1 / 159

Chapter 7

Chapter 7. The Basic Concepts of Algebra. Chapter 7: The Basic Concepts of Algebra. 7.1 Linear Equations 7.2 Applications of Linear Equations 7.3 Ratio, Proportion, and Variation 7.4 Linear Inequalities 7.5 Properties of Exponents and Scientific Notation

octavius-julius
Télécharger la présentation

Chapter 7

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 7 The Basic Concepts of Algebra 2012 Pearson Education, Inc.

  2. Chapter 7: The Basic Concepts of Algebra • 7.1 Linear Equations • 7.2 Applications of Linear Equations • 7.3 Ratio, Proportion, and Variation • 7.4 Linear Inequalities • 7.5 Properties of Exponents and Scientific Notation • 7.6 Polynomials and Factoring • 7.7 Quadratic Equations and Applications 2012 Pearson Education, Inc.

  3. Section 7-1 • Linear Equations 2012 Pearson Education, Inc.

  4. Linear Equations • Solving Linear Equations • Special Kinds of Linear Equations • Literal Equations and Formulas • Models 2012 Pearson Education, Inc.

  5. Algebraic Expression An algebraic expression involves only the basic operations of addition, subtraction, multiplication, or division (except by 0), or raising to powers or taking roots on any collection of variables and numbers. Algebraic expressions 2012 Pearson Education, Inc.

  6. Equation An equation is a statement that two algebraic expressions are equal. A linear equation in one variable involves only real numbers and one variable. Linear equations 2012 Pearson Education, Inc.

  7. Linear Equation in One Variable An equation in the variable x is linear if it can be written in the form Ax + B = C where A, B, and C are real numbers, with 2012 Pearson Education, Inc.

  8. Terminology A linear equation in one variable is also called a first-degree equation. If the variable in an equation is replaced by a real number that makes the statement of the equation true, then that number is a solution of the equation. An equation is solved by finding its solution set, the set of all answers. Equivalent equations are equations with the same solution set. 2012 Pearson Education, Inc.

  9. Addition Property of Equality For all real numbers A, B, and C, the equations A = B and A + C = B + C are equivalent. (The same number may be added to both sides of an equation without changing the solution set.) 2012 Pearson Education, Inc.

  10. Multiplication Property of Equality For all real numbers A, B, and C, the equations A = B and AC = BC are equivalent. (Both sides of an equation may be multiplied by the same nonzero number without changing the solution set.) 2012 Pearson Education, Inc.

  11. Solving a Linear Equation in One Variable Step 1 Clear fractions. Multiply both sides by a common denominator. Step 2Simplify each side separately. Clear parentheses and combine like terms. Step 3Isolate the variable terms on one side. Use the addition property of equality. 2012 Pearson Education, Inc.

  12. Solving a Linear Equation in One Variable Step 4 Transform so that the coefficient of the variable is 1. Use the multiplication property of equality. Step 5Check. Substitute the solution into the original equation. 2012 Pearson Education, Inc.

  13. Example: Solve a Linear Equation using the Distributive Property Solve 3(k + 1) + 2k = 13. Solution 3k + 3 + 2k = 13 Distributive property. 5k + 3 = 13 Combine like terms. 5k + 3 – 3 = 13 – 3 Subtract 3. 5k = 10 Combine like terms. 2012 Pearson Education, Inc.

  14. Example: Solve a Linear Equation using the Distributive Property Solution(continued) Divide by 5. k = 2 Check that the solution set is {2} by substituting 2 for k in the original equation. 2012 Pearson Education, Inc.

  15. Example: Solve a Linear Equation with Fractions Solve Solution Multiply by LCD, 6. Distributive property. Multiply. 2012 Pearson Education, Inc.

  16. Example: Solve a Linear Equation with Fractions Solution(continued) x + 7 + 6x – 24 = –24 Distributive property. 7x – 17 = –24 Combine like terms. Add 17. 7x – 17 + 17 = –24 + 17 7x = –7 Combine like terms. Divide by 7. x = –1 Now check. 2012 Pearson Education, Inc.

  17. Types of Linear Equations 2012 Pearson Education, Inc.

  18. Example: Contradiction Solve 3k + 4 – 2k = k. Solution k + 4 = k Combine like terms. k – k + 4 = k – k Subtract k. 4 = 0 Combine like terms. Because the result is false, the equation has no solution. The solution set is so the original equation is a contradiction. 2012 Pearson Education, Inc.

  19. Example: Identity Solve 3k + 4 – 2k = k + 4. Solution k + 4 = k + 4 Combine like terms. k – k + 4 = k – k Subtract k. 4 = 4 Combine like terms. Because the result is true, the solution set is {all real numbers} and the original equation is an identity. 2012 Pearson Education, Inc.

  20. Literal Equations and Formulas An equation involving variables (or letters), such as y = mx + b is called a literal equation. Formulas are literal equations in which more than one letter is used to express a relationship. It may be necessary to solve for one of the variables in a formula. This process is called solving for a specified variable. 2012 Pearson Education, Inc.

  21. Solving for a Specified Variable Step 1Transform the equation so that all terms containing the specified variable are on one side of the equation and all terms without the variable are on the other side. Step 2 If necessary, use the distributive property to combine the terms with the specified variable. 2012 Pearson Education, Inc.

  22. Solving for a Specified Variable Step 3 Divide both sides by the factor that is multiplied by the specified variable. 2012 Pearson Education, Inc.

  23. Example: Solving for a Specified Variable Solve the formula P = 2L + 2W for L. Solution P – 2W = 2L + 2W – 2W Subtract 2W. P – 2W = 2L Combine like terms. Divide by 2. 2012 Pearson Education, Inc.

  24. Models A mathematical model is an equation (or inequality) that describes the relationship between two quantities. A linear model is a linear equation. 2012 Pearson Education, Inc.

  25. Example: Temperature Conversion The relationship between degrees Celsius (C) and degrees Fahrenheit (F) is modeled by the linear equation F = 1.8C + 32. What Celsius degree corresponds to a Fahrenheit reading of 50°? 2012 Pearson Education, Inc.

  26. Example: Temperature Conversion Solution Because F = 50, the equation becomes 50 = 1.8C + 32 500 = 18C + 320 Multiply by 10. 180 = 18C Subtract 320. Divide by 18. C = 10. Therefore, a reading of 50 degrees Fahrenheit corresponds to a reading of 10 degrees Celsius. 2012 Pearson Education, Inc.

  27. Section 7-2 • Applications of Linear Equations 2012 Pearson Education, Inc.

  28. Applications of Linear Equations • Translating Words into Symbols • Guidelines for Applications • Finding Unknown Quantities • Mixture and Interest Problems • Monetary Denomination Problems • Motion Problems 2012 Pearson Education, Inc.

  29. Translating Words into Symbols Usually there are key words and phrases in a verbal problem that translate into mathematical expressions involving addition, subtraction, multiplication, and division. 2012 Pearson Education, Inc.

  30. Translation to Mathematical Expressions: Addition 2012 Pearson Education, Inc.

  31. Translation to Mathematical Expressions: Subtraction 2012 Pearson Education, Inc.

  32. Translation to Mathematical Expressions: Multiplication 2012 Pearson Education, Inc.

  33. Translation to Mathematical Expressions: Division 2012 Pearson Education, Inc.

  34. Translation to Mathematical Expressions: Equality The symbol of equality, =, is often indicated by the word is. Words that indicate the idea of “sameness” translate as =. 2012 Pearson Education, Inc.

  35. Guidelines for Applications Solving an Applied Problem Step 1 Read the problem carefully until you understand what is given and what is to be found. Step 2 Assign a variable to represent the unknown value, using diagrams or tables as needed. If necessary, express any other unknown values in terms of the variable. 2012 Pearson Education, Inc.

  36. Guidelines for Applications Step 3 Write an equation using the variable expression(s). Step 4 Solve the equation. Step 5 State the answer. Does it seem reasonable? Step 6 Check the answer in the words of the original problem. 2012 Pearson Education, Inc.

  37. Example: Application A fenced area contains a total of 74 hens and roosters. If there are 24 more hens than roosters, find the number of each. Solution Step 1 We are looking to find the number of hens and the number of roosters. Step 2 Let r = the number of roostersThen r + 24 = the number of hens 2012 Pearson Education, Inc.

  38. Example: Application Step 3 r + (r + 24) = 74 Step 4 2r + 24 = 74 2r = 50 r = 25 so h = 25 + 24 = 49 Step 5 There are 25 roosters and 49 hens Step 6 The sum of roosters and hens: 25 + 49 = 74 There are 24 more hens than roosters. 49 – 25 = 24. 2012 Pearson Education, Inc.

  39. Mixture and Interest Problems Percents often are used in problems involving mixing different concentrations of a substance or different interest rates. Mixing Problems base x rate (%) = percentage b x r = p Interest Problems (annual) principal x rate (%) = interest P x r = I 2012 Pearson Education, Inc.

  40. Example: Mixture A chemist plans to mix 10 liters of a 40% acid solution with some 70% to obtain a 50% solution. How much of the 70% solution should be used? Solution Step 1 We are looking for the amount of 70% solution. Step 2 Let x = the number of liters of 70% solution. 2012 Pearson Education, Inc.

  41. Example: 40% 10 L 70% x L 50% (10+ x) L 2012 Pearson Education, Inc.

  42. Example: Mixture Sum must equal Step 3 .40(10) + .70x = .50(10 + x) Step 4 4 + .7x = 5 + .5x .2x = 1 x = 5 2012 Pearson Education, Inc.

  43. Example: Mixture Step 5 The chemist should use 5 liters of the 70% solution. Step 6 5 liters of 70% and 10 liters of 40% is 5(.7) + 10(.4) = 7.5 liters of acid. 10 + 5 liters of 50% is 15(.5) 7.5 liters of acid. 2012 Pearson Education, Inc.

  44. Monetary Denomination Problems Problems that involve different denominations of money or items with different monetary values are similar to mixture and investment problems. Money Problems number x value of one = total value For example, if you have 9 quarters, the monetary value of the coins is 9x$.25 = $2.25 2012 Pearson Education, Inc.

  45. Example: Monetary Denomination For a bill totaling $5.65, a cashier received 25 coins consisting of nickels and quarters. How many of each type of coin did the cashier receive? Solution Step 1 We are looking for the number of nickels and quarters. Step 2 Let x = the number of nickels; then 25 – x = number of quarters. 2012 Pearson Education, Inc.

  46. Example: Mixture Totals Step 3 .05x + .25(25 – x) = 5.65 Step 4 5x + 625 – 25x = 565 –20x = –60 x = 3 2012 Pearson Education, Inc.

  47. Example: Mixture Step 5 There are 3 nickels and 22 quarters. Step 6 3($.05) + 22($.25) = $5.653 + 22 = 25 coins 2012 Pearson Education, Inc.

  48. Motion Problems The basic relationship between distance, rate, and time: distance = rate x time. Forms 2012 Pearson Education, Inc.

  49. Example: Motion Problem Jim can walk to work in ¾ hour. By bike the trip takes ¼ hour. If he bikes 8 mile per hour faster than he walks, how far is it to his workplace? Solution Step 1 We are looking for the distance to the workplace. Step 2 Let x = rate when walking; then x + 8 = rate biking. 2012 Pearson Education, Inc.

  50. Example: Motion Use d = rt. Same Step 3 Step 4 3x = x + 8 x = 4 2012 Pearson Education, Inc.

More Related