Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 9 - Section 9.1 - Solving Systems of Linear Equations in Three Variables and Problem Solving - Exercise Set - Page 640: 39

Answer

88 three pointers, 590 free throws, 766 field goals

Work Step by Step

2386 points free throws: 7*3P -26 3 pointers: field goals (2 points each): 176 +FT $F = 7P-26$ $G = 176+F$ P = 3 pointers F = free throws $2386 = (7P-26)*1+2*(176+F)+3P$ $2386=7P-26+2*(176+7P-26)+3P$ $2386=7P-26+2*(150+7P)+3P$ $2386=7P-26+300+14P+3P$ $2386=24P +274$ $2112 = 24P$ $2112/24 = 24P/24$ $P=88$ $F=7*88-26$ $F=616-26$ $F=590$ $G=176+F$ $G=176+590$ $G=766$
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