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Slide 1- 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

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Equations, Inequalities, and Problem Solving 2.1Solving Equations 2.2Using the Principles Together 2.3Formulas 2.4Applications with Percent 2.5 Problem Solving 2.6Solving Inequalities 2.7Solving Applications with Inequalities 2

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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Equations Equations and Solutions The Addition Principle The Multiplication Principle Selecting the Correct Approach 2.1

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Slide 2- 4 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution of an Equation Any replacement for the variable that makes an equation true is called a solution of the equation. To solve an equation means to find all of its solutions.

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Slide 2- 5 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine whether 8 is a solution of x + 12 = 21. Since the left-hand and right-hand sides differ, 8 is not a solution. 20 21 False 8 + 12 | 21 Substituting 8 for x Solution x + 12 = 21 Writing the equation Example

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Slide 2- 6 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equivalent Equations Equations with the same solutions are called equivalent equations. The Addition Principle For any real numbers a, b, and c, a = b is equivalent to a + c = b + c.

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Slide 2- 7 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve: 8.3 = y 17.9 Solution 8.3 = y 17.9 8.3 + 17.9 = y 17.9 + 17.9 9.6 = y Check: 8.3 = y 17.9 8.3 | 9.6 17.9 8.3 = 8.3 The solution is 9.6. Example

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Slide 2- 8 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Multiplication Principle For any real numbers a, b, and c with c 0, a = b is equivalent to a c = b c. Solve: Multiplying both sides by 4/3. Solution Example

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Slide 2- 9 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve: 7x = 84 Solution 7x = 84 Dividing both sides by 7. The solution is –12. Check: 7x = 84 7( 12) | 84 84 = 84 Example

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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Using the Principles Together Applying Both Principles Combining Like Terms Clearing Fractions and Decimals Contradictions and Identities 2.2

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Slide 2- 11 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve: 9 + 8x = 33 Solution 9 + 8x = 33 9 + 8x 9 = 33 9 9 + ( 9) + 8x = 24 8x = 24 x = 3 Check: 9 + 8x = 33 9 + 8(3) | 33 9 + 24 | 33 33 = 33 The solution is 3. Subtracting 9 from both sides Dividing both sides by 8 Example

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Slide 2- 12 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve: Adding 5 to both sides Multiplying both sides by 3/2. Simplifying Example

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Slide 2- 13 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve: 36 t = 17 Solution 36 t = 17 36 t 36 = 17 36 t = 19 ( 1) t = ( 19)( 1) t = 19 Check: 36 t = 17 36 19 | 17 17 = 17 The solution is 19. Subtracting 36 from both sides Multiplying both sides by 1 Example

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Slide 2- 14 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Combining Like Terms If like terms appear on the same side of an equation, we combine them and then solve. Should like terms appear on both sides of an equation, we can use the addition principle to rewrite all like terms on one side.

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Slide 2- 15 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve. 5x + 4x = 36 Solution5x + 4x = 36 9x = 36 x = 4 Check: 5x + 4x = 36 5(4) + 4(4) | 36 20 + 16 | 36 36 = 36 The solution is 4. Combining like terms Dividing both sides by 9 Simplifying Example

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Slide 2- 16 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve. 4x + 7 6x = 10 + 3x + 12 Solution: 4x + 7 6x = 10 + 3x + 12 2x + 7 = 22 + 3x 2x + 7 7 = 22 + 3x 7 2x = 15 + 3x 2x 3x = 15 + 3x 3x 5x = 15 x = 3 Combining like terms Subtracting 7 from both sides Simplifying Subtracting 3x from both sides Dividing both sides by 5 Example

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Slide 2- 17 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Check: 4x + 7 6x = 10 + 3x + 12 4( 3) + 7 6( 3) | 10 + 3( 3) + 12 12 + 7 + 18 | 10 9 + 12 13 = 13 The solution is 3.

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Slide 2- 18 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve. Solution: Example

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Slide 2- 19 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley An Equation-Solving Procedure 1. Use the multiplication principle to clear any fractions or decimals. (This is optional, but can ease computations. 2.If necessary, use the distributive law to remove parentheses. Then combine like terms on each side. 3.Use the addition principle, as needed, to isolate all variable terms on one side. Then combine like terms. 4.Multiply or divide to solve for the variable, using the multiplication principle. 5.Check all possible solutions in the original equation.

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Slide 1- 20 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Contradictions and Identities An identity is an equation that is true for all replacements that can be used on both sides of the equation. A contradiction is an equation that is never true. A conditional equation is sometimes true and sometimes false, depending on what the replacement is.

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Slide 1- 21 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The equation is true regardless of the choice for x, so all real numbers are solutions. The equation is an identity. Example Solution:

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Slide 1- 22 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The equation is false for any choice of x, so there is no solution for this equation. The equation is a contradiction. Example Solution:

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Slide 1- 23 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley There is one solution, 7. For all other choices of x, the equation is false. The equation is a conditional equation. Example Solution:

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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Formulas Evaluating Formulas Solving for a Variable 2.3

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Slide 2- 25 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Many applications of mathematics involve relationships among two or more quantities. An equation that represents such a relationship will use two or more letters and is known as a formula.

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Slide 2- 26 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The formula can be used to determine how far M, in miles, you are from lightening when its thunder takes t seconds to reach your ears. If it takes 5 seconds for the sound of thunder to reach you after you have seen the lightening, how far away is the storm? Solution We substitute 5 for t and calculate M. The storm is 1 mile away. Example

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Slide 2- 27 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Circumference of a circle. The formula C = d gives the circumference C of a circle with diameter d. Solve for d. Solution C = d d Example

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Slide 2- 28 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To Solve a Formula for a Given Variable 1. If the variable for which you are solving appears in a fraction, use the multiplication principle to clear fractions. 2. Isolate the term(s), with the variable for which you are solving on one side of the equation. 3. If two or more terms contain the variable for which you are solving, factor the variable out. 4. Multiply or divide to solve for the variable in question.

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Slide 2- 29 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve for y: x = wy + zy 6 Solution x = wy + zy 6 x = wy + zy 6 x + 6 = wy + zy x + 6 = y(w + z) We want this letter alone. Adding 6 to both sides Factoring Dividing both sides by w + z Example

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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Applications with Percent Converting Between Percent Notation and Decimal Notation Solving Percent Problems 2.4

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Slide 2- 31 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Percent Notation n% means

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Slide 2- 32 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Convert to decimal notation: a) 34% b) 7.6% Solution a) 34% = 34 0.01 = 0.34 b) 7.6% = 7.6 0.01 = 0.076 Example

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Slide 2- 33 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To convert from percent notation to decimal notation, move the decimal point two places to the left and drop the percent symbol. Convert the percent notation in the following sentence to decimal notation: Marta received a 75% on her first algebra test. Solution Move the decimal point two places to the left. Example

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Slide 2- 34 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To convert from decimal notation to percent notation, move the decimal point two places to the right and write a percent symbol. Convert to percent notation: a) 3.24b) 0.2c) Solution a) We first move the decimal point two places to the right: 3.24. and then write a % symbol: 324% Example

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Slide 2- 35 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued b) We first move the decimal point two places to the right (recall that 0.2 = 0.20): 0.20. and then write a % symbol: 20% c) Note that 3/8 = 0.375. We move the decimal point two places to the right: 0.37.5 and then write a % symbol: 37.5%

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Slide 2- 36 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Key Word in Percent Translations Of translates to or. What translates to a variable. Is or Was translates to =. % translates to

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Slide 2- 37 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What is 13% of 72? Solution Translate: Thus, 9.36 is 13% of 72. The answer is 9.36. Example

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Slide 2- 38 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 9 is 12 percent of what? Solution Translate: Example

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Slide 2- 39 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley What percent of 60 is 27? Solution Translate: Example

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Slide 2- 40 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To complete her water safety course instruction, Catherine must complete 45 hours of instruction. If she has completed 75% of her requirement, how many hours has Catherine completed? Solution Rewording: What is 75% of 45? Translating: a = 0.75 45 a = 33.75 Catherine has completed 33.75 hours of instruction. Example

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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Problem Solving Five Steps for Problem Solving Applying the Five Steps 2.5

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Slide 2- 42 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Five Steps for Problem Solving in Algebra 1. Familiarize yourself with the problem. 2. Translate to mathematical language. (This often means writing an equation.) 3. Carry out some mathematical manipulation. (This often means solving an equation.) 4. Check your possible answer in the original problem. 5. State the answer clearly, using a complete English sentence.

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Slide 2- 43 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To Become Familiar with a Problem 1. Read the problem carefully. Try to visualize the problem. 2. Reread the problem, perhaps aloud. Make sure you understand all important words. 3. List the information given and the question(s) to be answered. Choose a variable (or variables) to represent the unknown and specify what the variable represents. For example, let L = length in centimeters, d = distance in miles, and so on. 4.Look for similarities between the problem and other problems you have already solved.

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Slide 2- 44 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To Become Familiar with a Problem (continued) 5.Find more information. Look up a formula in a book, at the library, or online. Consult a reference librarian or an expert in the field. 6.Make a table that uses all the information you have available. Look for patterns that may help in the translation. 7.Make a drawing and label it with known and unknown information, using specific units if given. 8.Think of a possible answer and check the guess. Note the manner in which the guess is checked.

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Slide 2- 45 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The apartments in Wandas apartment house are consecutively numbered on each floor. The sum of her number and her next door neighbors number is 723. What are the two numbers? Solution 1. Familiarize. The apartment numbers are consecutive integers. Let x = Wandas apartment Let x + 1 = neighbors apartment Example

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Slide 2- 46 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate. Rewording: Translating: 3. Carry out. x + (x + 1) = 723 2x + 1 = 723 2x = 722 x = 361 If x is 361, then x + 1 is 362.

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Slide 2- 47 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4. Check. Our possible answers are 361 and 362. These are consecutive integers and the sum is 723. 5. State. The apartment numbers are 361 and 362.

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Slide 2- 48 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Digicon prints digital photos for $0.12 each plus $3.29 shipping and handling. Your weekly budget for the school yearbook is $22.00. How many prints can you have made if you have $22.00? Solution 1. Familiarize. Suppose the yearbook staff takes 220 digital photos. Then the cost to print them would be the shipping charge plus $0.12 times 220. $3.29 + $0.12(220) which is $29.69. Our guess of 220 is too large, but we have familiarized ourselves with the way in which the calculation is made. Example

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Slide 2- 49 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate. Rewording: Translating: 3. Carry out. 4. Check. Check in the original problem. $3.29 + 155(0.12) = $21.89, which is less than $22.00. 5. State. The yearbook staff can have 155 photos printed per week.

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Slide 2- 50 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley You are constructing a triangular kite. The second angle of the kite is three times as large as the first. The third angle is 10 degrees more than the first. Find the measure of each angle. Solution 1. Familiarize. Make a drawing and write in the given information. 2. Translate. To translate, we need to recall that the sum of the measures of the angles in a triangle is 180 degrees. 3x3x x x + 10 Example

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Slide 2- 51 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate (continued). 3. Carry out. The measures for the angles appear to be: first angle: x = 34 second angle: 3x = 3(34) = 102; third angle: x + 10 = 34 + 10 = 44

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Slide 2- 52 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4. Check. Consider 34, 102 and 44 degrees. The sum of these numbers is 180 degrees and the second angle is three times the first angle. The third angle is 10 degrees more than the first. These numbers check. 5. State. The measures of the angles are 34, 44 and 102 degrees.

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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Inequalities Solutions of Inequalities Graphs of Inequalities Set Builder and Interval Notation Solving Inequalities Using the Addition Principle Solving Inequalities Using the Multiplication Principle Using the Principles Together 2.6

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Slide 2- 54 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solutions of Inequalities An inequality is a number sentence containing > (is greater than), < (is less than), (is greater than or equal to), or (is less than or equal to). Determine whether the given number is a solution of x < 5: a) 4 b) 6 Solution a) Since 4 < 5 is true, 4 is a solution. b) Since 6 < 5 is false, 6 is not a solution. 6-3135-404-2-42 Example

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Slide 4- 55 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solutions of Inequalities An inequality is any sentence containing Any value for a variable that makes an inequality true is called a solution. The set of all solutions is called the solution set. When all solutions of an inequality are found, we say that we have solved the inequality. Examples:

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Slide 4- 56 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Determine whether 5 is a solution to We substitute to get 3(5) + 2 > 7, or 17 >7, a true statement. Thus, 5 is a solution. Example

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Slide 4- 57 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The graph of an inequality is a visual representation of the inequalitys solution set. An inequality in one variable can be graphed on a number line. Graph x < 2 on a number line. Solution Note that in set-builder notation the solution is Example 45 3 10 -5 -6 -4 -3 -2 2 6 )

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Slide 4- 58 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Interval Notation We can write solutions of an inequality in one variable using interval notation. Interval notation uses parentheses, ( ), and brackets, [ ]. If a and b are real numbers such that a < b, we define the open interval (a, b) as the set of all numbers x for which a < x < b. Thus, (a, b) a b )(

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Slide 4- 59 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Interval Notation The closed interval [a, b] is defined as the set of all numbers x for which Thus, a b [a, b] [ ]

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Slide 4- 60 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Interval Notation There are two types of half-open intervals, defined as follows: (a, b] a b ( ] [a, b) [)

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Slide 4- 61 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Interval Notation We use the symbols to represent positive and negative infinity, respectively. Thus the notation (a, ) represents the set of all real numbers greater than a, and (, a) represents the set of all numbers less than a. The notations (–, a] and [a, ) are used when we want to include the endpoint a. a a ( )

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Slide 2- 62 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Addition Principle for Inequalities For any real numbers a, b, and c: a < b is equivalent to a + c < b + c; a b is equivalent to a + c b + c; a > b is equivalent to a + c > b + c; a b is equivalent to a + c b + c.

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Slide 4- 63 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Solve and graph x – 2 > 7. x – 2 > 7 x – 2 + 2 > 7 + 2 x > 9 The solution set is Example 5 6 7 8 9 10 11 12 13 14 15 16 17 (

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Slide 2- 64 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve 4x 1 x 10 and then graph the solution. Solution 4x 1 x 10 4x 1 + 1 x 10 + 1 4x x 9 4x x x x 9 3x 9 x 3 The solution set is {x|x 3}. Adding 1 to both sides Dividing both sides by 3 Subtracting x from both sides Simplifying Example

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Slide 2- 65 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Multiplication Principle for Inequalities For any real numbers a and b, and for any positive number c: a < b is equivalent to ac < bc, and a > b is equivalent to ac > bc. For any real numbers a and b, and for any negative number c: a bc, and a > b is equivalent to ac < bc. Similar statements hold for and.

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Slide 2- 66 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve and graph each inequality: a) b) 4y < 20 Solution a) The solution set is {x|x 28}. The graph is shown below. 3015255201030 Multiplying both sides by 4 Simplifying Example ]

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Slide 2- 67 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley b) 4y < 20 The solution set is {y|y > 5}. The graph is shown below. At this step, we reverse the inequality, because 4 is negative. Dividing both sides by 4 4-5-313-6-22-4-60 (

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Slide 2- 68 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution 3x 3 > x + 7 3x 3 + 3 > x + 7 + 3 3x > x + 10 3x x > x x + 10 2x > 10 x > 5 The solution set is {x|x > 5}. Solve. 3x 3 > x + 7 Adding 3 to both sides Simplifying Subtracting x from both sides Simplifying Dividing both sides by 2 Simplifying 81357-20482 6 Example (

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Slide 2- 69 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution 15.4 3.2x < 6.76 100(15.4 3.2x) < 100( 6.76) 100(15.4) 100(3.2x) < 100( 6.76) 1540 320x < 676 320x < 676 1540 320x < 2216 x > 6.925 The solution set is {x|x > 6.925}. Solve. 15.4 3.2x < 6.76 Remember to reverse the symbol! Example

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Slide 2- 70 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution 5(x 3) 7x 4(x 3) + 9 5x 15 7x 4x 12 + 9 2x 15 4x 3 2x 15 + 3 4x 3 + 3 2x 12 4x 2x + 2x 12 4x + 2x 12 6x 2 x The solution set is {x|x 2}. Solve: 5(x 3) 7x 4(x 3) + 9 Using the distributive law to remove parentheses Simplifying Adding 2x to both sides Dividing both sides by 6 Adding 3 to both sides 2-7-5-31-8-6-22-4-80 Example ]

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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Applications with Inequalities Translating to Inequalities Solving Problems 2.7

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Slide 2- 72 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Important WordsSample SentenceTranslation is at leastBrian is at least 16 years old b 16 is at mostAt most 3 students failed the course s 3 cannot exceedTo qualify, earnings cannot exceed $5000 e $5000 must exceedThe speed must exceed 20 mphs > 20 is less thanNicholas is less than 60 lb.n < 60 is more thanChicago is more than 300 miles away. c > 300 is betweenThe movie is between 70 and 120 minutes. 70 < m < 120 no more thanThe calf weighs no more than 560 lb. w 560 no less thanCarmon scored no less than 9.4. c 9.4

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Slide 2- 73 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Translating at least and at most The quantity x is at least some amount q: x q. (If x is at least q, it cannot be less than q.) The quantity x is at most some amount q: x q. (If x is at most q, it cannot be more than q.)

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Slide 2- 74 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Lazer Line charges $65 plus $45 per hour for copier repair. Jonas remembers being billed less than $150. How many hours was Jonas copier worked on? Solution 1. Familiarize. Suppose the copier was worked on for 4 hours. The cost would be $65 + 4($45), or $245. This shows that the copier was worked on for less than 4 hours. Let h = the number of hours. 2. Translate. Rewording: Translating: Example

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Slide 2- 75 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3. Carry out. 65 + 45h 150 45h 85 4. Check. Since the time represents hours, we round down to one hour. If the copier was worked on for one hour, the cost would be $110, and if worked on for two hours the cost would exceed $150. 5. State. Jonas copier was worked on for less than two hours.

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Slide 2- 76 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Average, or Mean To find the average or mean of a set of numbers, add the numbers and then divide by the number of addends.

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Slide 2- 77 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Samantha has test grades of 86, 88, and 78 on her first three math tests. If she wants an average of at least 80 after the fourth test, what possible scores can she earn on the fourth test? Solution 1. Familiarize. Suppose she earned an 85 on her fourth test. Her test average would be This shows she could score an 85. Lets let x represent the fourth test score. Example

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Slide 2- 78 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate. Rewording: Translating: 3. Carry out.

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Slide 2- 79 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4. Check. As a partial check, we show that Samantha can earn a 68 on | the fourth test and average 80 for the four tests. 5. State. Samanthas test average will not drop below 80 if she earns at least a 68 on the fourth test.

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