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Solving Word Problems Involving Cans and Bottles with Equations

This document provides a focused approach to solving word problems involving cans and bottles using algebraic equations. It illustrates how to express relationships between the quantities of cans and bottles through mathematical expressions and to formulate relevant equations based on given conditions. Examples include determining the total money earned from cans and bottles and calculating their respective quantities based on the described parameters. The solution process is step-by-step, enabling better understanding and application of algebra to real-world scenarios.

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Solving Word Problems Involving Cans and Bottles with Equations

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  1. VocabularyHomework 6-27 Three less than n n-3 Three times as many as n 3n One more than n n+1 One more than twice as many as n 2n+1

  2. Example One Can money plus botttle money is 218 cents 10Can plus 12botttle is 218 10C + 12B = 218 The number of cans is one more than four times the number of bottles The number of cans is one more than four times the number of bottles She gets 10 cents for each can and 12 cents for each bottle The total money she got was $2.18 52B = 208 C = 4B + 1 (4B + 1) • ÷52 ÷52 10 C + 12B = 218 B = 4 40B + 10 + 12B = 218 4 bottles C = 4B + 1 52B + 10 = 218 C = 4(4) + 1 -10 -10 C = 16 + 1 = 17 cans 52B = 208

  3. Classwork One How many cans and bottles? The number of cans is three more than two times the number of bottles She gets 6 cents for each can and 4 cents for each bottle The total money she got was $1.24 The number of cans is two more than five times the number of bottles She gets 8 cents for each can and 14 cents for each bottle The total money she got was $4.18 The number of cans is five more than six times the number of bottles • Cans wiegh15 onces each and bottles wiegh 95 onces each • The total wieght was 405

  4. Classwork Two 6-23

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