MATH 1

Below is the summary of the processes involved in solving example # 3. You mayverify your answers with the following: 2x – 6 = 12 2x – 6 + 6 = 12 + 6 Addition Property of Equality 2x + 0 = 18 Inverse Property for Addition 2x = 18 Identity Property Addition ½(2x) = ½(18) Multiplication Property of Equality (1)x = 18/2 Inverse Property for Multiplication x= 9 Identity Property for Multiplication To check: It’s correct! 2(9) – 6 = 12 18 – 6 = 12 12 = 12 For example # 5, the reasons are being given to you. You need to figure out theresulting equation for every indicated reason.Example 5: Solve for x in 12 – 4x = 21 – 7xSteps to be taken Resulting equation ReasonAdd 7x to both sides (12 – 4x) + 7x = (21 – 7x) + 7x APE ________Regroup the addends 12 + (-4x + 7x) = __________ ________on both sides ________Add –7x and 7x on the right _______________________ ________side of the equation ________ ________Perform addition on the right ________side of the equation _______________________ ________Add –4x + 7x _______________________ ________Interchange the addends on ________ ________the left side of the equation _______________________ ________Add (-12) to both sides _______________________Regroup the addends on the leftside of the equation _______________________Add 12 and (-12) _______________________Perform addition on the left sideof the equation _______________________Perform addition on the right sideof the equation _______________________Multiply both sides by 1/3 _______________________Perform multiplication on the leftside of the equation _______________________ 17

Perform multiplication on the rightside of the equation _______________________ ________You may verify your answers below:Steps to be taken Resulting equation ReasonAdd 7x to both sides (12 – 4x) + 7x = (21 – 7x) + 7x APERegroup the addends 12 + (-4x + 7x) = 21 + (-7x + 7x) Associative P for + on both sides 12 + (-4x + 7x) = 21+ 0 Inverse P for +Add –7x and 7x on the right 12 + (-4x + 7x) = 21 Identity P for + side of the equationPerform addition on the right 12 + 3x = 21 Closure P for + side of the equation 3x + 12 = 21 Commutative P for +Add –4x + 7x (3x + 12) + (-12) = 21 + (-12) APEInterchange the addends on 3x + [12 + (-12)] = 21 + (-12) Associative P for + the left side of the equation 3x + 0 = 21 + (-12) Inverse P for +Add (-12) to both sidesRegroup the addends on the 3x = 21 + (-12) Identity P for + left side of the equation 3x = 9 Closure P for +Add 12 and (-12) (3x) (1/3) = (9) (1/3) MPEPerform addition on the left x = (9) (1/3) Inverse P of x side of the equationPerform addition on the right x=3 Closure P for x side of the equationMultiply both sides by 1/3Perform multiplication on the left side of the equationPerform multiplication on the right side of the equationTo check: 12 – 4x = 21 – 7x It’s correct! 12 – 4(3) = 4 – 2(2) 12 – 12 = 21 – 21 0=0On your own, try to solve for x in 2(3x – 6) = 4 – 2x.______________________________ ______________________________________________________ ______________________________________________________ ________________________If your solution set is {2}, then your answer is correct! 18

Self-check 3A. The solution 3(3 – 2) = 5(x + 12) is given. Just supply the missing part.Solution: 3(x – 2) = 5(x + 12) 1. _______________ 3x – 6 = 5x + 60 Apply commutative property on both sides of the equation. Then add -5x to both sides (APE). 2. _______________ Combine 3x – 5x. 3. _______________ -2x – 6 = 60 Inverse Property for Addition -2x – 6 + 6 = 60 + 6 4. _________and _______ Multiply both sides by -1/2 (MPE). -2x + 0 = 66 Inverse Property for Multiplication -2x = 66 Identity Property for Multiplication 5. _______________ (1)x = -33 x = -33B. Solve the following equations. Write the letter corresponding to the equation on the box(es) containing its solution to reveal the message. Message in the BoxesH: x + 5 = -3 N: 2y – 4 = y - 4L: x – 20 = -11 T: 4 + 3y = 16G: x – 18 = -5 R: 6z – 5 = 2z + 3M: x + 5 = 19 I : 3(3z – 2) = 4z + 9E: y + 54 = 81 A: 5(z – 2) = 4(2z + 5)S: y + 75 = 28 Y: 9 + 5z = 3(z – 5)9 27 -10 2 0 3 0 13 14 -10 4 -83 -47 !27 -10 -47 -12 Answer Key on page 29 19

Lesson 4 Different Properties of Inequality and the Solutions of Inequalities in One Variable This lesson focuses on solving first-degree inequalities in one variable. It would behelpful for you to recall the different properties of real numbers and of equality as wediscuss the different properties of inequality. Properties of Inequalities A. Transitive Property of Inequality If Jose is younger than Celia, and Celia is younger than Minda, how will you comparethe age of Jose and Minda? _____________________Jose Celia Minda Moreover, if Minda is taller than Celia, and Celia is taller than Jose, how will youcompare the height of Minda and Jose? _________________________. You are correct if your answers are “Jose is younger than Minda” and “Minda is tallerthan Jose” respectively. The above examples illustrate the transitive property of inequality. In real numbers,we may have the following examples that would illustrate such property. 1) If 3n > 5p, and 5p > 9r, then 3n > 9r. 2) If (6b - 5c) > (9b + 1), and (9b + 1) > (7c – 12), then (6b – 5c) > (7c –12). 3) If 5y < 7w, and 7w < 2m, then 5y < 2m. 4) If (21g – 13) < (2k + 5), and (2k + 5) < (4f + 17), then (21g – 13) < (4f + 17). Following the pattern above, fill in the blanks to complete the statement. 5) If (12p – 9) > 25m, and 25m > (17n + 13), then __________________. 6) If (17w + 3) < (2y – 1), and (2y – 1) < (3d – 10), then _________________. 20

You are correct if your answers are (12p – 9) > (17n + 13) and (17w + 3) < (3d –10) respectively. In your own words, describe the transitive property of inequality. _____________________________________________________. You may verify your answer with the statements below. Given a, b, c ∈ R. If a > b and b > c, then a > c If a < b and b < c, then a < c B. Addition Property of Inequality (API) The discussion here is very similar to the discussion of Addition Property of Equality.Recall that, for any real number a, b, c, when a = b, then a + c = b + c. This time, we deal with inequality. Suppose that Rodora has P7000 in a bank while Mavic has P5000. Who has agreater amount of deposit? _______________ Then, both of them deposit P2000 each. How much would be the deposit of eachnow? Who has a greater amount of deposit? ______________________ Then, suppose after two months, both of them withdrew P1000 each. How muchwould be the deposit of each now? Who has a greater amount of deposit?__________________ You are correct if you conclude that after both had deposited P2000 each, Rodora’sdeposit is greater than Mavic’s. Similarly, after both had withdrawn P1000 each, Rodora’sdeposit is greater than Mavic’s. The above examples illustrate the addition property of inequality. The statementsbelow illustrate this property of inequality as well. 1) If (6b - 5c) > (9b + 1), then (6b – 5c) -5 > (9b + 1) - 5. 2) If 5y < 7w, then 5y + 3m < 7w + 3m. 3) If (21g – 13) < (2k + 5), then (21g – 13) – 8f < (2k + 5) – 8f. Following the pattern above, fill in the blanks to complete the statement. 4) If (12p – 9) > 25m, then (12p –9) + 5y __ _______________. 21

5) If (17w + 3) < (2y – 1), then (17w + 3) – 10n ___ ______________. You are correct if your answers are > 25 m + 5y and < (2y - 1) – 10n respectively. In your own words, can you now describe the addition property of inequality? ___________________________________________________________. You may verify your answer with the statement below. Given a, b, c ∈ R. If a > b and a + c > b + c If a < b and a + c < b + c C. Multiplication Property of Inequality Suppose that we have the inequality 8 > 5.a) Multiply both sides by any positive number, say 4. 8 (4) > 5 (4) 32 > 20 Trueb) Multiply both sides by any negative number, say (-1). 8 (-1) > 5 (-1) (-8) > (-5) False. What did you observe? ______________________________________ You are correct if you were able to observe that the direction of the inequality doesnot change when both sides are multiplied to a positive number. However, the direction ofthe inequality changes when both sides are multiplied to a negative number. In notations,we have: If a > b, and c > 0, then ac > bc. Also, if a < b, and c > 0, then ac < bc. If a > b, and c < 0, then ac < bc. Also, if a < b, and c < 0, then ac > bc. Now, could you figure out if both sides of the inequality were multiplied to zero?_________________________________. 22

Yes, you are correct if your answer is: both products will be zero, and hence areequal to each other. At this point, you are now ready to solve first-degree inequalities in one variable. Thesteps involved are very similar to the steps we consider in solving equalities. However,instead of using the properties of equality, we now use the properties of inequality.Example 1: Solve x + 5 > 12 Add (-5) to both sides of the inequality (API) (x + 5) + (-5) > 12 + (-5) Associative Property, and Closure Property x + [5 + (-5)] > 7 Inverse Property for Addition x+0>7 Identity Property for Addition x>7 Therefore, the solutions are all real numbers greater than 7. To check, youtake several values such as 8, 9, and 10 and substitute these in the originalinequality. To illustrate:If x = 8 → 8 + 5 > 12; If x = 9 → 9 + 5 > 12; If x = 10 → 10 + 5 > 1213 > 12.Correct! 14 > 12.Correct! 15 > 12.Correct!You illustrate the solutions on the number line, thus,1 23 4 5 6 7 8 9 10 The hollow dot or unshaded circle indicates that 7 is not included in thesolution set.Example 2: Solve 4x – 3 ≤ 9Solution: 4x – 3 ≤ 9(4x – 3) + 3 ≤ 9 + 3 Add 3 to both sides of the inequality (API)4x + [(-3) + 3] ≤ 12 Associative and Closure Property 4x + 0 ≤ 12 Inverse Property for Addition 4x ≤ 12 Identify Property for Addition ¼(4x) ≤ ¼(12) Multiply both sides of the inequality by 1(x) ≤ 3 ¼ (MPI) x≤3 Associative and Inverse Property for x Identity Property for MultiplicationThe solution set consist of all numbers less than or equal to 3. 23

The representation of the solution set on the number line is shown below:-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 The solid dot or shaded circle indicates that 3 is included in the solution set.Example 3: Solve 3x – 6 > 5x + 4Solution: 3x – 6 > 5x + 4 (-6) + 3x > 4 + 5x[(-6) + 3x) + (-5x) > [4 + 5x] + (-5x)(-6) + [ 3x + (-5x)] > 4 + [ 5x + (-5x)] (-6) + (-2x) > 4 + 0 (-2x) + (-6) > 4[(-2x) + (-6)] + 6 > 4 + 6-2x + [ (-6) + 6] > 10 -2x + 0 > 10 -2x > 10 (-2x) (-1/2) < (10) (-1/2) Why? X < -5To check, you take x = 6 Correct! 3(-6) – 6 > 5(-6) + 4 18 – 6 > - 30 + 4 -24 > -26The graph of the solution set is shown below.-9 -8 -7 -6 - 5 -4 -3 -2 -1 0 24

Example: Solve 3(2x + 4) ≤ 2(15 – 6x) Solution: 3(2x + 4) ≤ 2(15 – 6x) 6x + 12 ≤ 30 – 12x 6x + 12x + 12 ≤ 30 – 12x + 12x 18x + 12 ≤ 30 18x + 12 + (-12) ≤ 30 + (-12) 18x ≤ 18 1/8(18x) ≤ 1/18(18) x≤1Check and represent the solution set on the number line.Self-check 3 Solve each of the following inequalities. Represent each solution set on thenumber line.1. x + 2 < 7 4. y – 13 ≥ 3 – 7y2. 10 + x > 8 5. 8z + 13 > -3 + 10z3. 12 – y ≤ -4 + 3y Answer Key on page 29 Let’s summarizeLook back!  Properties of Real Numbers A. Closure Property If a, b, c ∈ R, then a + b ∈R. If a, b ∈ R, then ab ∈ R. 25

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) = a E. Inverse Property a + (-a) = 0 and (-a) + a = 0 a · 1/a = 1 and 1/a · a = 1 F. Distributive Property a(b + c) = ab + ac (b + c)a = ba + ca G. Properties of Multiplication 0(a) = 0 and a(0) = 0 Properties of Equality A. Reflexive Property of equality a = a, a ∈ R 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 26

E. Multiplication Property of Equality If a = b, then ac = bc Properties of Inequalities A. Transitive Property of Inequality If a > b and b > c, then a > c for a, b, c, ∈ R. If a < b and b < c, then a < c for a, b, c, ∈ R. B. Addition Property of Inequality If a > b, then a + c > b + c for a, b, c, ∈ R. If a < b, then a + c < b + c for a, b, c, ∈ R. C. Multiplication Property of Inequality If a > b and c > 0, then ac > bc. If a < b and c > 0, then ac < bc. If a > b and c < 0, then ac < bc. If a < b and c < 0, then ac > bc. To solve first degree equations and inequalities in one variable algebraically is to apply the properties of real number and the properties of equality and inequality.What to do after (Posttest)A. Matching type: For #1-5: Match the number sentence in Column A to the property of real numbers it demonstrates found in Column B. Column A Column B1) 5 • (2m + 7n) = (5 • 2m) + (5 • 7n) a) Associative Property2) (12p + 19q) + 0 = 12p + 19q b) Commutative Property3) [(-8) + 8) + 15y = 0 + 15y c) Closure Property4) 9 • (11 • 23) = (11 • 23) • 9 d) Identity Property5) (8 • 12) • 1/12 = 8 • (12 • 1/12) e) Inverse Property f) Distributive Property 27

B. Matching type: For # 6-10: 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 3m + 2n = 14p, then g) Addition Property of Inequality14p = 3m + 2n. h) Multiplication Property of Equality7) If (m-n)< (p+q) and (p+q) < 0, i) Multiplication Property of Inequalitythen (m-n)r < 0. j) Reflexive Property8) If m > 7n, then m + p > 7n + p. k) Symmetric Property9) If q+ r =15, then q + r –13m = 15 –13m. l) Transitive Property10) If 15y < 75, then -3y > -15.C. Multiple Choice. Choose the letter of the correct answer.11. What is the value of x in 2 (x + 3) = 5x + 7?a. –1/3 c. 1/3b. 1/3 d. 312. What is the solution of 3(2x + 4) = 2(24 – 6x)?a. x = 0 c. x = 2b. x = 1 d. x = 313. Which of the following inequality has {y/y > 1} the solution set?a. -5 + 3y < 8 c. -5 + 3y < -8b. -5 – 3y < 8 d. -5 – 3y < -814. Which of the following graphs represents the solution set of 5z – 1 ≤ 20 + 2z? a. -1 0 1 2 3 4 5 6 7 8 b. -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. What is the solution set of 2 (3y +10) < 7 (2x –4)?a. y = 6 c. y = -6b. y > 6 d. y < 6 Answer Key on page 31 28

Answer KeyPretest page 291. Distributive Property 9. Symmetric Property2. Closure Property 10. Multiplication Property of Equality3. Commutative Property 11. a4. Distributive Property 12. d5. Inverse Property 13. c6. Transitive property 14. b7. Multiplication Property of Inequality 15. a8. Addition Property of EqualityLesson 1 Self-check 1 page 81. Closure property 6. Identity Property for Multiplication2. Commutative Property 7. Inverse Property for Addition3. Closure Property 8. Associative Property4. Distributive Property 9. Multiplication Property of Zero5. Commutative Property 10. Identity Property for AdditionLesson 2 Self-check 2 page 13 1. Transitive Property of Equality 2. Reflexive Property of Equality 3. Symmetric Property of Equality 4. Multiplication Property of Equality 5. Addition Property of Equality 6. Reflexive Property of Equality 7. Symmetric Property of Equality 8. Symmetric Property of Equality 9. Multiplication Property of Zero 10. Identity PropertyLesson 3 Self-check 3 page 25A. 1. Distributive Property 2. (-6) + 3x + (-5x) = 60 + 5x + (-5x) 3. Addition Property of Equality 4. Inverse Property and Identity Property for Addition 5. - 1/2(-2x) = - 1/2(66)B. H: x + 5 = -3 G: x – 18 = -5 E: y + 54 = 81 x = -3 – 5 x = 18 – 5 y = 81 – 54 29

x = -8 x = 13 y = 27L: x – 5 = -11 M: x + 5 = 19 S: y + 75 = 28 x = -11 + 20 x = 19 – 5 y = 28 – 75 x=9 x = 14 y = -47N: 2y – 4 = y – 4 I: 3(3z – 2) = 4z + 9 2y – 4 = 4 – 4 6z – 6 = 4z + 9 y=0 9z – 4z = 9 + 6 5z = 15T: 4 + 3y = 16 z=3 3y = 16 –4 3y = 12 A: 5(z – 2) = 4(2z + 5) y=4 5z – 10 = 8z + 20 5z – 8z = 20 + 10R: 6z – 5 = 2z + 3 6z – 2z = 3 + 5 3z = 30 4z = 8 z = 10 z=2 Y: 9 + 5z = 3(z – 5) 9 + 5z = 3z – 15 5z – 3z = -15 - 9 2z = -24 z = 12 Message in the BoxesL EARN I NG MA T H9 27 -10 2 0 3 0 13 14 -10 4 -8IS EASY3 -47 27 -10 -47 -12Exploration 4 -3 -2 -1 0 1 2 3 4 5 6 1. x + 2 < 7 x<7–2 x<52. 10 + x > 8 x > 8 – 10 30

x > -2 -5 -4 -3 -2 -1 0 1 2 3 4 53. 12 – y ≤ -4 + 3y 5678 -y – 3y ≤ -4 – 12 -4y ≤ -16 y ≥4 -2 -1 0 1 2 3 44. y – 13 ≥ 3 – 7y -1 0 12345678 y + 7y ≥ 3 + 13 8y ≥ 3 + 13 8y ≥ 16 y≥2 -25. 8z + 13 > -3 + 10z 8z – 10z > -3 – 13 -2z > -16 z<8 - 2 -1 0 1 2 3 4 5 6 7 8Posttest page 27 9. Addition Property of Equality 10. Multiplication Property of Equality1. Distributive Property 11. a2. Identity Property 12. c3. Inverse Property 13. d4. Commutative Property 14. c5. Associative Property 15. b6. Symmetric property7. Transitive Property8. Addition Property of Inequality END OF MODULE 31

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. 32

Module 12 The Real Thing of XYZ What this module is all about Teachers usually compute their students’ final grades by taking the average of thequarter grades. Suppose your grades in the first three quarters were 85, 90, and 87. Whatminimum grade should you get in the fourth quarter to stay in the honor’s group? Practical problems like the one above will be studied in this module. Here, we willlearn some methods and techniques on how to solve problems that involve first degreeequations and inequalities in one variable. What you are expected to learn At the end of this module, the student is expected to apply first degree equations andinequalities in one variable to solve various word problems in mathematics. We classifythese problems according to their type, and some methods and techniques on how to solveeach kind of problem are presented in the following lessons: Lesson 1 Solving Number Problems Lesson 2 Solving Geometry Problems Lesson 3 Solving Coin Problems Lesson 4 Solving Distance Problems Lesson 5 Solving Age Problems Lesson 6 Solving Problems Involving Inequalities in One Variable 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)Directions: Solve the problem and choose the letter that corresponds to the correct answer.1. One number is five less than another. If their sum is 101, what is the smaller number?a. 43 b. 45 c. 46 d. 482. The difference between two numbers is 19. The larger number is 23 less than three times the smaller number. What is the smaller number?a. 21 b. 22 c. 23 d. 243. The sum of three consecutive odd numbers is 39. What is the middle number?a. 11 b. 13 c. 15 d. 174. I am thinking of two numbers. The larger number is 18 more than the smaller number. Two times the smaller number is 5 less than the larger number. What is the larger number?a. 13 b. 15 c. 31 d. 335. Every afternoon, Roger deposits in a box whatever remains of his daily allowance. At the end of one month, he found out that the number of 1-peso coins was twice the number of the 25-centavo coins, and the number of 5-peso coins was two more than the number of the 25-centavo coins. He counted the coins and had a total of P82.50. How many 1- peso coins did Roger have?a. 20 b. 15 c. 10 d. 56. I only have 1-peso and 5-peso coins in my pocket. Their total value is P35.00. If the number of the 5-peso coins is 1 more than the number of the 1-peso coins, how many 5- peso coins do I have?a. 3 b. 4 c. 5 d. 67. The length of a rectangle is three less than twice its width. If the perimeter of the rectangle is 36 units, what is its length?a. 13 b. 12 c. 11 d. 10 3

8. The perimeter of a square is 100 cm. What is the length of its side?a. 20 b. 25 c. 35 d. 509. What is the largest counting number that satisfies this condition: the sum of 5 less than the number and 3 more than the number is less than 62?a. 30 b. 31 c. 32 d. 3310. Richard’s grades in Math during the first three quarters were 78, 87, and 83. What is the least grade Richard must get during the last quarter so that he can have an average grade of at least 85 for all quarters?a. 90 b. 91 c. 92 d. 93 Answer Key on page 21 What you will do Read the following lessons carefully in order to enhance your skill in solving wordproblems.Lesson 1 Solving Number Problems The following steps are recommended in solving word problems in mathematics:READ, KNOW, PLAN, SOLVE, and CHECK. There may be many ways to solve a problem,but for our purpose, we will adhere to these steps as much as we can. If we do this, solvingword problems becomes easier. Later, when you have mastered these steps, you maydevelop your own more effective way of solving a problem. 4

1. READ Do I know the • Read the problem carefully. Get a meaning of all the general idea of what is required. words?2. KNOW What am I • Determine what is asked and what are asked to the given facts in the problem. Think find? Which what previously learned formula or key facts do I concept may be helpful to solve the need to problem. consider? 3. PLAN How do I represent the • Represent the unknown/s using a4. SOLVE unknown, and what variable and translate phrases into operation, strategy, or symbols. technique should I use to solve the problem? What is the • Set up the equation and solve for equation and the unknown variable. what is the required answer?5. CHECK Is my answer • Check the value obtained by reasonable? Does it substituting it back into the original answer the question equation. Also, verify if the answer correctly? satisfies all conditions in the problem.We illustrate these steps of solving word problems as follows: 5

Lesson 1 Solving Number Problems The first kind of word problems that we will solve is number problems. In the twoexamples that follow, we illustrate in detail the steps in solving word problems as outlinedearlier. On your own, explore how these steps are undertaken.ExplorationExample 1. The sum of two numbers is 119. The second number is eight more than twice the first number. What are the numbers?To solve the problem, we follow these steps:a. Read. Analyze the problem carefully and get a general idea of what is required.b. Know. Determine what is asked and what are given in the problem. The problem asked for two numbers. The sum of the two numbers is 112. The second number is eight more than twice the first.c. Plan. Make representation of the unknown. Let x = the first number 2x + 8 = the second numberd. Solve. Set up the equation and solve for the unknown. x + 2x + 8 = 119Solution: x + 2x + 8 = 119 (APE) 3x + 8 – 8 = 119 – 8 3x = 111 (MPE) → the first number 1 (3x) = 1 (111) → the second number 33 x = 372x + 8 = 2(37) + 8 = 82d. Check. We only check whether the sum of the two numbers is 119. ? 37 + 82= 119 119 = 119 which is true! 6

Therefore, the required numbers are 37 and 82. The next problem requires the concept of even and odd numbers. In this case, weare actually dealing with positive integers or the counting numbers. These are numbers inthe set {1, 2, 3, 4, 5, …}. What are examples of even numbers? 2, 4, 6, 8, … are even numbers. All of them can be divided by 2. What are examples of odd numbers? 1, 3, 5, 7,… are odd numbers. Their remainder is 1 when divided by 2. Before we solve the next number problem, we first practice our skill of translatingwords into symbol. Did you know? 1. Consider the even number 20. What is the next even number after 20? It is 22. How do we get 22? By adding 2 to 20, that is, 20 + 2=22. What is the next even number after 22? It is 24. What number do you add to 20 to get 24? It is 4 because 20 + 4=24. 2. If the first even number is x, what will be the next even number after x? It is x + 2. What is the next even number after x + 2? It is (x + 2 )+ 2 or x + 4. We are now ready to solve the next problem.Example 2. If the sum of three consecutive even numbers is 48, find the numbers. We now solve the problem by following the enumerated steps given earlier. a. Read. Analyze the problem carefully and get a general idea of what is required. b. Know. Determine what is asked and what are given in the problem. Three consecutive even numbers whose sum is 48. 7

c. Plan. Make representation of the unknown. let x = the first even number x + 2 = the second even number x + 4 = the third even numbere. Solve. Set up the equation and solve for the unknown. x + x + 2 + x + 4 = 48 Solution: x + x + 2 + x + 4 = 48 3x + 6 = 48 3x + 6 – 6 = 48 – 6 3x = 42 1 3x = 1 (42) 3 3  x = 42 3 x = 14 the first even number x + 2 = 14 + 2 = 16 the second even number x + 4 = 14 + 4 = 18 the third even numberf. Check. We check whether the three numbers add up to 48. ? 14 + 16 + 18 = 48 48 = 48 which is correct! Therefore, the required numbers are 14, 16, and 18. Self-check 1 Solve the following word problems. 1. The sum of two numbers is 91. If one number is 9 less than the other number what are the numbers? 2. The larger of two numbers is 2 more than twice the smaller number. The difference between the two numbers is 8. Find the numbers. 3. The sum of 2 consecutive odd numbers is 104. What are the numbers? 8

4. The sum of three consecutive even numbers is 102. Find the middle number. 5. Twice the sum of two consecutive odd numbers is 30 more than twice the smaller number. Find the numbers. Answer Key on page 21Lesson 2 Solving Geometric Problems Exploration In Lesson 1, we solved problems involving counting numbers. This time, let us try tosolve some geometric problems. First, let us explore the properties of some geometricfigures. The perimeter of a geometric figure is the distance around it. The perimeter isdenoted by P. A square is a closed figure with four equal sides and four right angles. s s s P = 4s s A rectangle is a closed figure with four sides that form four right angles. Thisdefinition means that its opposite sides are also equal and parallel. The length l of arectangle is the measure of the longer side, while its width w is the measure of a shorterside. l P = 2l + 2w ww l 9

Now, we are ready to consider the following problem in Geometry.Example 3. The length of a rectangle is one less than three times its width. If the perimeter of the rectangle is 46 units, find the length and width of the rectangle.We solve the problem, by following the following steps.a. Read. Analyze the problem carefully and get a general idea of what is required.b. Know. Determine what is asked and what are given in the problem. The perimeter of the rectangle is 46. The length is one less than three times its width.c. Plan. Make a representation of the unknown. let w = width 3w– 1 = lengthd. Solve. Set up the equation and solve for the unknown. We recall that the formula for the perimeter of a rectangle is P = 2l + 2w where lis the length and w is the width. By substitution, we have the following equation: 2(3w – 1) + 2w =46Solution: 2(3w – 1) + 2w = 46 6w – 2 + 2w = 46 8w – 2 + 2 = 46 + 2 Distributive Property 8w = 48 APE 1 (8w) = 1 (48) MPE 88 w = 48 8 w=6 → the width 3w – 1 = 3(6) – 1 = 17 → the lengthe. Check. We verify whether the perimeter of the rectangle is 46. ? 46 = 2(6) + 2(17) ? 46 = 12 + 34 46 = 46 which is correct! 10

Self-check 2 Directions: Read each of the following problems carefully. Then do the tasks that follow.A. A rope 25 meters long is cut into 2 pieces such that the length of the longer piece is 13 metres more than twice the length of the shorter piece. Answer the following questions based on the given problem. 1. What is asked in the problem? __________________________________________________________________ 2. How long is the rope? __________________________________________________________________ 3. If x represents the length of the shorter piece, how do you represent the length of the longer piece? __________________________________________________________________ 4. Form the equation from the questions 2 and 3. __________________________________________________________________ 5. Solve the equation then check the solution. __________________________________________________________________ 6. What is now the length of the longer piece? __________________________________________________________________B. Solve the following problems. Check your solution. 1. The width of the playground is 45 metres less than its length. The perimeter is 594 meters. What are the dimensions of the playground? 2. A rope is 10 metres long. It is cut into 2 pieces such that the length of the longer piece is one more than twice the length of the shorter piece. Find the length of the shorter piece. Answer Key on page 21 11

Lesson 3 Solving Coin Problems Exploration In explorations 1 and 2, you were exposed to a very detailed process in solving wordproblems. Now, try to make your solution short. To do this, study the following illustrativeexamples. The problems that follow are coin problemsProblem: Every afternoon, Jowell keeps in a box whatever is left from his daily allowance. At the end of one week, he found out that he had twice as many as 10 centavos and the twenty-five centavo coins were two more than the five centavo coins. He counted a total of P4.50. How many of each kind did he have?To solve the problem, you have to determine: Unknowns: a. number of 5-centavo coins b. number of 10-centavo coins c. number of 25-centavo coins Given: The value of the money is P4.50 Representation: Let x = the number of 5-centavo coins 2x = the number of 10-centavo coins x + 2 = the number of 25-centavo coins 0.05(x) = the value of the 5-centavo coins 0.1(2x) = the value of the 10-centavo coins 0.25(x + 2) = the value of the 25-centavo coins Equation: 0.05x + 0.10(2x) + 0.25(x + 2) = 4.50 0.25x + 0.25x + 0.50 = 4.50 0.50x + 0.50 – 0.50 = 4.50 – 0.50 0.50x = 4.00 12

1/0.50(0.50x) = 1/0.50(4.00) x = 4/0.5 x=8 5-centavo coin 2x = 2(8) = 16 10-centavo coin x + 2 = 8 + 2 = 10 25-centavo coinCheck: It’s correct! 8 x 0.05 = 0.40 16 x 0.10 = 1.60 10 x 0.25 = 2.50 P 4.50Self-check 3Solve the following problems. Check your solutions.1. Mark has 22 coins. It is composed of 1-peso coins and 5-peso coins. If the number of 5-centavo coins is four more than the number of 1-peso coins, how much money does he have?2. Jimmy’s money is P24.00. It is composed of 25-centavo coins and 1-peso coins. If the number of 25-centavo coins is four times the number of 1-peso coins, how many 1-peso coins does he have? Answer Key on page 22Lesson 4 Solving Distance Problems ExplorationConsider the following problem. This is a problem on uniform motion. 13

Problem: Two cars leave Manila at the same time and travel in opposite directions. After 3 hours, they are 396 km apart. If the average speed of one car is 12 km/hr less than the other. What is the average speed of each? You will note that the problem mentioned above is about distance. Do you remember the distance formula?Did you know?D = rt where d = the distance r = the rate/average speed t = the timeTo solve the problem in this lesson, you determine the following;Unknown: average speed of each carKnown/ Given: 396 km – the total distance 3 hours – timeRepresentation: for the speed Let x = the average speed of the first car x – 12 = the average speed of the second carfor the distance 3x = the distance covered by the first car after 3 hours3(x – 12) = the distance covered by the second car after 3 hoursEquation:distance traveled by the first car + distance traveled by the second car = total distance3x + 3(x – 12) = 3963x + 3x – 36 = 3966x – 36 + 36 = 396 + 36 6x = 432 1/6(6x) = 1/6(432) x = 432/6 x = 72 km/h → average speed of the first car x – 12 = 72 – 12 = 60 km/h → average speed of the second car ?Check: 3(72) + 3(60 ) = 396 14

? 216 + 180 = 396 396 = 396 It’s correct!Self-check 4Solve the following problems.1. Two trucks traveled the same distance. The first truck traveled at 60 km/h and the second truck at 50 km/h. It took truck B one hour longer to make the trip. How long did it take each truck to make the trip?2. Two boys start running at the same time and at the same place. They run in opposite directions. The average speed of the first boy is 2 m/s less than twice the average speed of the second boy. After 15 seconds, the two boys are 150 metres apart. Find the average speed of each boy. Answer Key on page 22Lesson 5 Solving Age Problems Exploration Take a look at this problem and solve. This is a problem on age.Problem: Six years ago, Mrs. Dela Cruz was five times as old as her daughter. How old is Sheila now if her age is one-third of her mother’s present age?Unknown: the ages of Sheila and her mother now 15

Known: the mother’s age was 5 times of Sheila’s age six years ago Sheila’s age now is one-third the age of her mother’s present ageRepresentation: Let x = the age of Mrs. Dela Cruz now 1/3x = the age of Sheila nowWhat is the age of each six years ago? x – 6 = the age of Mrs. Dela Cruz six years ago1/3x – 6 = the age of Sheila six years agoEquations:How do you relate the age of Sheila and her mother six years ago?[Mrs. Dela Cruz age was 5 times Sheila’s age six years ago] x – 6 = 5(1/3 x – 6) x – 6 = 5/3 x – 30x – 5/3 x – 6 = 5/3 x – 5/3 x – 30-2/3 x – 6 = -30-2/3 x – 6 + 30 = -30 + 6 -2/3 x = -24-3/2(-2/3 x) = - 3/2(-24) x = 72/2 x = 36 → Mrs. Dela Cruz’s age 1/3 x = 1/3(36) = 36/3 = 12 → Sheila’s ageCheck: x – 6 = 5(1/3 x – 6) ? 36 – 6 = 5[1/3(36) – 6] ? 30 = 5(12 – 6) ? 30 = 5(6) 30 = 30 Self-check 51. Roger is 8 years older than Jose. If five years ago the age of Roger was 1 year more than twice the age of Jose, how old is he?2. The sum of Japi’s age and Jimmy’s age is 40. Japi is 4 year older than Jimmy. Hoe old is Japi? Answer Key on page 23 16

Lesson 6 Solving Problem Involving Inequalities in One VariableExploration 6 In explorations 1 to 5, you were introduced to five types of word problems where youmade use of your skill in solving first degree equations in one variable applying theproperties of real numbers and equality. Let us now solve problems involving first degreeinequalities in one variable.Consider the following problems:1. Find the largest counting number such that the sum of twice the number and thrice the number is less than 65.Representation:Let x = the counting number 2x = twice the counting number 3x = thrice the counting numberInequality: 2x + 3x < 65 5x < 65 1/5(5x) < 1/5(65) x < 13Check: Try x = 12 2x + 3x < 65 It’s true!2(12) + 3(12) < 65 24 + 36 < 65 60 < 65 Note that x could be any counting number less than 13. Among these countingnumbers, 12 is the largest. Therefore, the answer is 12.2. Janno’s score on the first three of four 50-item tests were 44, 46, and 41. Determine what should be the largest score of Janno on the fourth test so he could have an average score of at least 43 for all the tests. 17

The word “at least” means in Mathematics “not less than” which implies “more than or equal to” while “at most” means “not more than” which also implies “less than or equal to”.Representation: Let x = the score in the fourth testInequality: 44 + 46 + 41 + x / 43 4 131 + x / 43 4 4(131 + x) / 4(43) 4 131 + x / 172 131 + (-131) + x / 172 + (-131) x / 41Check: 44 + 46 + 41 + 41 / 43 4 172 / 43 4 43 / 43 Note that x could be greater than or equal to 41 but not exceeding 50 since it is a 50-item test. Therefore, the score is 50. 18

Self-check 6 Solve the following problems. 1. The three Garcia girls were born in consecutive years. The sum of their ages is more than 39 decreased by the age of the youngest. What are the least possible ages of the girls? 2. The average of two numbers is greater than 35. One number is 4 more than twice the other. Find the smallest number. Answer Key on page 24 Let’s summarize First degree equations and inequalities is very useful in solving problems on relations among numbers, geometry, uniform motion, money problems, age problems and others. To solve problems, you will undertake the following steps: 1. READ – Read and understand carefully the problem. 2. KNOW – Determine what is asked 3. PLAN – Make a representation. 4. SOLVE – Form an equation and solve. 5. CHECK – Substitute the answer to the original equation to check whether or not you arrived at the correct answer. 19

What to do after (Posttest)Multiple Choice. Choose the letter of the correct answer.1. The sum of 2 numbers is 98. If one number is eight less than the other, find the largernumber.a. 50 b. 51 c. 52 d. 532. The sum of three consecutive odd numbers is 99. Find the largest number.a. 33 b. 35 c. 37 d. 393. A car traveled 711 km in 3 hrs. What was his average speed?a. 217 b. 227 c. 237 d. 2474. The sum of the ages of Jose and his father is 43. if his father’s age is one more than five times the age of Jose, how old is his father?a. 36 b. 37 c. 38 d. 395. What is the smallest number such that the sum of thrice of it and five times of it is greater than 80?a. 11 b. 12 c. 13 d. 14 Answer Key on page 24 20

Answer KeyPretest page 3 1. d 4. b 2. a 5. c 3. cLesson 1 Self-Check 1 page 81. Representation: Let x = the first number x – 9 = the second number Equation: x + x – 9 = 91 Solve: x + x – 9 = 91 2x = 92 + 9 2x = 100 x = 50 the first number x – 9 = 41 the second number Check: 50 + 41 ≟ 91  91 = 91 It’s correct!2. Representation: Let x = the first odd number x + 2 = the second numberEquation: x + x + 2 = 104 2x = 104 – 2 2x = 102 x = 51 the first odd number x + 2 = 53 the second numberCheck: 51 + 53 ≟104  104 = 104Lesson 2 Self-Check 2 page 11A. 1. the length of the shorter piece 2. 25 metres 3. x + 13 4. x + x + 13 = 25 5. x + x + 13 = 25 2x + 13 = 25 2x = 25 – 13 2x = 12 x = 6 metres length of the shorter piece x + 13 = 19 metres length of the longer piece 21

Check: 6 + 19 = 25 25 = 25 It’s correct! 6. 6 metresLesson 3 Self-Check 3 page 131. Representation: Let x = the number of 1-peso coinx + 4 = the number of 5-peso coin1(x) = the value of 1-peso coins5(x + 4) = the value of 5-peso coinsEquation: x + x + 4 = 222x = 22 – 42x = 18x = 9 the number of 1-peso coinsx + 4 = 13 number of 5-peso coins9(1) = P 9.0013(5) = 65.00 P 74.00 the amount of money he hasCheck: 9(1) + 13(5) ≟ 74 9 + 65 = 74 74 = 74 It’s correct!9 + 13 ≟ 2222 = 22 It’s correct! 2. Representation: Let x = the number of 1-peso coins 4x = the number of 25-centavo coins 1(x) = the value of the 1-peso coins 0.25(4x) = the value of the 25-centavo coins Equation: 1(x) + 0.25(4x) = P24.00 1x + 1 = 24 2x = 24 x = 12 the number of 1-peso coins 4x = 48 the number of 25-centavo coins Check: 12(1) + 0.25(48) ≟ P 24  P12 + P12 = 24  24 = 24Lesson 4 Self-Check 4 page 151. Representation: Let x = the number of hours for A to make the trip x + 1 = the number of hours for B to make the same trip 60(x) = distance covered by truck A 22

50(x + 1) = distance covered by truck B Equation: 60x = 50(x + 1), since they traveled the same distance 60x = 50x + 50 -60x – 50x = 50 10x = 50 x = 5 hours for truck A x + 1 = 6 hours for truck B Check: 60(5) ≟ 50(5 + 1)  300 = 300 2. Let x = the average speed of the second boy 2x – 2 = the average speed of the first boy 15(x) = the distance covered by the second boy 15(2x – 2) = the distance covered by the first boy 15x + 15(2x – 2) = 150 15x + 30x – 30 = 150 45x = 150 + 30 45x = 180 x = 4 m/s — second boy’s average speed 2x – 2 = 2(4) – 2 = 6 m/s — first boy’s average speedLesson 5 Self-Check 5 page 16 1. Let x = Jose’s age x + 8 = Roger’s age x – 5 = Jose’s age 5 years ago x + 8 – 5 = Roger’s age 5 years ago x + 8 – 5 = 2(x - 5) + 1 x + 3 = 2x – 10 + 1 2x – 10 + 1 = x + 3 2x – x = 9 + 3 x = 12 yrs — the age of Jose x + 8 = 20 yrs — the age of Roger 2. Let x = the age of Jimmy x + 4 = the age of Jopi x + x + 4 = 40 2x = 40 – 4 2x = 36 x = 18 — age of Jimmy x + 4 = 22 — age of Jopi 23

Lesson 6 Self-Check 6 page 191. Let x = age of the first girl x + 1 = the age of the second girl x + 2 = the age of the third girl x + x + 1 + x + 2 > 39 – x 3x + 3 > 39 – x 3x + x > 39 – 3 4x > 36 x>9The least possible ages of the girls are 9, 10, and 11.2. Let x = the first number 2x + 4 = the second number x + 2x + 4 > 35 2 3x + 4 > 70 3x > 70 – 4 3x > 66 x > 22The smallest possible number is 23.Posttest page 201. d 4. a2. b 5. a3. c END OF MODULE BIBLIOGRAPHYFuller, G. (1977). College algebra. (4th ed). New York, NY: Van Nostrand Company.Leithold, L. (1989). College algebra. USA: Addison-Wesley Publishing Company, Inc. 24

Mckeague, C.P. (1990). Intermediate slgebra: A text/workbook. (3rd ed). USA: Harcourt Brace Jovanovich, Inc. 25

Module 13 Point to PointWhat this module is all about Would you like to find the approximate distance between your house and yourschool? To do this, you need to draw the map of your barangay, town, or city on a piece ofpaper that has been divided into small, same-sized squares. By taking your house and yourschool as points on this paper, this module can help you compute the distance between thetwo. In this module, we will study the parts of the number line and the Cartesiancoordinate plane, locate points in one and two-dimensional spaces, determine the absolutevalue of a number, and find the distance and the midpoint of two points.This module contains three lessons, namely:Lesson 1 Plotting Points on the Number Line and Finding the Absolute Value of a NumberLesson 2Lesson 3 Plotting Points on the Cartesian Plane Finding the Distance and the Midpoint of Two Points What you are expected to learnAfter studying this module, you are expected to:a) determine the coordinate of a point on the number line;b) find the absolute value of a number;c) compute the distance between two points on the number line;d) simplify expressions containing the absolute value sign;e) describe the parts of the Cartesian plane; 1

f) give the (x, y)-coordinates of a point;g) plot a point given its (x, y)-coordinates;h) determine the quadrant where a given point is located;i) find the distance between two points; andj) determine the coordinates of the midpoint of two given points. 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. 2

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 next Study this module module/STOP. Take the PosttestWhat to do before (Pretest)Direction. Read each item and choose the letter of the correct answer.1. What is the absolute value of -3?a. -3 b. 3 c. 1 d. − 1 332. On the number line, what is the distance between -2 and 7?a. 14 b. 9 c. 5 d. -3.53. What is the value of − 4 + 2 − 2 7 − 5 ?a. 18 b. 6 c. -2 d. -4 3

4. What is the abscissa or the x-coordinate of the point (-3, 3)?a. -3 b. − 1 c. 3 d. 1 3 35. A point has positive abscissa and negative ordinate. In which quadrant does it lie?a. I b. II c. III d. IV6. A point lies 4 units to the left of the x-axis and 5 units below the y-axis. What arethe coordinates of the point?a. (4, 5) c. (-4, -5)b. (-4, 5) d. (4, -5)7. The ordinate of a point on the x-axis is alwaysa. negative c. positiveb. zero d. undefined8. What is the distance between the points (1, 3) and (4, 7)?a. 15 c. 7b. 5 d. 79. What is the distance between the points (0, 2) and (-2, 0)?a. 12 c. 8b. 4 d. 810. What are the coordinates of the midpoint of the points (2,5) and (-6,3)?a. (-2,8) c. (-8,-16)b. (-2,4) d. (8,2) Answer Key on page 20 What you will do Read carefully the questions that follow, answer the questions asked, and then dothe suggested activities to enhance your competence in mathematics. 4

Lesson 1 Plotting Points on the Number Line and Finding the Absolute Value of a Number Since our elementary years, we have seen the following diagram called the number line. -5 -4 -3 -2 -1 0 1 2 3 4 5 The number line is a horizontal line having two arrow heads with 0 at the middle. Themiddle number 0 is called the starting point or the origin. Numbers on the right side of 0 arecalled positive numbers, while those on the left are called negative numbers. Did you know? By definition, to every point on the number line we associate a unique number calledits coordinate. We denote points by capital letters. If P is a point, we denote its coordinateby the corresponding small letter p.Example 1 Identify the coordinate of each point: a) A b) B c) C A BC -5 -4 -3 -2 -1 0 1 2 3 4 5Answer: a) A has coordinate -4. b) B has coordinate 0. c) C has coordinate 3. The absolute value of a number is its distance from the origin. Since distance is apositive quantity, the absolute value of a number cannot be less than 0. For a number x, wedenote the absolute value of x by x . 5

Example 2 a) What is the absolute value of -5? The answer is 5 because -5 is 5 units from 0. 5 units -5 -4 -3 -2 -1 0 1 2 3 4 5b) What is the absolute value of 4? The answer is 4 because 4 is 4 units from 0. 4 units -5 -4 -3 -2 -1 0 1 2 3 4 5 c) What is the absolute value of 0? The answer is 0 because it is 0 unit from 0. 0 unit -5 -4 -3 -2 -1 0 1 2 3 4 5The distance between two points on the number line is given by d = a − b , where a is thecoordinate of the first point and b is the coordinate of the second point. AB abThe distance between the points A and B is d = a − b .Example 3 Find the distance between the following pair of points:a) P and Q b) Q and R C) Q and S. PQ RS -5 -4 -3 -2 -1 0 1 2 3 4 5 6

Answers: a) Since p =-5 and q = -3, the distance between P and Q is d = − 5 − (−3) = − 5 + 3 = − 2 = 2 units. b) Since q = -3 and r = 1, the distance between Q and R is d = − 3 −1 = − 4 = 4 units. c) Since q = -3 and s = 4, the distance between Q and S is d = − 3 − 4 = − 7 = 7 units.Example 4 Simplify the following expressions:a) − 7 − 2 + 17 − (−1) b) 6 − (− 3) − 5 − (−10)Answers: a) − 7 − 2 + 17 − (−1) = − 9 + 17 + 1 = 9 + 18 = 27b) 6 − (− 3) − 5 − (−10) = 6 + 3 − 5 + 10 = 9 −15 = −6Self-check 1Answer the following.1. Find the coordinate of the following points on the number line. a) B b) A c) O B AO -5 -4 -3 -2 -1 0 1 2 3 4 52. Find the distance between the given pair of points.a) A and T b) T and S c) A and I A TI S -5 -4 -3 -2 -1 0 1 2 3 4 53. Simplify the following expressions.a) 5 − 9 − 4 b) 2 7 − (−3) − 4c) 10 − 2 + 18 − (− 3) d) 6 − 2 + 21 − (−1) 7

e) 3 − 5 −1 − 2 6 − 2 f) − 4 − 6 − 35 − (−1) + 7 − 4 Answer Key on page 20Lesson 2 Plotting Points on the Cartesian Plane Did you know? In Lesson 1, we learned that any point on the number line is associated to a numbercalled its coordinate. Since the location of every point on the number line can be describedby a single coordinate only, the number line is said to be a 1-dimensional space. P p Point P on the number line has coordinate p. A point may not lie on the number line. To describe the location of the point, weconstruct another number line that is vertical to and passing through 0, the origin of theoriginal number line. The resulting figure is called the Cartesian coordinate plane or simplythe Cartesian plane. The horizontal line in the Cartesian plane is called the x-axis, while the vertical line iscalled the y-axis. The intersection of the axes is called the origin, also denoted by O. Every point in the Cartesian plane is associated with two numbers x and y, called itscoordinates. The coordinates of a point are written using ordered pair (x, y). Also, if (x, y)are the coordinates of a point P, we write P(x, y). Note that the origin has coordinates (0, 0). Since two numbers are associated to every point in the Cartesian plane, this plane issaid to be a 2-dimensional space. 8

y-axisThe Cartesian Plane 5 • P(x, y) 4x 3 2y origin 1 x-axis-5 -4 -3 -2 -1 01 2 3 45 -1 -2 Point P has coordinates (x, y). -3 -4 -5Example 5 Use the diagram below to find the coordinates of the following points. a) T b) E c) A d) M y-axis 5 4•T 3 2 •E 1 x-axis-5 -4 -3 -2 -1 01 2 3 45 -1 M• -2 -3 -4 • A -5 9