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2. S E C T I O N 3.1 Introduction to Equations

3. 33 A.To determine whether a given number is a solution of an equation B.To solve an equation of the form x + a = b C.To solve an equation of the form ax = b D.To solve application problems using the basic percent equation E.To solve uniform motion problems Objectives

4. 44 OBJECTIVE A To determine whether a given number is a solution of an equation

5. 55 An equation expresses the equality of two mathematical expressions. The expressions can be either numerical or variable expressions. 9 + 3 = 12 3x – 2 = 10 y 2 + 4 = 2y – 1 z = 2 To determine whether a given number is a solution of an equation Equations

6. 66 The equation given below is true if the variable is replaced by 5. x + 8 = 13 5 + 8 = 13 The equation is false if the variable is replaced by 7. 7 + 8 = 13 A solution of an equation is a number that, when substituted for the variable, results in a true equation. 5 is a solution of the equation x + 8 = 13. 7 is not a solution of the equation x + 8 = 13. To determine whether a given number is a solution of an equation A false equation A true equation

7. 77 How To Is –2 a solution of 2x + 5 = x 2 – 3? Yes, –2 is a solution of the equation. To determine whether a given number is a solution of an equation Replace x by –2. Evaluate the numerical expressions. If the results are equal, –2 is a solution of the equation. If the results are not equal, –2 is not a solution of the equation.

8. 88 Is –4 a solution of 5x – 2 = 6x + 2? Solution: Yes, –4 is a solution. Example 1

9. 99 OBJECTIVE B To solve an equation of the form x + a = b

10. 10 To solve an equation means to find a solution of the equation. The simplest equation to solve is an equation of the form variable = constant, because the constant is the solution. The solution of the equation x = 5 is 5 because 5 = 5 is a true equation. The solution of the equation given below is 7 because 7 + 2 = 9 is a true equation. x + 2 = 9 7 + 2 = 9 To solve an equation of the form x + a = b

11. 11 Note that if 4 is added to each side of the equation x + 2 = 9, the solution is still 7. x + 2 = 9 x + 2 + 4 = 9 + 4 x + 6 = 137 + 6 = 13 If –5 is added to each side of the equation x + 2 = 9, the solution is still 7. x + 2 = 9 x + 2 + (–5) = 9 + (–5) x – 3 = 47 – 3 = 4 To solve an equation of the form x + a = b

12. 12 Equations that have the same solution are called equivalent equations. The equations x + 2 = 9, x + 6 = 13, and x – 3 = 4 are equivalent equations; each equation has 7 as its solution. These examples suggest that adding the same number to each side of an equation produces an equivalent equation. This is called the Addition Property of Equations. To solve an equation of the form x + a = b

13. 13 Addition Property of Equations The same number can be added to each side of an equation without changing its solution. In symbols, the equation a = b has the same solution as the equation a + c = b + c. In solving an equation, the goal is to rewrite the given equation in the form variable = constant. The Addition Property of Equations is used to remove a term from one side of the equation by adding the opposite of that term to each side of the equation. To solve an equation of the form x + a = b

14. 14 How To Solve: x – 4 = 2 x – 4 = 2 x – 4 + 4 = 2 + 4 x + 0 = 6 x = 6 To solve an equation of the form x + a = b The goal is to rewrite the equation in the form variable = constant. Add 4 to each side of the equation. Simplify. The equation is in the form variable = constant.

15. 15 Check: The solution is 6. Because subtraction is defined in terms of addition, the Addition Property of Equations also makes it possible to subtract the same number from each side of an equation without changing the solution of the equation. To solve an equation of the form x + a = b A true equation

16. 16 Solve: Solution: Example 3 Subtract from each side. Rewrite and with a common denominator.

17. 17 The solution is. Example 3 – Solution cont’d

18. 18 OBJECTIVE C To solve an equation of the form ax = b

19. 19 The solution of the equation given below is 3 because 2  3 = 6 is a true equation. 2x = 6 2  3 = 6 Note that if each side of 2x = 6 is multiplied by 5, the solution is still 3. 2x = 6 5(2x) = 5  6 10x = 30 10  3 = 30 To solve an equation of the form ax = b

20. 20 If each side of 2x = 6 is multiplied by –4, the solution is still 3. 2x = 6 (–4)(2x) = (–4)6 –8x = –24 –8  3 = –24 The equations 2x = 6, 10x = 30, and –8x = –24 are equivalent equations; each equation has 3 as its solution. These examples suggest that multiplying each side of an equation by the same nonzero number produces an equivalent equation. To solve an equation of the form ax = b

21. 21 Multiplication Property of Equations Each side of an equation can be multiplied by the same nonzero number without changing the solution of the equation. In symbols, if c  0, then the equation a = b has the same solutions as the equation ac = bc. The Multiplication Property of Equations is used to remove a coefficient by multiplying each side of the equation by the reciprocal of the coefficient. To solve an equation of the form ax = b

22. 22 How To Solve: To solve an equation of the form ax = b The goal is to rewrite the equation in the form variable = constant. Multiply each side of the equation by.

23. 23 1  z = 12 z = 12 The solution is 12. Because division is defined in terms of multiplication, each side of an equation can be divided by the same nonzero number without changing the solution of the equation. To solve an equation of the form ax = b Simplify. The equation is in the form variable = constant.

24. 24 When using the Multiplication Property of Equations, multiply each side of the equation by the reciprocal of the coefficient when the coefficient is a fraction. Divide each side of the equation by the coefficient when the coefficient is an integer or a decimal. To solve an equation of the form ax = b

25. 25 Solve: Solution: The solution is –12. Example 4

26. 26 OBJECTIVE D To solve application problems using the basic percent equation

27. 27 An equation that is used frequently in mathematics applications is the basic percent equation. Basic Percent Equation Percent  Base = Amount P  B = A In many application problems involving percent, the base follows the word of. To solve application problems using the basic percent equation

28. 28 How To 20% of what number is 30? P  B = A 0.20B = 30 B = 150 The number is 150. To solve application problems using the basic percent equation Use the basic percent equation. P = 20% = 0.20, A = 30, and B is unknown. Solve for B.

29. 29 The simple interest that an investment earns is given by the simple interest equation I = Prt, where I is the simple interest, P is the principal, or amount invested, r is the simple interest rate, and t is the time. How To A $1500 investment has an annual simple interest rate of 7%. Find the simple interest earned on the investment after 18 months. The time is given in months but the interest rate is an annual rate. Therefore, we must convert 18 months to years. To solve application problems using the basic percent equation

30. 30 18 months = years = 1.5 years To find the interest, solve I = Prt for I. I = Prt I = 1500(0.07)(1.5) I = 157.5 The investment earned $157.50. To solve application problems using the basic percent equation P = 1500, r = 0.07, t = 1.5

31. 31 The amount of a substance in a solution can be given as a percent of the total solution. For instance, if a certain fruit juice drink is advertised as containing 27% cranberry juice, then 27% of the contents of the bottle must be cranberry juice. The method for solving problems involving mixtures is based on the percent mixture equation Q = Ar, where Q is the quantity of a substance in the solution, A is the amount of the solution, and r is the percent concentration of the substance. To solve application problems using the basic percent equation

32. 32 In most cases, you should write the percent as a decimal before solving the basic percent equation. However, some percents are more easily written as a fraction. For example, To solve application problems using the basic percent equation

33. 33 The data in the table below show the number of households (in millions) that downloaded music files for a three-month period in a recent year. (Source: NPD Group) For the three-month period, what percent of the files were downloaded in May? Round to the nearest percent. Example 7

34. 34 To find the percent: Find the total number of files downloaded for the three-month period. Use the basic percent equation. B is the total number of files downloaded for the three-month period; A = 12.7, the number of files downloaded in May; P is unknown. Example 7 – Strategy

35. 35 14.5 + 12.7 + 10.4 = 37.6 P  B = A P(37.6) = 12.7  0.34 Approximately 34% of the files were downloaded in May. Example 7 – Solution Use the basic percent equation. B = 37.6, A = 12.7

36. 36 OBJECTIVE E To solve uniform motion problems

37. 37 Any object that travels at a constant speed in a straight line is said to be in uniform motion. Uniform motion means that the speed and direction of an object do not change. For instance, a car traveling at a constant speed of 45 mph on a straight road is in uniform motion. To solve uniform motion problems

38. 38 The solution of a uniform motion problem is based on the uniform motion equation d = rt, where d is the distance traveled, r is the rate of travel, and t is the time spent traveling. For instance, suppose a car travels at 50 mph for 3 h. Because the rate (50 mph) and time (3 h) are known, we can find the distance traveled by solving the equation d = r t for d. d = r t d = 50(3) d = 150 The car travels a distance of 150 mi. To solve uniform motion problems r = 50, t = 3

39. 39 How To A jogger runs 3 mi in 45 min. What is the rate of the jogger in miles per hour? Strategy Because the answer must be in miles per hour and the given time is in minutes, convert 45 min to hours. To find the rate of the jogger, solve the equation d = rt for r. To solve uniform motion problems

40. 40 Solution To solve uniform motion problems d = 3, t =

41. 41 4 = r The rate of the jogger is 4 mph. To solve uniform motion problems Multiply each side of the equation by the reciprocal of.

42. 42 If two objects are moving in opposite directions, then the rate at which the distance between them is increasing is the sum of the speeds of the two objects. For instance, in the diagram below, two cars start from the same point and travel in opposite directions. The distance between them is changing at 70 mph. To solve uniform motion problems

43. 43 Similarly, if two objects are moving toward each other, the distance between them is decreasing at a rate that is equal to the sum of the speeds. The rate at which the two planes shown below are approaching one another is 800 mph. To solve uniform motion problems

44. 44 If a motorboat is on a river that is flowing at a rate of 4 mph, then the boat will float down the river at a speed of 4 mph when the motor is not on. Now suppose the motor is turned on and the power adjusted so that the boat would travel 10 mph without the aid of the current. Then, if the boat is moving with the current, its effective speed is the speed of the boat using power plus the speed of the current: 10 mph + 4 mph = 14 mph. (See the figure below.) To solve uniform motion problems

45. 45 However, if the boat is moving against the current, the current slows the boat down. The effective speed of the boat is the speed of the boat using power minus the speed of the current: 10 mph – 4 mph = 6 mph. (See the figure below.) To solve uniform motion problems

46. 46 Two cyclists start at the same time at opposite ends of an 80-mile course. One cyclist is traveling 18 mph, and the second cyclist is traveling 14 mph. How long after they begin cycling will they meet? Strategy: The distance is 80 mi. Therefore, d = 80. The cyclists are moving toward each other, so the rate at which the distance between them is changing is the sum of the rates of each of the cyclists. The rate is 18 mph + 14 mph = 32 mph. Therefore, r = 32. To find the time, solve the equation d = r t for t. Example 10

47. 47 d = rt 80 = 32t 2.5 = t The cyclists will meet in 2.5 h. Example 10 – Solution d = 80, r = 32 Solve for t.