College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises - Page 139: 79

Answer

The original number of members is $50$ person

Work Step by Step

$\textbf{i}\hspace{0.3cm}$ If the original number of people is $x$. So the cost per person is $900 / x$. $\textbf{ii}\hspace{0.3cm}$ Let the new number of people is $x-5$. So cost per person is $900 / (x-5)$ . $\textbf{iii}\hspace{0.3cm}$ Hence the cost per person after $5$ people withdrew is $\$2$ more than the original cost $\Rightarrow 900 / (x - 5) = (900 / x ) + 2$ $\Rightarrow x(x-5)\bigg(900 / (x - 5)\bigg) = x(x-5)\bigg(900 / x \bigg) + 2x(x-5)$ $\Rightarrow 900x = 900(x-5) + 2x(x-5)$ $\Rightarrow 900x = 900x-4500 + 2x^2-10x$ $\Rightarrow 2x^2-10x-4500 =0$ $\Rightarrow x^2-5x-2250 =0$ $\Rightarrow (x+45)(x-50) =0$ $\Rightarrow (x+45) =0$ or $(x-50) =0$ $\Rightarrow x=-45 \hspace{0.3cm}{\color{red}{reject}}$ or $x=50$ i.e The original number of members is $50$ person
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