equation 2y – 7 = 3 and thus, {5} isthe solution set of that equation. 6

Example 3. 3z = 18. Can you give a value which will make that equation true? Write youranswer below and show how you check if your answer is correct. 3z = 18If z = _____, then 3(_ ) = 18 ___ = 18 If your answer is 6, then you are correct. Is that the only value which willmake the equation true? _____ Yes, 6 is the only value that will make 3z = 18 atrue sentence. Thus, 6 is the solution of 3z = 18 and {6} is its solution set. From examples 1 – 3 discussed above, how many solutions does a firstdegree equation in one variable have? ___How do you define the solution of a first-degree equation in one variable?________________________________________________________________What about the solution set of a first-degree equation in one variable?________________________________________________________________Let us now summarize. A first-degree equation in one variable has only one solution. The solution of a first-degree equation in one variable is the value that makes the equation a true statement. The solution set of a first-degree equation in one variable is the set of all its solution. In order to apply what we have summarized let us consider otherexamples.Example 4. x – 2 = 10. What is the solution of the given equation? _____ Why? _______The solution of the given equation is 12 because if x is replaced by 12, theequation is true. 7

What is the solution set of the given equation? ______If your answer is {12 }, then you are correct.Example 5. 4a + 1 = 3 What is the solution of the given equation? _____ What is its solutionset? ____If a is replaced by ½, the given equation is true. Thus, the solution of the givenequation is ½ and its solution set is { ½ }.Self-check!I. Determine if the given value of x is a solution of the first degree equation 14 = 3x + 2.1. x = 2 2. x = 4 3. x = 5II. Use guess-and check to give the solution set of each of the following first-degree equations. Tell what number you choose as your first guess and why you chose it.1. x + 1 = 8 3. 4z = 20 5. 2b - 1 = 32. 9a + 2 = 5 4. 2y + 1 = 9Lesson 2 We learned in the previous discussion that a first-degree equation in onevariable has only one solution. What about if we have first-degree inequalities inone variable? Let us discover this using the following examples.Example 1. z > 5. If z = 1, is the inequality true? ______. No, the inequality is false if z = 1because 1 is not greater than 5. Can you give a value that will make the inequality true? _______. 8

Is that the only value that will make the equation true? Can you give othervalues?________________________________________________________________If your answers are all greater than 5, say 5.1, 5.5, 6, 6.75, 6.9, 7, 8, etc., thenyou are correct. Those values that can be replaced to make the inequality trueare called solutions of the given inequality. How many solutions does the inequality z > 5 have? ____________.You are correct. That inequality has many solutions. Some solutions of thegiven inequality are 5.1, 5.5, 6, 6.75, 6.9, 7, and 8. Based on your observations in the discussions above, when is theinequality true? ______________________________________ Correct, theinequality is true if the value of z is greater than 5. Thus, the solution setconsists of all real numbers which are greater than 5. In symbols, the solutionset can be written as {z/z > 5} and is read as “the set of all z’s such that z isgreater than 5”. This is the replacement set of z > 5 with the greatest number ofelements.Example 2. y + 7 < 11 What are some values of y that will make the inequality true? Write youranswers below and show your checking. y + 7 < 11 y + 7 < 11If y = 3 , ( _ ) + 7 < 11 If y = 1 , ( _ ) + 7 < 11 ___ < 11 ___ < 11 y + 7 < 11 y + 7 < 11If y = __ , ( _ ) + 7 < 11 If y = __ , ( _ ) + 7 < 11 ___ < 11 ___ < 11 Give some values which will make the given inequality a true statement.________________________________________________________________Some solutions are -1, -2, -3.5, 0, ¼, 1/3, ½, 1, 2, 3, 3.2, 3.5 and 3.75. In general, what values will make the inequality true? _______ If youranswer is all real numbers less than 4, then you are right. What is the solution set of the given inequality and how do you write thatin symbols? The solution set is the set of all ___________________________.In symbols, this is written as {y/y < 4}Example 3. 3a > 1 9

Can you give some values of the variable a which will make the inequalitytrue? Enumerate your answers below and show your checking. 3a > 1 3a > 1If a = ½ , 3( _ ) > 1 If a = __ , 3( _ ) > 1 ___ > 1 ___ > 1 3a > 1 3a > 1If a = __ , 3( _ ) > 1 If a = __ , 3( _ ) > 1 ___ > 1 ___ > 1If your answers are all greater than 1/3, then you are correct. What is thesolution set of the given inequality? __________________________________The solution set is the set of all real numbers greater than 1/3.How do you write the solution set in symbols? ___________________________In symbols, the solution set is {a/a > 1/3}We remember the following. A first-degree inequality in one variable has many solutions. A solution of a first-degree equation in one variable is a valuethat makes the inequality a true statement. The solution set of a first-degree equation in one variable isthe set of all its solution.Let us have other examples.Example 4. h + 2 > 5. What are some values that will make the inequality true? Write youranswers and checking below. h+2>5If h = 3.5 , ( _ ) + 2 > 5 ___ > 5h +2>5 h +2>5 10

If h = __ , ( _ ) + 2 > 5 If h = __ , ( _ ) + 2 > 5 ___ > 5 ___ > 5 h +2>5If h = __ , ( _ ) + 2 > 5 ___ > 5 What is the solution of the inequality? ____________________________The solution of the inequality consists of all real numbers greater than 3. What is the solution set written in symbols? In symbols, the solution set is{h/h > 3}.Example 5. 2r + 1 ≤ 3 Give some values that will make the inequality true. Write your answersand checking below.If r = __ , 2r + 1 ≤ 3 If r = __ , 2r + 1 ≤ 3 2( _ ) + 1 ≤ 3 2( _ ) + 1 ≤ 3 ___ ≤ 3 ___ ≤ 3If r = __ , 2r + 1 ≤ 3 2( _ ) + 1 ≤ 3 ___ ≤ 3 What is the solution set of the given inequality? ____________________.The solution set is {r/r ≤ 1}Self-check!I. Determine whether the inequality 3a < -6 is true or false for the given value of a. 1. a = -2 2. a = -3 3. a = -4II. Give the solution set of each of the following inequalities and check for 3 values in the solution set. 11

1. p + 3 < 0 6. –6z > 182. 4s < 0 7. 2c – 1 ≤ 13. m + 1 > 0 8. q – 5 ≥ 114. 2y > -2 9. 2r + 1 ≥ 55. w + 2 ≤ 12 10. 2b - 1 ≥ -3Lesson 3 Consider the number line below and the position of the numbers –5, –3, 1,1½ , and 6. •• •• • -5 -4 -3 -2 -1 0 1 2 3 4 5 6What do you notice with the points and the real numbers on the number line?________________________________________________________________You are correct. Every real number corresponds to a point on the number lineand every point on the number line corresponds to a real number. The realnumber corresponding to a point on the number line is called the coordinate ofthe point. Let us represent the solution set of each equation that we had inexploration 1 using a number line. Equations Solution Set 1. 5 + x = 12 {7} 2. 2y – 7 = 3 {5} 3. 3z = 18 {6} 4. x – 2 = 10 {12} 5. 4a + 1 = 3 {½} x=71. • -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 y=52. • -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 12

z=63. • -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 x = 124• -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 a=½5. • -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 The representation of the solution of an equation on a number lineis called the graph of the equation. What do you notice with the graph ofthe equations given above?__________________________________________________________You are correct. The graph of a first-degree equation in one variable isjust a point.What if we are to graph the inequalities that we had in exploration 2?Inequalities Solution Set z>5 {z/z > 5} {a/a > 5} 3a > 1 {h/h > 3}h+2>5 {y/y < 4}y + 7 < 11 {r/r ≤ 1}2r + 1 ≤ 3 z>51. ° -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 a>5°2. -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 h>33. ° -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 y<4 13

°4. -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 r≤15. • -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12What do you observe with the graph of a first-degree inequality in onevariable?___________________________________________________________Right. The graph of a first-degree inequality in one variable is a ray. Theray can be extended to the right or to the left.When is the ray extended to the left? _____________________________The ray is extended to the left if the solution set consists of real numberswhich are less than a given value.When is the ray extended to the right?On the other hand, the ray is extended to the right if the solution setconsists of real numbers which are greater than a given value.We remember the following. The graph of a first degree equation in one variableis a point. The graph of a first degree inequality in one variableis a ray as illustrated below.inequality graphx<a ° ax≤a • a°x > a ax≥a • 14

aSelf-CheckI. Represent the following solution sets on a number line.1. x = 3.5 6. a > 12.52. y = -6 7. d > -93. h < -2 8. k ≥ 04. r ≤ 5 9. m ≥ -45. z ≤ -8 10. n ≥ 10II. Write the equation or inequality described by each graph. Use the variable x.1. • -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 122. • -1 0 1 23. • -2 -1 0 1 24. • 1 -2 -1 0°5. -2 -1 012 Lesson 4 In lessons 1 and 2, we learned what are the solutions and the solutionsets of equations and inequalities. In lesson 3, we studied how to represent anequation or inequality on a number line. This time, we will learn how to find thesolution of a given equation or inequality from a given replacement set. The set 15

of real numbers is considered the largest replacement set from which the solutionset of a first-degree equation or a first-degree inequality may be obtained. Recall that a first degree equation or inequality in one variable may be trueor false when replaced by a specific value. The value(s) that would replace thevariable will be taken from a replacement set. You should always rememberthat the solution set taken from a given replacement set is not necessarily thesolution set of an equation or inequality.We consider the following examples.Example 1. Given the replacement set {1, 2, 3, 4, 5}, find the solution set of theequation 3x + 1 = 16.Solution: If x = 1 → 3(1) + 1 = 16 If x = 2 → 3(2) + 1 = 16 4 = 16 false 7 = 16 false If x = 3 → 3(3) + 1 = 16 If x = 4 → 3(4) + 1 = 16 10 = 16 false 13 = 16 false If x = 5 → 3(5) + 1 = 16 16 = 16 true Thus, the solution of the equation 3x + 1 = 16 is 5 and the solution set is{5}.Example 2. Given the replacement set {-1, 0, 1}. Find solution(s) of the inequality a – 5 < -4Solution:If a = -1 → -1 – 5 < -4 If a = 0 → 0 – 5 < -4 -6 < -4 true -5 < -4 trueIf a = 1 → 1 – 5 < -4 -4 < -4 falseThus, some solutions are –1 and 0. Finally, from the given replacement set, thesolution set of a – 5 < -4 is {-1, 0}.However, you should note that {-1, 0} is not the solution set of the inequality a –5 < -4 when the replacement set is the set of all real numbers. {-1, 0} is just thesolution set taken from the given replacement set which is {-1, 0, 1}. 16

Example 3. Given the replacement set {1, 1½, 2}, find the solution set of –2>t – 3.Solution: If t = 1 → -2 > 1 – 3 Is this true or false? __________You are correct. The resulting sentence is false, because –2 = -2. If t = 1½ → -2 > 1½ – 3 Is this true or false? _______The resulting sentence is false because –2 < -1½ . If t = 2 → -2 > 2 – 3 Is this true or false? _______The resulting sentence is false because –2 < -1.What does the above discussions imply?________________________________The discussions above implies that none from the replacement set makes theinequality true. Hence, from the given replacement set, the solution set is anempty set or φ (read as null set).Again, it may be noted that when the replacement set is the set of all realnumbers, the solution set of the inequality –2 > t – 3 is not the empty set.What do you observe from examples 1 –3? How do we get the solution set ofan equation or inequality from a given replacement set?________________________________________________________________Let us now summarize what we discussed. To find the solution set of an equation or inequality from a given replacement set, we substitute all the elements of the given replacement set. The set of value(s) which makes the equation or inequality a true statement is the solution set. If no element from the replacement set makes the equation or inequality true, the solution set is an empty set. The solution set of an equation or inequality taken from a given replacement set is not necessarily the solution set of the given equation or inequality when the replacement set is the set of all real numbers. 17

Self-CheckFind the solution set of each of the given sentence when the replacement set is{-2, -1, 0, 1, 2}.1. x + 1 = 2 6. 2r + 4 ≥ 82. 2y + 1 = 9 7. a – 2 < 63. 4z = 4 8. 2c + 1 < 84. y + 2 > 2 9. b – 1 ≤ -35. h – 1 > -2 10. 2f + 1 ≤ -2Lesson 5 In the previous lessons, we learned how to find the solution set of firstdegree equations or first degree inequalities. When you are asked to solve anequation or inequality, it means that you are to find its solution set. The method of solving equations or inequalities that was illustrated in theprevious explorations is by inspection. There are several ways of doinginspection, namely: guess-and-check, cover-up and working backwards. Suchmethods will be discussed in this exploration.Example 1. Consider the equation 2x + 6 = 14.Let us illustrate how this equation is solved using three methods.Method 1. Guess-and-Check This is the method that we used in the previous explorations. In thismethod, one guesses and substitutes the value to see if a true equation results. 2x + 6 = 14 If x = 2 → 2x + 6 = 14If x = 1 → 2 (1) + 6 = 14 This is false since 2 (_) + 6 = 14This is false since 8 ≠ 14 10 ≠ 14 2x + 6 = 14 2x + 6 = 14If x = 3 → 2 (_) + 6 = 14This is false since __ ≠ 14 If x = __ → 2 (_) + 6 = 14 This is now ____since __ = 14 18

From the discussions above, what is the solution of the given equation? ______We see that 4 is the solution and {4} is the solution set of the given equation.Method 2. Cover-upIn this method, we cover-up the term with the variable. 2x + 6 = 14 + 6 = 14What number should be added to 6 to get 14? _______If your answer is 8, then you are correct. 2x + 6 = 14 → 8 + 6 = 14 2 + 6 = 14What number should be multiplied by 2 to give 8? ________Your answer must be 4.This means that x = 4 and {4} is the solution set.Method 3. Working backwardsIn this method, the reverse procedure is used.The equation 2x + 6 = 14 shows the following. x is multiplied by 2. The product is added to 6 to get 14.Thus, if the process is reversed, we have 6 is subtracted from 14. (The answer is 8) The result is divided by 2. (The answer is 4)This means that x = 4 and the solution set is {4}.Example 2. Solve the inequality y + 3 < 5Method 1. Guess-and-CheckIf y = 5 → y +3<5 y+3<5 (5) + 3 < 5 If y = 3 → (3) + 3 < 5This is false since This is false since 6 > 5 8 >5 19

If y = 2 → y+3<5 (2) + 3 < 5This is false since 5 =5 At this point, you should see that we can now substitute a value that isless than 2 to make the inequality true. Some examples are shown below. y+3<5 y+3<5If y = 1 → (1) + 3 < 5 If y = ½ → (½ )+ 3 < 5This is true since 4 < 5 This is true since 3 ½ < 5Thus, what is the solution of the given inequality? ________________________You are correct. The solution is the set of all real numbers which are less than 2.What is the solution set? ____________________________________________The solution set is { y/y < 2 }.Method 2. Cover-up y +3<5 +3<5What number should be added to 3 to get a value less than 5? _______If your answer is less than 2, then you are correct.This means that y < 2 and {y/y < 2}is the solution set.Method 3. Working BackwardsThe inequality y + 3 < 5 shows the following. y is added to 3. The sum is less than 5.Thus, if the process is reversed, we have 3 is subtracted from 5. (The answer is 2) The result is less than 2.This means that y < 2 and the solution set is {y/y < 2}. 20

Remember that To solve an equation/inequality using 1. Guess-and-Check We guess and substitute some values to see if a true equation/inequality results. The value/s which make/s the equation/inequality true is the solution of the equation/inequality. 2. Cover-up We cover-up the term with the variable and solve for value/s which will make the resulting open sentence true. 3. Working Backwards We reverse the procedure in the given equation/inequality.Self-CheckI. Solve the following equations by inspection.1. 23 = 7 + F 4. 4 + 2p = 102. 144 = 12a 5. 86 = b – 23. ½ h = 9II. Represent the solution sets of the equations given in I on a number line.III. Find the solution set of the following inequalities by inspection and graph the solution set on a number line.1. t + 7 < 8 4. 2x – 2 > 62. 2a ≤ 6 5. 3y ≥ -93. ½ b > 4 21

PosttestMultiple Choice. Choose the letter of the correct answer.1. The following are true EXCEPTa. 19 – x = 25, if x = 6 c. 21 – 4m = 13, if m = 2b. 7a + 2 = 16, if a = 2 d. –5 = -b + 4, if b = 92. Which of the following is true if a = -1?a. a – 7 > 0 c. 3a ≥ -1b. 2a < 0 d. 2 + a ≤ -33. What equation has {-1} as solution set?a. y + 1 = 0 c. y > -1b. y - 1 = -1 d. y < -14. Determine the open sentence with -2 as one of its solutions.a. x > 1 c. x < -2b. x < 1 d. x > -25. Which of the following equations represents the graph below?c. M + 5 = -11 •d. 11 = A - 5 6 c. T + 5 = 11 d. H – 5 = -11 22

6. Illustrate the graph of a = -½.a. • -2 -1b. • -1 0c. • 01 d. • 127. Represent the graph below by an open sentence. ° -5a. x = -5 b. x > -5 c. x ≤ -5 d. x ≥ -58. Which of the following graphs illustrates the solution set of a ≥ 2?a. ° 2b. • 2c. ° 2d. • 2 23

9. Given the replacement set {1, 2, 3}, find the solution set of 2(n + 1) = 10. a. {1} b. {2} c. {3} d. {4}10. What is the solution set of 24 – x = 30? a. {6} b. {-6}c. {54} d. {-54}11. The following equations has –3 as solution EXCEPT a. 23 – x = 26 c. –y = 3 b. 6x = -18 d. y – 3 = 612. Determine which sentence has 7 as the only solution. a. b + 2 = 9 c. r ≥ 7 b. h/2 = 14 d. w ≤ 713. Taken from the replacement set {-2, -1, 0, 1}, the solution set of the inequality c – 5 < 5 is a. {0} b. {-2, -1} c. {-2, -1, 0, 1} d. { }14. What open sentence is satisfied by all elements of the replacement set {½, 1}? a. 4x = 2 c. 2n > 1 b. j – 0 = 1 d. 3t < 415. Find the solution set of the inequality 3n > 21. a. {n/n > 0} c. {n/n < 0} b. {n/n > 7} d. {n/n < 7} Answer KeyPretest 4. b 7. c 10. b 13. c 5. b 8. c 11. c 14. a1. c 6. b 9. a 12. d 15. d2. a3. b 24

Exploration 1 Self-Check I. 1. 14 = 3(2) + 2 False, thus not a solution. 2. 14 = 3(4) + 2 True, thus a solution. 3. 14 = 3(5) + 2 False, thus, not a solution II. 1. The solution set is {7} because 7 + 1 = 8. 2. {1/3} is the solution set because 9 (1/3) + 2 = 5 3. The solution set is {5} because 4(5) = 20. 4. {4} because 2(4) + 1 = 9 5. {2} because 2(2) – 1 = 3Exploration 2 Self-CheckI. 1. 3 (-2) < -6 false 2. 3 (-3) < -6 true 3. 3 (-4) < -6 trueII. If p = -6, -6 + 3 < 0 1. {p/p < -3} If p = -4, -4 + 3 < 0 -3 < 0 -1 < 0 If s = -3, 4(-3) < 0 If p = -5, -5 + 3 < 0 -12 < 0 -2 < 0 If m = 2, 2 + 1 > 0 2.. {s/s < 0} If s = -1, 4(-1) < 0 3>0 -4 < 0 If y = 2, 2 (2) > -2 If s = -2, 4(-2) < 0 4 > -2 -8 < 0 If z = -6, -6 (-6) > 18 3. {m/m > -1} If m = 0, 0 + 1 > 0 36 > 18 1>0 If m = 1, 1 + 1 > 0 2>0 4. {y/y > -1} If y = 0, 2 (0) > -2 0 > -2 If y = 1, 2 (1) > -2 0 > -2 5. {z/z < -3} If z = -4, -6 (-4) > 18 24 > 18 If z = -5, -6 (-5) > 18 30 > 1825

6. {w/w ≤ 10} If w = 10, 10 + 2 ≤ 12 12 ≤ 12 If w = 11, 11 + 2 ≤ 12 If w = 12, 12 + 2 ≤ 12 13 ≤ 12 14 ≤ 127. {c/c ≤ 1} If c = 1, 2 (1) – 1 ≤ 1 If c = -1, 2 (-1) – 1 ≤ 1 1≤1 -3 ≤ 18. {q/q ≥ 16} If c = 0, 2 (0) – 1 ≤ 1 If q = 18, 18 – 5 ≥ 11 -1 ≤ 1 13 ≥ 11 If q = 16, 16 – 5 ≥ 11 If r = 4, 2 (4) + 1 ≥ 5 11 ≥ 11 9 ≥5 If q = 17, 17 – 5 ≥ 11 If b = 1, 2 (1) - 1 ≥ -3 12 ≥ 11 1 ≥ -39. {r/r ≥ 2} If r = 2, 2 (2) + 1 ≥ 5 5≥5 If r = 3, 2 (3) + 1 ≥ 5 7≥510. {b/b ≥ -1} If b = -1, 2 (-1) - 1 ≥ -3 -3 ≥ -3 If b = 0, 2 (0) - 1 ≥ -3 -1 ≥ -3Exploration 3 Self-CheckI. 1. • 2. 0 3.5 • 0 -63. ° -2 26

4. 55. -86. ° 12.57. ° -98. 09. -4 10. 10II. 1. x = 4 2. x = -½ 3. x = 2/3 4. x = -1/4 5. x = -1 1/3Exploration 4 Self-Check1. If x = 1, x + 1 = 2, 1 + 1 = 2 True Solution set: {1}2. If all elements of the replacement set are substituted for y, the resulting statements are all false. Thus, the solution set is { }.3. If z = 1, 4z = 4, 4(1) = 4 True Solution set: {1}4. If y = 1, y + 2 > 2, 1 + 2 > 2 True Solution set: {1, 2} If y = 2, y + 2 > 2, 2+ 2 > 2 True Solution set: {0, 1, 2} 0 – 1 > -2 True5. If h = 0, h – 1 > -2, 1 – 1 > -2 True If h = 1, h – 1 > -2, 2 – 1 > -2 True If h = 2, h – 1 > -2, 27

6. If r = 2, 2r + 4 ≥ 8, 2(2) + 4 ≥ 8 True Solution set: {2}7. If a = -2, a – 2 < 6, -2 – 2 < 6 TrueIf a = -1, a – 2 < 6, -1 – 2 < 6 TrueIf a = 0, a – 2 < 6, 0 – 2 < 6 TrueIf a = 1, a – 2 < 6, 1 – 2 < 6 True If a = 2, a – 2 < 6, 2 – 2 < 6 True Solution set: {-2, -1, 0, 1,2}8. If c = -2, 2c + 1 < 8, 2(-2) + 1 < 8 TrueIf c = -1, 2c + 1 < 8, 2(-1) + 1 < 8 TrueIf c = 0, 2c + 1 < 8, 2(0) + 1 < 8 TrueIf c = 1, 2c + 1 < 8, 2(1) + 1 < 8 TrueIf c = 2, 2c + 1 < 8, 2(2) + 1 < 8 True Solution set: {-2, -1, 0, 1,2}9. If all elements of the replacement set are substituted for b, the results are all false. Thus, the solution set is { }.10. If f = -2, 2f + 1 ≤ -2, 2(-2) + 1 ≤ -2 True Solution set: {-2}Exploration 5 Self-CheckI.1. 23 = 7 + F → 23 = 7 + 16 → F = 162. 144 = 12a → 144 = 12 (12) → a = 123. ½ h = 9 → ½ (18) = 9 → h = 184. 4 + 2p = 10 → 4 + 2 (3) = 10 → p = 35. 86 = b – 2 → 86 = 88 – 2 → b = 88II.1. • 0 162. • 0 12 28

3. • 0 184. • 035. 0 • solution set: {t/t < 1}III. 881. t + 7 < 8 °2. 2a ≤ 6 solution set: {a/a ≤ 3} 13. ½ b > 4 solution set: {b/b > 8} 34. 2x – 2 > 6 solution set: {x/x > 4} °5. 3y ≥ -9 solution set: {y/y ≥ -3} 8Posttest 6. b 7. b ° 1. a 8. c 2. b 9. d 4 3. a 10. c 4. b -3 5. c 11. d 12. a 13. c 14. d 15. b 29

Module 11 The Way to XYZ What this module is all about This module discusses the different properties of equality and inequality and itsapplication in solving first degree equations and inequalities in one variable. As mentionedin Module 10, equations and inequalities are useful in industry and other fields like sciences.To illustrate, a certain investor might be interested in the set of values where the operationalexpenses are minimized while the benefits are maximized. Your previous knowledge in theproperties of real numbers and skills on the four fundamental operations on monomials willhelp much in learning this module. This module has four lessons: Lesson 1 Properties of Real Numbers Lesson 2 Properties of Equality Lesson 3 Solving Linear Equations in One Variable Lesson 4 Different Properties of Inequality and the Solutions of Inequalities in One Variable What you are expected to learnAfter going through this module, you are expected to:• review basic properties of real numbers;• state and illustrate the different properties of equality;• determine the solution set of first degree equations in one variable by applying the properties of equality;• determine the solution set of first degree inequalities in one variable; and,• visualize solutions of simple mathematical inequalities on a number line. 1

How to learn from this moduleThis is your guide for the proper use of the module: 1. Read the items in the module carefully. 2. Follow the directions as you read the materials. 3. Answer all the questions that you encounter. As you go through the module, you will find help to answer these questions. Sometimes, the answers are found at the end of the module for immediate feedback. 4. To be successful in undertaking this module, you must be patient and industrious in doing the suggested tasks. 5. Take your time to study and learn. Happy learning! The following flowchart serves as your quick guide in using this module. Start Take the Pretest Check your paper and count your correct answers. Is your score Yes Scan the items you 80% or above? missed. No Proceed to the nextStudy this module module/STOP.Take the Posttest 2

What to do before (Pretest)A. Matching type: Match the number sentence in Column A to the property of real numbers it demonstrates found in Column B. Column A Column B1) (2+3)+7 = 2 + (3+7) a) Associative Property2) (5 • 12) • 7 = 60 • 7 b) Commutative Property3) (8 + 10) + 15 = 15 + (8 + 10) c) Closure Property4) 9 • (11 + 23) = (9•11) + (9•23) d) Identity Property5) 8 • (12 • 1/12) = 8 • 1 e) Inverse Property f) Distributive PropertyB. Matching type: Match the number sentence in Column C to the property of equality or inequality it demonstrates found in Column D. Column C Column D6) If 8 + 2 < 14 and 14 < 20, g) Addition Property of Equality h) Multiplication Property of Equality then 8 + 2 < 20. i) Multiplication Property of Inequality7) If (m-n)< (p+q) and r > 0, j) Reflexive Property k) Symmetric Property then (m-n)r < (p+q)r. l) Transitive Property8) If m=n, then m + p = n + p.9) If q + r = 15, then 15 = q +r.10) If 15y = 75, then 3y = 15.11) What is the solution of the equation 2(m – 3) = 4(3 + m)?a. -9 c. -3b. -6 d. 012) What is/are the value/s of n that will make the equation true in (3n - 7) = 1/5 (21n + 3)?a. n ≥ 8 c. 8b. n ≤ -8 d. -813) Which of the following is the solution set of 10 + 5 < 3 - 2z?a. z < -5 c. z < 6b. z < -6 d. z < 5 3

14) Which of the following graphs represents the solution set of 8y – 5 ≥ 10 + 3y?a. -1 0 1 2 3 4 5 6 7 8b. -1 0 1 2 3 4 5 6 7 8c. -1 0 1 2 3 4 5 6 7 8d. -1 0 1 2 3 4 5 6 7 815) Which of the following is the solution set of 3q – 19 > 16 – 2q?a. q > 7 c. q > 35b. q < 7 d. q < 35 Answer Key on page 29 What you will do Read the following lessons carefully. Then do the suggested activities patiently.Lesson 1 Properties of Real Numbers In the previous module, you were taught how to find solutions to first-degreeequations and inequalities in one variable using a given set of values and by inspection. 4

In this module, you will learn another way of solving first-degree equations andinequalities in one variable. Let’s take a moment to review the properties of real number that you learned before. Properties of Real Number1. Closure Propertya) Closure Property for Addition: The sum of any pair of real numbers is also a realnumber. In notation, we have :If a, b ∈ R, then a + b ∈ R. (“∈ reads as element”)Ex. 1. 2 + 8 = 10 ; 2. 5 ÷ 18 = 23b) Closure Property for Multiplication: The product of any pair of real numbers is alsoa real number. In notation, we have:If a, b ∈ R, then ab ∈ R.Ex. 1. 2(6) = 12 ; 2. 8(7) = 562. Commutative Propertya) Commutative Property for Addition: The sum of two real numbers is the same,no matter what order the numbers are added. In notation, we have:a+b=b+a1. 8 + 5 ? 5 + 8; 2. 12.5 + 3.8 ? 3.8 +12.5 13 = 13 16.3 = 16.3b). Commutative Property for Multiplication: The product of two real numbers isthe same, no matter what order the numbers are multiplied. In notation, we have:ab = baThis means that multiplying two real numbers will give the same productregardless of the order in which the numbers are multiplied.Ex. 1. 3 • 7 ? 7 • 3 2. 2/3 • 1/5 ? 1/5 • 2/3 21 = 21 2/15 = 2/15 5

3. Associative Propertya) Associative Property for Addition. The sum of three or more real numbers is thesame, no matter how the numbers are grouped. In notation, we have:a + b + c = (a + b) + cEx. 1. (6 + 5) + 9 ? 6 + (5 + 9)11 + 9 ≟ 6 + 14 20 = 20b) Associative Property for Multiplication. The product of three or more realnumbers is the same, no matter how the numbers are grouped. In notation,(ab)c = a(bc)Ex. ( 20 • ¼ ) • 2 ? 20 • ( ¼ • 2 ) 5 • 2 ≟ 20 • ½ 10 = 104. Identity Propertya) Identity Property for Addition. The sum of any real number and zero is equal to the given number. The number 0 is called the additive identity.a + 0 = a and 0 + a = aEx. 1. 0 + 8 = 8 2. 25m + 0 = 25mb) Identity Property for Multiplication. The product of any real number and 1 is equal to the given number. The number 1 is called the multiplicative identity.a • 1 = a and 1 • a = a Ex. 1. 37 • 1 = 37 2. 1 • 19n = 19n5. Inverse Propertya) Inverse Property for Addition. The sum of a real number and its opposite is 0. Thenumber opposite the given real number is called its additive inverse.a + (-a) = 0 and (-a) + (a) = 0Ex. 1. 2/3 + ( -2/3 ) = 0 2. –5q + (5q) = 0 6

b) Inverse Property for Multiplication. The product of a real number and itsreciprocal is 1. The reciprocal of any given real number is called its multiplicativeinverse. Note also that zero does not have a multiplicative inverse.a • 1/a = 1 and 1/a • a = 1Ex. 1. 2/3 • 3/2 = 1 2. (-5) • ( - 1/5) = 115)Distributive Property. Multiplication is distributive with respect to addition.a(b + c) = ac and (b + c)a = ba + caEx. 1. 2(3 + 5) = 2(3) + 2(5) ; 2. (7n+ 6)5 = 7n(5) + 6(5)16)Multiplicative Property of Zero. The product of a given real number and 0 is 0.a · 0 = 0 and 0 · a = 0Ex. 1. 2(0) = 0 2. 0(9m + 2) = 0Before you proceed to take the self-Check, remember the following:In each statement that follows, a, b, and c, are all real numbers.A. Closure Property : If a, b, c ∈ R, then a + b ∈ R. If a, b ∈ R, then ab ∈ R.B. Commutative Property: a + b = a + b a(b) = b(a)C. Associative Property: (a + b) + c = a + (b + c) (ab)c = a(bc)D. Identity Property: a + 0 = a and 0 + a = a a(1) = a and 1(a) = aE. Inverse Property: a + (-a) = 0 and (-a) + a = 0 a · 1/a = 1 and 1/a · a = 1F. Distributive Property: a(b + c) = ab + ac (b + c)a = ba + caG. Properties of Multiplication: 0(a) = 0 and a (0) = 0. 7

Self-check 1Name the property illustrated in each statement.1. 2 + 3 = 5 _______________________2. 3 + 8 = 8 + 3 _______________________3. 5(7) = 35 _______________________4. 2(5 + 7) = 10 + 14 _______________________5. 2/5 + 3/7 = 3/7 + 2/5 _______________________6. 18 x 1 = 18 _______________________7. 8 + (-8) = 0 _______________________8. 3/5 + (2/5 + 4/5) = (3/5 + 2/5) + 4/5 _______________________9. 1000(0) = 0 _______________________10. 1,000,000 + 0 = 1,000,000 _______________________ Answer Key on page 29Lesson 2 Properties of Equality In Lesson 1, you have reviewed the different properties of real numbers. Theseproperties are very helpful in solving first-degree equations and inequalities in one variable.Another set of properties, the Properties of Equality will be useful in solving equations. Here are the properties of equality. Properties of Equality1. Reflexive Property of Equality (RPE) Observe the following: 8 =8 y+4=y+4 5m + 3 = 5m + 3 20n – 7 = 20 n – 7 8

These equations demonstrate the reflexive property of equality. How should we complete the equation below to show the reflexive property of equality? 8p – 12 = ________? If your answer is 8p – 12, then you are correct. The above equations are examples that demonstrate reflexive property of equality. In your own words, explain the reflexive property. _______________________________________________ Then verify your notion with the statement that follows: The reflexive property of equality means that any number is equal to itself.In symbol, we write: a = a, a ∈R 2. Symmetric Property of Equality (SPE) Observe the next set of expressions, and try to establish a pattern. 1. If 3 + 5 = 8, then 8 = 3 + 5. 2. If 20 = 4(5), then 4(5) = 20. 3. If 15 = 2m + 3, then 2m + 3 = 15. 4. If 2w – 7y = 5z, then 5z = 2w – 7y. Now, using the pattern that you have observed, what do you think should be written on the blanks so that the next set of expressions is written in a similar form as those of the expressions written above? 5. If 8p – 12 = 30, then __________. 6. If 5c + 2d = 7f, then _____________. If your answers are 30 = 8p – 12, and 7f = 5c + 2d, then your pattern is correct. The above examples demonstrate the symmetric property of equality. Try to generate your own notion of symmetric property, then, verify your notion with the statement that follows: The symmetric property of equality means that when two quantities are equal, the equality will hold true, no matter in what side of the equation is each of them is written.In symbol, we write: If a = b, then b = a; a, b ∈ R. 3. Transitive Property of Equality 9

Observe the third set of expressions, and try to establish a pattern. 1. If 2 + 3 = 5 and 5 = 1 + 4, then 2 + 3 = 1 + 4. 2. If 4(8) = 32 and 32 = 2(16), then 4(8) = 2(16). 3. If 4m + 7 = 9n, and 9n = 45, then 4m + 7 = 45. 4. If 2w – 7y = 5z and 5z = 9y + 3, then 2w – 7y = 9y + 3. Now, using the pattern that you have observed, what do you think should be writtenon the blanks so that the next set of expressions is written in a similar form as those ofthe expressions written above? 5. If 8p – 12 = 7q and 7q = 15 p, then _____________. 6. If 5c + 2d = 7f and 7f = 10 c -25, then _____________. If your answers are 8p – 12 = 15p, and 5c + 2d = 10c - 25, then your answers arecorrect. The above examples demonstrate the transitive property of equality. To further illustrate, supposed that the price of 5 t-shirts is the same as the price oftwo pairs of pants; and the price of these two pairs of pants is the same as the price of7 pairs of shorts. Then, how do you compare the prices of the 5 t-shirts and the 7 pairsof shorts? ________________ If you were able to conclude that the two prices are equal, then you are correct! Can you generate now your own notion of transitive property? Try it, then, verify yournotion with the statement that follows: The transitive property of equality means that when the first two quantities are equal to the same quantity, then the first two given quantities are equal. In symbols, we write: If a = b and b = c, then a = c where a,b,c ∈ R. 4. Addition Property of Equality (APE) Observed the following figures below.Figure 1 Figure 2 Figure 3 10

In Figure 1, the scale is in balance and each side holds 50 g each. In the second figure, 20 g is added to only one side of the scales. What happened to the scale now? ____________ Yes, you are correct if you observed that the scale is no longer in balance. That will always happen, if the two sides do not hold equal mass. In the third figure, 20 g were added to both sides. What happened to the scale? ________________________ Yes, you are correct if you observed that the scale maintains its balance. That is, adding equal amounts to both sides of an equation maintains the equality of both sides.This time, observe the next set of expressions, and try to establish a pattern. 1. If 2 + 3 = 5, then (2 + 3) + 7 = 5 + 7. 2. If 4(8) = 32, then 4(8) + (-12) = 32 + (-12). 3. If 4m + 7 = 9n, then (4m + 7) + 3 = 9n + 3. 4. If 2w – 7y = 5z, then (2w – 7y) + (-21) = 5z + (-21). Now, using the pattern that you have formulated, what do you think should be writtenon the blanks so that the next set of expressions is written in a similar form as those ofthe expressions written above? 5. If 8p – 12 = 7q, then (8p –12) + 12 = _____________. 6. If 5c + 2d = 7f, then (5c + 2d) + (-19) = _____________. If your answers are 7q + 12, and 7f + (-19), then your answers are correct. The aboveexamples demonstrate the addition property of equality. Can you generate now yourown notion of addition property of equality? Try it, then, verify your notion with thestatement that follows: The addition property of equality means that when two quantities are equal and the same quantity is added to each of the two quantities, then, the sums are equal. In symbol, we write: If a = b then a + c = b + c, where a,b,c ∈ R. 5. Multiplication Property of Equality (MPE) Observe the figures below. 11

Figure 4 Figure 5 In Figure 4, the scale is in balance. Multiplying each weight by the same numberdoes not tip the balance to one side as shown in Figure 5.This principle illustrates the multiplication property of equality.Furthermore, observe the fifth set of expressions, and try to establish a pattern.1. If 2 + 3 = 5, then (2 + 3) • 7 = 5 • 7.2. If 4(8) = 32, then 4(8) • (-12) = 32 • (-12).3. If 4m + 7 = 9n, then (4m + 7) • 3 = 9n • 3.4. If 2w – 7y = 5z, then (2w – 7y) • (-21) = 5z • (-21). Now, using the pattern that you have formulated, what do you think should be writtenon the blanks so that the next set of expressions is written in a similar form as those ofthe expressions written above?5. If 8p – 12 = 7q, then (8p –12) • 12 = _________________.6. If 5c + 2d = 7f, then (5c + 2d) • (-19) = ________________. If your answers are 7q • 12, and 7f • (-19), then your answers are correct. The aboveexamples demonstrate the multiplication property of equality. Can you generate nowyour own notion of multiplication property of equality? Try it, then, verify your notionwith the statement that follows:The multiplication property of equality means that when two quantities areequal and the same quantity is multiplied to each of the two quantities, then,the products are equal.In symbol, we write:If a = b then a • c = b • c where a,b,c ∈ R.Remember the following: In each statement a, b, c are real numbers.A. Reflexive Property of equality: a = a. 12

B. Symmetric Property of Equality: If a = b, then b = a.C. Transitive Property of Equality: If a = b and b = c, then a = c.D. Addition Property of Equality: If a = b, then a + c = b + c.E. Multiplication Property of Equality: If a = b, then ac = bc.Self-check 2Identify the property illustrated in each of the following: _______________ _______________1. If 6 + 2 = 8 and 8 = 7 + 1, then 6 + 2 = 7 + 1 _______________2. 16 – 5 = 16 – 5 _______________3. If 2a + 3 = a + 5, then a + 5 = 2a + 3 _______________4. If 3(5) = 15, then 3(5)(1/5) = 15(1/5) _______________5. If 3x – 5 = 4, then 3x – 5 + 5 = 4 + 5 _______________6. 13m – 5n = 13m – 5n _______________7. If 5 = 2 + 3, then 2 + 3 = 5 _______________8. If 10 –2 = 4x, then 4x = 10 –2 _______________9. 18(0) = 010. (25 + 8) + 0 = (25 + 8) Answer Key on page 29Lesson 3 Solving Linear Equations in One Variable In Lesson 1, we reviewed the properties of real numbers. In Lesson 2, we discussedthe properties of equality. In this lesson, we will use these properties of real numbers andequality in solving first-degree equations. Suppose we are asked to solve for the value of x in the equation: 4x –2 = x + 7. We will use the following algebraic tiles to solve this equation. = x = -x = 1 = -1 13

Step 1: Represent the equation using the algebraic tiles. 4x – 2 = x+7Step 2: Add one black rectangular tile and two white square tiles to both sides. (Recall:What would happen to a scale in balance if the same amount were added to both sides?)Step 3: Simplify. A white and a black rectangle will cancel out. Similarly, a white and a blacksquare will cancel out.Step 4: Divide the number of squares into three groups.How many white squares correspond to each rectangle? ___________________There are three squares that correspond to each rectangle. Hence, x = 3.To verify, replace x with 3 and check if the equation holds true.Thus, 4x - 2 = x + 7 4 (3) - 2 = 3 + 7 It’s correct! 10 = 10 14

To summarize, we applied addition property of equality by adding (-x) and 2 to bothsides. By closure property, the equation became 3x = 9. Then, we applied multiplicationproperty of equality by multiplying 1/3 to both sides. By closure property, the equation endedup with x = 3.Ex. 2. This time, we try to solve the equation without the algebraic tiles. Suppose we are asked to solve for the value of x in the equation: x + 15 = 37.Our goal here is apply series of operations so that only the variable x will be left on one sideof the equation. What do you think should be done so that the left side of the equation willonly have the variable x? _________________________________________ Verify your solution with the steps that followed. If you were not able to generate yourown solution, try to follow the discussion below. Steps are being given on the left column.Try to execute these steps on the space provided on the right column.Step 1: Add (-15) to both sides of the equation. ____________________ Why can we do that? ____________You are correct if your answer is[(x + 15)] + (-15) = 37 + (-15).We can do this, because of APE.Step 2: Regroup the addends on the left side of the equation. ____________________ Why can we do that? __________________You are correct if your answer isx + [15 + (-15)] = 37 + (-15).We can do this, because of AssociativeProperty of Addition.Step 3: Perform the indicated operation on both sides. _____________________ What property of real numbers is used in each operation? __________________You are correct if your answer is x + 0 = 22.The left side used the Inverse Property of Additionwhile the right side used Closure Property forAddition.Step 4: Add x + 0. What property of real number ____________________ Is used? _______________You are correct if your answer is x = 22.That is because of identity property of addition.Hence, the solution of the equation is 22. The solution set is {22} 15

To verify, replace x with 22 and check if the equation holds true.Thus, x + 15 = 37 22 + 15 = 37 It’s correct! 37 = 37 In the next example, the steps to solve for the value of x in the equation 10 = -25 + x,are being executed on the left column. Provide the reason for each step on the spaceprovided on the right column.Ex. 3. Solve for x in the equation 10 = -25 + xSolution: 10 = -25 + x Given 1) ____________________ -25 + x = 10 2) ____________________ 3) ____________________ x + (-25) = 10 4) ____________________ 5) ____________________ [x + (-25)] + 25 = 10 + 25 6) ____________________ 7) ____________________ x + [(-25) + 25] = 10 + 25 x+0 = 10 + 25 x = 10 + 25 x = 35 You are correct if what you have listed as the reasons are the same with what islisted here: (1) Symmetric Property of Equality, (2) Commutative Property of Addition, (3)Addition Property of Equality, (4) Associative Property of Addition, (5) Inverse Property ofAddition, (6) Identity Property of Addition, and (7) Closure Property of Addition.To check, we substitute 35 for the value of x, and verify if the equality holds true:10 = -25 + x It’s correct!10 = -25 + 3510 = 10Ex. 4. Solve for the value of x in 2x – 6 = 12. Solution: By addition property of equality, add 6 to both sides of the equation . So 2x – 6 = 12 becomes ______________. Then, apply inverse property of equality so that the equation becomes ______. Next, by applying the identity property of addition, the equation becomes __________. By multiplication property of equality, multiply both sides of the equation by ½ (the reciprocal of 2). The equation now becomes ________. Then, apply the inverse property of multiplication so that the equation becomes ___________. Lastly, apply the identity property of multiplication to have the equation _____________. 16