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3.3 Solving Application Problems with One Unknown Quantity
Chapter 3 Solving Application Problems
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3.3 Solving Application Problems with One Unknown Quantity
Objectives Translate word phrases into algebraic expressions. Translate sentences into equations. Solve application problems with one unknown quantity. Copyright © 2010 Pearson Education, Inc. All rights reserved.
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Objective 1: Translate word phrases into algebraic expressions.
Write each phrase as an algebraic expression. Use x as the variable. Example Words Algebraic Expression A number plus 2 The sum of 8 and a number 5 more than a number A number increased by 6 9 less than a number A number subtracted from 3 A number decreased by 4 10 minus a number Copyright © 2010 Pearson Education, Inc. All rights reserved.
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Objective 1: Translate word phrases into algebraic expressions.
Write each phrase as an algebraic expression. Use x as the variable. Example Words Algebraic Expression 8 times a number The product of 12 and a number Double a number (meaning “2 times”) The quotient of –6 and a number A number divided by 10 15 subtracted from 4 times a number The result is Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec
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Objective 2: Translate sentences into equations.
If 5 times a number is added to 11, the result is 26. Find the number. Example Let x represent the unknown number. The number is 3. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec
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Objective 2: Translate sentences into equations.
(continued) Example Check We see that 5 • = = 26. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec
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Objective 3: Solve application problems with one unknown quantity.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec
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Objective 3: Solve application problems with one unknown quantity.
Heather put some money aside in an envelope for small household expenses. Yesterday she took out $20 for groceries. Today a friend paid back a loan and Heather put the $34 in the envelope. Now she has $43 in the envelope. How much was in the envelope at the start? Example Step 1 Read the problem: Unknown: amount of money in the envelope at the start Known: took out $20; put in $34; ended up with $43 Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec
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Objective 3: Solve application problems with one unknown quantity.
(continued) Example Step 2(a) Assign a variable: Let m represent the money at the start. Step 3 Write an equation: Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec
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Objective 3: Solve application problems with one unknown quantity.
(continued) Example Step 4 Solve the equation: Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec
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Objective 3: Solve application problems with one unknown quantity.
(continued) Example Step 5 State the answer: There was $29 in the envelope. Step 6 Check the answer: Started with $29 in the envelope Took out $20, so $29 – $20 = $9 in the envelope Put in $34, so $9 + $34 = $43 Now has $43 Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec
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Objective 3: Solve application problems with one unknown quantity.
Michael has completed 5 less than three times as many lab experiments as David. If Michael has completed 13 experiments, how many experiments has David completed? Example Step 1 Read the problem: Unknown: number of experiments David did Known: Michael did 5 less than 3 times the number David did; Michael did 13. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec
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Objective 3: Solve application problems with one unknown quantity.
(continued) Example Step 2(a) Assign a variable: Let n represent the number of experiments David did. Step 3 Write an equation: Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec
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Objective 3: Solve application problems with one unknown quantity.
(continued) Example Step 4 Solve the equation: Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec
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Objective 3: Solve application problems with one unknown quantity.
(continued) Example Step 5 State the answer: David did 6 experiments. Step 6 Check the answer: 3 times David’s number • 6 = 18 Less – 5 = 13 Michael did 13 Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec
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