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Chapter 9Basic Algebra   © 2010 Pearson Education, Inc.   All rights reserved.
9.8 Using Equations to Solve Application Problems                             Objectives 1. Translate word phrases into ex...
ParallelExample 1          Translating Word Phrases into          Expressions with Variables Write each word phrase in sym...
ParallelExample 2          Translating Word Phrases into            Expressions with Variables Write each word phrase in s...
ParallelExample 2          Translating a Sentence into an           EquationIf 8 times a number is added to 13, the result...
Copyright © 2010 Pearson Education, Inc. All rights reserved.   Slide 9.8- 6
ParallelExample 5            Solving an Application Problem with            One Unknown   Frankie has washed 6 less than t...
ParallelExample 5                Solving an Application Problem withcontinued       One Unknown   Frankie has washed 6 les...
ParallelExample 5                Solving an Application Problem withcontinued       One Unknown   Frankie has washed 6 les...
ParallelExample 6            Solving an Application Problem with            Two Unknowns   On a shopping spree, Yoshi spen...
ParallelExample 6                 Solving an Application Problem withcontinued        Two Unknowns On a shopping spree, Yo...
ParallelExample 6            Solving an Application Problem withcontinued   One UnknownStep 5 State the answer. The amount...
ParallelExample 7            Solving a Geometry Application            Problem   The length of a rectangle is 3 inches mor...
ParallelExample 7                   Solving a Geometry Applicationcontinued          ProblemThe length of a rectangle is 3...
ParallelExample 7            Solving a Geometry Applicationcontinued   ProblemStep 5 State the answer.       The width is ...
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Lesson 9.8 (word problems)

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Transcript of "Lesson 9.8 (word problems)"

  1. 1. Chapter 9Basic Algebra © 2010 Pearson Education, Inc. All rights reserved.
  2. 2. 9.8 Using Equations to Solve Application Problems Objectives 1. Translate word phrases into expressions with variables. 2. Translate sentences into equations. 3. Solve application problems. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 2
  3. 3. ParallelExample 1 Translating Word Phrases into Expressions with Variables Write each word phrase in symbols, using x as the variable. Words Algebraic Expression A number plus nine x + 9 or 9 + x 7 more than a number x + 7 or 7 + x −12 added to a number −12 + x or x + (−12) 3 less than a number x–3 A number decreased by 1 x–1 14 minus a number 14 – x Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 3
  4. 4. ParallelExample 2 Translating Word Phrases into Expressions with Variables Write each word phrase in symbols, using x as the variable. Words Algebraic Expression 3 times a number 3x Twice a number 2x The quotient of 8 and a 8 number x A number divided by 15 x 15 The result is = Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 4
  5. 5. ParallelExample 2 Translating a Sentence into an EquationIf 8 times a number is added to 13, the result is 45.Find the number.Let x represent the unknown number. 8 times a number added to 13 is 45 8x + 13 = 45 Next, solve the equation. Check: 8x + 13 − 13 = 45 − 13 8x + 13 = 45 8x = 32 8(4) + 13 = 45 8 x 32 = 45 = 45 8 8 x=4 The solution is 4. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 5
  6. 6. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 6
  7. 7. ParallelExample 5 Solving an Application Problem with One Unknown Frankie has washed 6 less than twice as many windows washed as Rita. If Frankie has washed 14 windows, how many windows has Rita washed?Step 1 Read. The problem asks for the number of windows that Rita has washed.Step 2 Assign a variable. There is only one unknown, Rita’s number of windows washed. Step 3 Write an equation. The number Frankie 6 less than twice washed. Rita’s number. 14 = 2x – 6 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 7
  8. 8. ParallelExample 5 Solving an Application Problem withcontinued One Unknown Frankie has washed 6 less than twice as many windows washed as Rita. If Frankie has washed 14 windows, how many windows has Rita washed?Step 4 Solve. 14 = 2x – 6 14 + 6 = 2x – 6 + 6 20 = 2x 20 2 ×x = 2 2 10 = xStep 5 State the answer. Rita washed 20windows. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 8
  9. 9. ParallelExample 5 Solving an Application Problem withcontinued One Unknown Frankie has washed 6 less than twice as many windows washed as Rita. If Frankie has washed 14 windows, how many windows has Rita washed?Step 6 Check. 14 = 2x – 6 14 = 2(10) – 6 14 = 14 So 10 is the correct solution because it “works” in the original problem. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 9
  10. 10. ParallelExample 6 Solving an Application Problem with Two Unknowns On a shopping spree, Yoshi spent $54 more than Lowell. The total spent by them both was $276. Find the amount spent by each person.Step 1 Read. The problem asks for the amount spent by each person.Step 2 Assign a variable. There are two unknowns. Let x represent the amounts spent by Lowell and x + 54 be the amount spent by Yoshi.Step 3 Write an equation. Amount spent by Amount spent Yoshi. by Lowell x + x + 54 = 276 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 10
  11. 11. ParallelExample 6 Solving an Application Problem withcontinued Two Unknowns On a shopping spree, Yoshi spent $54 more than Lowell. The total spent by them both was $276. Find the amount spent by each person.Step 4 Solve. x + x + 54 = 276 2x + 54 = 276 2x + 54 − 54 = 276 − 54 2x = 222 1 2 x 222 = 2 2 1 x = 111 Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 11
  12. 12. ParallelExample 6 Solving an Application Problem withcontinued One UnknownStep 5 State the answer. The amount Lowellspent is x, so Lowell spent $111. The amount Yoshispent is x + 54, so Yoshi spent $165.Step 6 Check. Use the words in the original problem. Yoshi’s $165 is $54 more dollars than Lowell’s $111, so that checks. The total spent is $111 + $165 = $276 which also checks. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 12
  13. 13. ParallelExample 7 Solving a Geometry Application Problem The length of a rectangle is 3 inches more than the width. The perimeter is 78 inches. Find the length and width.Step 1 Read. The problem asks for the length and width of the rectangle.Step 2 Assign a variable. There are two unknowns, length and width. Let x represent the width and x + 3 represent the length.Step 3 Write an equation. Use the formula for P = 2l + 2w perimeter of a rectangle. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 13
  14. 14. ParallelExample 7 Solving a Geometry Applicationcontinued ProblemThe length of a rectangle is 3 inches more than the width. The perimeter is 78inches. Find the length and width.Step 4 Solve. P = 2l + 2w 78 = 2(x + 3) + 2 ∙ x 78 = 2x + 6 + 2x 78 = 4x + 6 78 – 6 = 4x + 6 – 6 72 = 4x 72 14 ×x = 4 41 18 = x Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 14
  15. 15. ParallelExample 7 Solving a Geometry Applicationcontinued ProblemStep 5 State the answer. The width is x, so the width is 18 inches. The length is x + 3, so the length is 21 inches.Step 6 Check. Use the words in the original problem. The original problem says that the perimeter is 78 inches. P = 2 ∙ 18 in. + 2 ∙21in. P = 36 in. + 42 in. P = 78 in. checks Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 15
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