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CHAPTER 2 Solving Equations and Inequalities Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 2.1Solving Equations: The Addition Principle 2.2Solving Equations: The Multiplication Principle 2.3Using the Principles Together 2.4Formulas 2.5Applications of Percent 2.6Applications and Problem Solving 2.7Solving Inequalities 2.8Applications and Problem Solving with Inequalities

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OBJECTIVES 2.6 Applications and Problem Solving Slide 3Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. aSolve applied problems by translating to equations.

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1. Familiarize yourself with the problem situation. 2. Translate the problem to an equation. 3. Solve the equation. 4. Check the answer in the original problem. 5. State the answer to the problem clearly. 2.6 Applications and Problem Solving Five Steps for Problem Solving in Algebra Slide 4Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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If a problem is given in words, read it carefully. Reread the problem, perhaps aloud. Try to verbalize the problem as if you were explaining it to someone else. Choose a variable (or variables) to represent the unknown and clearly state what the variable represents. Be descriptive! Make sure you understand all important words. For example, let L = length in centimeters, d = distance in miles, and so on. 2.6 Applications and Problem Solving To Familiarize Yourself with a Problem (continued) Slide 5Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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Make a drawing and label it with known information, using specific units if given. Also, indicate the unknown information. 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. 2.6 Applications and Problem Solving To Familiarize Yourself with a Problem (continued) Slide 6Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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Create a table that lists all the information you have available. Look for patterns that may help in the translation to an equation. Think of a possible answer and check the guess. Note the manner in which the guess is checked. 2.6 Applications and Problem Solving To Familiarize Yourself with a Problem Slide 7Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE Solution 1. Familiarize. Make a drawing. Noting the lengths. 3x3x x 480 in 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. AA 480-in. piece of pipe is cut into two pieces. One piece is three times the length of the other. Find the length of each piece of pipe. (continued) Slide 8Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 3. Solve. x + 3x = 480 4x = 4804 x = 120 inches 2. Translate. From the statement of the problem. One piece is three times the length of the other the total is 480 inches. x + 3x = Applications and Problem Solving a Solve applied problems by translating to equations. ASolving applied problems. (continued) Slide 9Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 4. Check. Do we have an answer to the problem? No, we need the lengths of both pieces of pipe. If x = 120 the length of one piece 3x = the length of the other piece. 3(120) = 360 inches Since = 480 our answer checks. 5. State. One section of pipe is 120 inches and the other section is 360 inches. 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. ASolving applied problems. Slide 10Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. BDigicon prints digital photos for $0.12 each plus $3.29 shipping and handling. Your weekly budget for the school yearbook is $ How many prints can you have made if you have $22.00? (continued) Slide 11Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 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. $ $0.12(220) which is $ Our guess of 220 is too large, but we have familiarized ourselves with the way in which the calculation is made. 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. BSolving applied problems. (continued) Slide 12Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 2. Translate. Rewording: Translating: 3. Carry out. 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. BSolving applied problems. (continued) Slide 13Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 4. Check. Check in the original problem. $ (0.12) = $21.89, which is less than $ State. The yearbook staff can have 155 photos printed per week. 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. BSolving applied problems. Slide 14Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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Consecutive integers: 16, 17, 18, 19, 20; and so on x, x + 1, x + 2; and so on. 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. Slide 15Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. Consecutive odd integers: 21, 23, 25, 27, 29, and so on x, x + 2, x + 4, and so on. Consecutive even integers: 16, 18, 20, 22, and so on x, x + 2, x + 4, and so on.

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EXAMPLE Solution 1. Familiarize. The apartment numbers are consecutive integers. Let x = Wandas apartment Let x + 1 = neighbors apartment 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. CThe 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? (continued) Slide 16Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 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 Applications and Problem Solving a Solve applied problems by translating to equations. CConsecutive integers. (continued) Slide 17Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 4. Check. Our possible answers are 361 and 362. These are consecutive integers and the sum is State. The apartment numbers are 361 and Applications and Problem Solving a Solve applied problems by translating to equations. CConsecutive integers. Slide 18Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. DYou 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. (continued) Slide 19Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 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 Applications and Problem Solving a Solve applied problems by translating to equations. DSolving applied problems. (continued) Slide 20Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 2. Translate (continued). 3. Carry out. 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. DSolving applied problems. (continued) Slide 21Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 2. Translate (continued). 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. DSolving applied problems. (continued) Slide 22Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. The measures for the angles appear to be: If x = 34: second angle: 3x = 3(34) = 102; first angle: x = 34 third angle: x + 10 = = 44

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EXAMPLE 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. 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. DSolving applied problems. Slide 23Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE 2.6 Applications and Problem Solving a Solve applied problems by translating to equations. EThe Hicksons are planning to sell their home. If they want to be left with $142,00 after paying 6% of the selling price to a realtor as a commission, for how much must they sell the house? (continued) Slide 24Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE Familiarize. Suppose the Hicksons sell the house for $150,000. A 6% commission can be determined by finding 6% of 150,000. 6% of 150,000 = 0.06(150,000) = Subtracting this commission from $150,000 would leave the Hicksons with 150,000 – 9,000 = 141, Applications and Problem Solving a Solve applied problems by translating to equations. EApplications of equations. (continued) Slide 25Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE Translate. Reword the problem and translate. Selling price less Commission is Amount remaining x – 0.06x = 142, Applications and Problem Solving a Solve applied problems by translating to equations. EApplications of equations. Slide 26Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE Solve. x – 0.06x = 142,000 1x – 0.06x = 142, x = 142, x = 151, Applications and Problem Solving a Solve applied problems by translating to equations. EApplications of equations. (continued) Slide 27Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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EXAMPLE Check. Find 6% of 151,063.83: 6% of 151, = (0.06)(151,063.83) = Subtract the commission to find the remaining amount. 151, – = 142,000. State. To be left with $142,000 the Hicksons must sell the house for $151, Applications and Problem Solving a Solve applied problems by translating to equations. EApplications of equations. Slide 28Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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