Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.7 Solving Quadratic Equations by Factoring - 5.7 Exercise Set - Page 351: 20


$x=\left\{ \dfrac{10}{3},20 \right\}$

Work Step by Step

Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (1-0.05x)(1-0.3x)=0 ,$ is \begin{array}{l}\require{cancel} 1-0.05x=0 \\\\ -0.05x=-1 \\\\ x=\dfrac{-1}{-0.05} \\\\ x=20 ,\\\\\text{OR}\\\\ 1-0.3x=0 \\\\ -0.3x=-1 \\\\ x=\dfrac{-1}{-0.3} \\\\ x=\dfrac{10}{3} .\end{array} Hence, $ x=\left\{ \dfrac{10}{3},20 \right\} .$
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