Algebra 1

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

Chapter 6 - Systems of Equations and Inequalities - 6-3 Solving Systems Using Elimination - Practice and Problem-Solving Exercises: 39


1 hour of Horseback Riding costs: 30.00 dollars 1 hour of parasailing costs: 51.00 dollars.

Work Step by Step

First lets take a look at the two packages offered by the hotelL: 1. Cost= 192 dollars, 3 hours of horseback riding, 2 hours of parasailing 2. Cost= 213 dollars, 2 hours of horseback riding, 3 hours of parasailing Then lets substitute: - the cost as (c) - the cost of 1 hour of horseback riding as (r) - the cost of 1 hour of parasailing as (p) Now we can create the two equations/systems to represent the problem. The coefficient of the variables "r" and "p" represent the number of hours of each activity. 1. 3r+2p=192 2. 2r+3p=213 Then lets multiply the first equation by a scalar of 2 and the second equation by a scalar of 3. Now our two equations look like: 1. 6r+4p=384 2. 6r+9p=639 Now lets subtract the second equation from the first equation now that we have the same coefficient of r and would allow us to eliminate all of the r-variables. We get the equation: -5p=-255 Divide by -5 on both sides to isolate "p" p=51. Now that we have the cost of 1 hour of parasailing lets substitute into our first original equation to find the cost of horseback riding. 3r+2(51)=192 3r+102=192 Subtract 102 from both sides and also divide by 3 on both sides of the equation after. We get: r=30. So the cost of 1 hour of horseback riding is 30.00 dollars and the cost of 1 hour of parasailing is 51.00 dollars.
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