Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 11 - Additional Topics - 11.2 - 3 x 3 Systems of Equations - Problem Set 11.2: 23

Answer

Peaches cost 1.29 per pound, pears cost .69 dollars per pound, and cherries cost .99 dollars per pound.

Work Step by Step

Let g be the cost per pound of peaches, o be the cost per pound of pears, and c be the cost per pound of cherries. Thus: $ 2g + c + 3o = 5.64 \\ g + 2c + 2o = 4.65 \\ 2g + 4c + o = 7.23$ In order to solve systems of three linear equations, we multiply the first and second equation by values that will cancel out a variable when they are added. Thus, we multiply the first equation by -2 and the second equation by 1 and add to obtain: $-3g -4o = -6.63$ We now multiply the first and third equation by values that will cancel out a variable when they are added. Thus, we multiply the first equation by -4 and the third equation by one and add to obtain: $ -6g - 11o = -15.33$ Plugging $-3g = -6.63 + 4o$ into this equation, we obtain: $-3o -13.26 = -15.33 \\ o = .69 $ Now, we plug this value into one of the equations that only has o and g in them to find: $ g = 1.29$ Finally, we plug the values of g and o into the first equation listed to find: $ c = .99$
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