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- 1. Chapter 9Basic Algebra © 2010 Pearson Education, Inc. All rights reserved.
- 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. 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. 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. 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. Copyright © 2010 Pearson Education, Inc. All rights reserved. Slide 9.8- 6
- 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. 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. 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. 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. 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. 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. 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. 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. 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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