Intermediate Algebra for College Students (7th Edition)

Published by Pearson

Chapter 3 - Section 3.1 - Systems of Linear Equations in Two Variables - Exercise Set - Page 193: 97

Answer

a) Sold = 150 and supplied =300 b) Sold = 250 and supplied =250

Work Step by Step

a) Since, we have $N= -5P+750$ and $N= 2.5P$ Thus, when $P=120$ Then $N= -5(120)+750=150$ and $N= 2.5(120)=300$ Therefore, Sold = 150 and supplied =300 b) Since, we have $N= -5P+750$ and $N= 2.5P$ Thus, when $-5P+750= 2.5P$ or, $P=\dfrac{750}{7.5}=100$ Now, when $P=100$ then $N= -5(100)+750=250$ and $N= 2.5(100)=250$ Therefore, Sold = 250 and supplied =250 Hence, a) Sold = 150 and supplied =300 b) Sold = 250 and supplied =250

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