Elementary Algebra Problem Solving Linear Equation,Line, Properties of real number
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The document discusses elementary algebra concepts, focusing on real numbers and linear equations. It includes problem-solving examples related to real-life scenarios, such as deposits, temperature changes, and growth over time, illustrating the application of linear equations in various contexts. Additionally, it explains methods for solving linear equations in two variables using substitution and elimination.
Elementary Algebra Problem Solving Linear Equation,Line, Properties of real number
- 1. Elementary Algebra Fatima S. Lara and Anafel F. Into
- 2. Table of Contents Linear Equation Real Numbers Linear Equation in Two Variable
- 4. The type of number we normally use, such as 1, 15.82, −0.1, 3/4, etc. Positive or negative, large or small, whole numbers or decimal numbers are all Real Numbers. They are called "Real Numbers" because they are not Imaginary Numbers. What is a Real Numer?
- 5. Properties of Real Numbers
- 6. Properties of Real Numbers
- 7. Word Problems involving Real Numbers Ana has a negative balance of 600php in his bank account. If she deposit 3550php into her account. How much money does Ana have in her account after the deposit? 1. What is the problem? How much Ana have in her account after the deposit? Problem 1: Steps in Problem Solving: 2. What is given? Negative 600php and deposit 3550php
- 8. 3. What is the solution? -600php + 3,550php = ? 5. Checking Therefore, Ana have 2,950php in her account. 4. Solving the Problem -600php + 3,550php = 2,950php
- 9. In Baguio City, Philippines the temperature was -14 °F in the morning. If the temperature dropped 7°F, what is the temperature now? 1. What is the problem? What is the present temperature? Problem 2: Steps in Problem Solving: 2. What is given? -14°F and -7°F
- 10. 3. What is the solution? -14°F + -7°F = ? 5. Checking Therefore, the temperature now is -21°F. 4. Solving the Problem -14°F + -7°F = -21°F
- 11. A submarine at -38feet dives 50feet. What is the submarine elevation after the dive? 1. What is the problem? What is the submarine elevation after the dive? Problem 3: Steps in Problem Solving: 2. What is given? -38feet and -50feet
- 12. 3. What is the solution? -38feet + -50feet = ? 5. Checking Therefore, the submarine elevation after the dive is -88feet. 4. Solving the Problem -38feet + -50feet = -88feet
- 13. Submarine was situated 800feet below sea level. If it ascends 250feet. What is the new position? 1. What is the problem? What is the new position of the submarine situated? Problem 4: Steps in Problem Solving: 2. What is given? -800feet and 250feet
- 14. 3. What is the solution? 800feet + 250feet =? 5. Checking -800feet + 250feet = -550feet 4. Solving the Problem -800feet + 250feet = -550feet
- 15. The Jurassic period was believed to have started about 190,000,000 B.C. and ended about 135,000,000 B.C. About how long was this? 1. What is the problem? How long Jurassic period last? Problem 5: Steps in Problem Solving: 2. What is given? 190,000,000 B.C. and 135,000,000 B.C.
- 16. 3. What is the solution? 190,000,000 B.C – 135,000,00 B.C. =? 5. Checking Therefore, the Jurassic period last for about 55,000,000 B.C. 4. Solving the Problem 190,000,000 B.C – 135,000,00 B.C. = 55,000,000 B.C.
- 17. Problem Solving involving Linear Equation Average
- 18. Linear Equation *Linear equations are equations of the first order. These equations are defined for lines in the coordinate system. An equation for a straight line is called a linear equation. The general representation of the straight-line equation is y=mx+b, where m is the slope of the line and b is the y-intercept. Linear equations are those equations that are of the first order. These equations are defined for lines in the coordinate system.
- 19. Linear Equation
- 21. 01 1. Ana created an instagram account and started with 3 following (her mom, her dad, and her sister). She plans to find 7 new followers each day. Create an equation that describes how many followers she has each day. How many followers will she have after 10days? 100 days? Steps in Problem Solving: 1. What is the problem? How much follower she have after 10days? 100days? 2. What is given? 3followers, 7followers, 10days and 100days 3. What is the solution? y = mx + b y = 7x + 3 4. Solving the Problem y = mx + b y = 7x + 3 x = 10 x = 100 y = 7(100) + 3 y = 7(10) + 3 y = 700 + 3 y = 70 + 3 y = 703 y = 73 5. Checking Therefore, after 10days Ana has 73 followers. Therefore, after 100days Ana has 703 followers.
- 22. 02 1. Mr. Lam wants to create a library of books in his classroom. He is starting with 12 books. He plans to buy 3 new books each week. Create an equation that describe the number of books Mr. Lam has each week. How many books does Mr. Lam have after 5 weeks? 10 weeks? Steps in Problem Solving: 1. What is the problem? How many books does Mr. Lam have after 5 weeks? 10 weeks? 2. What is given? 3 new books, 12 books, after 5weeks and after 10 weeks 3. What is the solution? y = mx + b y = 3x + 12 4. Solving the Problem y = 3x + 12 x = 5 x = 10 y = 3(5) + 12 y = 3(10) + 12 y = 15 + 12 y = 30 + 12 y = 27 y = 42 5. Checking Therefore, after 5 weeks Mr. Lam has 27 books. Therefore, after 10 weeks Mr. Lam has 42 books.
- 23. 03 Ms. Ruddy opens a retirement account and the starting is 3000php. She plans to add 1000php per month to the account. Make an equation that describes the amount of money in the account each month. How much money does Ms. Ruddy have after 6 months? 12 months? Steps in Problem Solving: 1. What is the problem? How much money does Ms. Ruddy have after 6 months? 12 months? 2. What is given? Starting is 3000php, 1000php per month, 6 months and 12 months. 3. What is the solution? y = mx + b y = 1000phpx + 3000php
- 24. 4. Solving the Problem y = 1,000phpx + 3000 x = 6 x = 12 y = 1,000php (6) + 3000php y = 1,000php (12) + 3000php y = 6000php + 3000php y = 12,000php + 3000php y = 9000php y = 15,000php 5. Checking Therefore, after 6 months Ms. Rudy has 9000php. Therefore, after 12 months Ms. Rudy has 15,000php.
- 25. 04 Mira wants to gain weight. She started with 45kg. Her goal is to gain 3kg every week. Make an equation that describes the weight Mira gain every week. How much weight does Mira gain after 2 months? 5 months? Steps in Problem Solving: What is the problem? How much weight does Mira gain after 2 months? 5 months? What is given? 45kg, 3kg every week, 2 months and 5 months What is the solution? y = mx + b y = 3kg x + 45kg
- 26. Solving the Problem x = 2 x = 5 y = 3kg (2) + 45kg y = 3kg (5) + 45kg y = 6kg + 45kg y = 15kg + 45kg y = 51kg y = 60kg Checking Therefore, after 2 months Mira gain 51kg. Therefore, after 5 months Mira gain 60kg.
- 27. 05 Ben opens a store. The starting capital is 50,000php. He plans to add 10,000 each month. Create and equation that describes the amount Ben added each month for his store. How much does Ben have added after 1 year? 2 years? Steps in Problem Solving: 1. What is the problem? How much does Ben have added after 1 year? 2 years? 2. What is given? Capital 50,000php, 10,0000php per month, 12 months and 24 months. 3. What is the solution? y = mx + b y = 10,000phx + 50,000php
- 28. 4. Solving the Problem y = 10,000php + 50,000php x = 12 y = 10,000php (12) + 50,000php y = 120,000php + 50,000php y = 170,000php x = 24 y = 10,000phpx + 50,000php y = 10,000php (24) + 50,000php y = 240,000php + 50,000php y = 290,000php 5. Checking Therefore, after 12 months Ben has added 170,000php. Therefore, after 24 months Ben has added 290,000php
- 29. Linear Equation in Two Variables using Elimination and Substitution.
- 30. What is Linear Equation in Two Variables?
- 31. What are the methods in Solving Linear Equation involving Two Variables? 1. Substitution 2. Elimination 3. Graphing 4. Matrix
- 38. What is a Linear Equation in two variables? A linear system of two equations with two variables is any system that can be written in the form. ax + by = p cx + dy = q A solution to a system of equations is a value of x and a value of y that, when substituted into the equations, satisfies both equations at the same time.
- 39. There are x chairs and y tables in a room. There are 8 chairs and tables in all. The number of tables is 2 less than the number of chairs. How many chairs and tables are there? 1. What is the problem? How many chairs and tables in the room ? Problem 1 : Steps in Problem Solving: 2. What is given? x = the no. of chairs y = the no. of tables 8 chairs and tables in all
- 40. 3. What is the solution? y = x - 2 -x + y = -2 x + y = 8 Solution using Elimination Method x + y =8 + -x + y =-2 2y = 6 2y/2 = 6/2 y = 3 no. of tables Substitute y = 3 x + y = 8 x + 3 = 8 x = 8 – 3 x = 5 no. of chairs 5. Checking Therefore, there are 5 chairs and 3 tables in the room. 4. Solving the Problem Solution using Substitution Method y = x - 2 -x + y = -2 x + y = 8 x + (x-2) = 8 2x - 2 = 8 2x = 8 + 2 2x = 10 2x/2 = 10/2 x = 5 no. of chairs y = x – 2 y = 5 – 2 y = 3 no. of tables
- 41. Your teacher is giving you a test worth 100 points containing 40 questions. There are two – points and four - points questions on the test. How many of each types of question are on the test? 1. What is the problem? How is the total two - points and four – points question on the test ? Problem 2 : Steps in Problem Solving : 2. What is given? x = total questions of two - points y = total questions of four – points 100 points and 40 questions
- 42. 3. What is the solution? x + y = 40 2x + 4y = 100 Solution using Elimination Method x + y = 40 -2(x + y = 40) + 2x + 4y = 100 + 2x + 4y = 100 -2x + 2y = -80 + 2x + 4y = 100 2y = 20 2y/2 = 20/2 y = 10 total questions of 4 points Substitute y = 10 x + y = 40 x + 10= 40 x = 40 – 10 x = 30 total questions of 2 points 5. Checking Therefore, there are 30 questions of 2 points and 10 questions of 4 points on the test. 4. Solving the Problem Solution using Substitution Method x + y = 40 y = 10 x + y = 40 2x + 4y = 100 x + 10 = 40 2( 40 – y ) + 4y = 100 x = 40 - 10 80 – 2y + 4y = 100 x = 30 total 80 + 2y = 100 questions 2y = 100 – 80 of 2 points 2y = 20 2y/2 =20/2 y = 10 total questions of 4 points
- 43. There are 25 students in a room. The number of females students is 9 more than the number of male students. How many female students are there? Male students? 1. What is the problem? How many female and male students are there in the room? Problem 3 : Steps in Problem Solving : 2. What is given? x = the number of male students y = the number of female students 25 students in the room
- 44. 3. What is the solution? x + y = 25 y = x + 9 Solution using Elimination Method x + y = 25 x + y =25 y = x + 9 - x + y = 9 2y = 34 2y/2 = 34/2 y = 17 total number of female students Substitute y = 17 x + y = 25 x + 17 = 25 x = 25 – 17 x= 8 total number of male students 5. Checking Therefore, there are 17 female and 8 male students in the room. 4. Solving the Problem Solution using Substitution Method x + y = 25 substitute: y = x + 9 y = x + 9 y = 8 + 9 x + y = 25 y = 17 total x + (x + 9) = 25 number of 2x + 9 = 25 females 2x = 25 – 9 students 2x = 16 2x/2 = 16/2 x = 8 total number of male students
- 45. Roman’s age is three times the sum of the ages of his tow son’s. After 5 years his age will be twice the sum of the ages of his two son’s. Find the age of Roman? 1. What is the problem? What is the age of Roman? Problem 4: Steps in Problem Solving : 2. What is given? x = Roman’s age 3times y = the sum of ages of his two son’s 5 + 5 , y+10 After 5 years = x + 5
- 46. 3. What is the solution? x + 5 = 2 (y + 10) x = 3y Solution using Elimination Method x + 5 = 2 ( y + 10) = x – 2y = 15 x = 3y = + -x + 3y = 0 y = 15 son’s age Substitute y = 15 x = 3y x= 3 (15) x= 45 Roman’s age 5. Checking Therefore, Roman’s age is 45years old. 4. Solving the Problem Solution using Substitution Method x + 5 = 2( y + 10) x + 5 = 2y + 20 x - 2y = 20 – 5 x – 2y = 15 Substitute x = 3y 3y – 2y = 15 y = 15 son’s ages Substitute y = 15 x = 3 ( 15) x = 45 Roman’s age
- 47. Michael and Tom are brothers. Their combined age is 20 and Tom is 4 years older than Michael. What are Michael and Tom’s ages? 1. What is the problem? What are Michael and Tom’s ages? Problem 5: Steps in Problem Solving : 2. What is given? x = Michael’s age y = Tom’s age is 4times older The combined ages is 20.
- 48. 3. What is the solution? x + y = 20 y = x + 4 Solution using Elimination Method x + y = 20 x + y = 20 y = x + 4 + -x + y = 4 2y = 24 2y/2 = 24/2 y = 12 Tom’s age Substitute y = 12 y = x + 4 12 = x + 4 12-4 = x x = 8 Michael’s age 5. Checking Therefore, Michael’s age is 8 and Tom’s age is 12. 4. Solving the Problem Solution using Substitution Method x + y = 20 x + y = 20 y = x + 4 8 + y = 20 x + (x + 4) = 20 y = 20 - 8 x + x + 4 = 20 y = 12 Tom’s age 2x + 4 = 20 2x = 20 – 4 2x = 16 2x/2 = 16/2 x = 8 Michael’s age
- 49. THANK YOU !!!
- 50. Elementary Algebra Fatima S. Lara and Anafel F. Into