Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 6 - Systems of Equations and Inequalities - 6-4 Application of Linear Systems - Practice and Problem-Solving Exercises - Page 387: 22


We need 2.1 oz of Perfume A and 0.9 oz of Perfume B

Work Step by Step

We are told the prices per ounce of Perfume A and Perfume B. The price per ounce of Perfume A is 15.00 dollars per ounce and the prince per ounce of Perfume B is 35.00 dollars per ounce. We want to create a 3oz perfume with the cost of 63.00 dollars. The question asks us for the number of ounces of each perfume we need to make this possible. First, lets represent the ounces of Perfume A as the variable (a) and the ounces of Perfume B as the variable (b). The coefficient of these two variables will be the price per ounce of that perfume. With this, we can create two equations to represent our situation: 1. 15a+35b=63 (This represents the cost of the perfume) 2. a+b=3 (This symbolizes the weight of the perfume solution) Now lets solve this by multiplying the second equation by 15. We get the two equations: 1. 15a+35b=63 2. 15a+15b=45 Now lets eliminate the a-terms by subtracting the second equation from the first equation. We get: 15a+35b=63 - 15a+15b=45 ------------ 20b=18 (Divide both sides by 20) b=0.9 oz Now that we know the value of b, lets substitute this into our second equation to find the value of a. a+0.9=3 (Subtract 0.9 from both sides) a=2.1 So, we need 2.1 oz of Perfume A and 0.9 oz of Perfume B in order to create a perfume solution that weighs 3oz and costs 63.00 dollars.
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