## Elementary Algebra

Published by Cengage Learning

# Chapter 10 - Quadratic Equations - Chapter 10 Review Problem Set - Page 468: 28

#### Answer

He bought 40 shares at 20 dollars per share.

#### Work Step by Step

We call s the original number of shares bought and p the original price they were bought at. This allows us to create a system of linear equations: $sp = 720 \\ 800 = (s-20)(p+8)$ Subbing in $p = \frac{720}{s}$ we find: $800 = sp -20p + 8s -160 \\ 800 = 720 -(\frac{20 \times720}{s}) + 8s - 160 \\ 0 = -240 - \frac{14400}{s} + 8s \\ 0 = \frac{-240s}{s} - \frac{14400}{s} + \frac{8s^2}{s} \\ 0 = 8s^2 - 240s - 14400 \\ 8(s-60)(s+30) \\s = 60$ Since 20 shares were not sold, this means that he sold 40 shares. Also: $p = \frac{720}{60} = 12$ Since they increased in price by 8 dollars, this means they were sold at 20 dollars per share.

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