Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 1 - Section 1.2 - Fractions in Algebra - Exercise Set - Page 30: 101


$4\displaystyle \frac{1}{3}$ is not a solution.

Work Step by Step

For the given number to be a solution of the equation, the equation must hold true when we substitute the variable with the given number. First, convert the number to an improper fraction: $4\displaystyle \frac{1}{3}=\frac{4\times 3+1}{3}=\frac{13}{3}$ Now, substitute into the equation: $LHS=\displaystyle \frac{13}{3}\div 6+\frac{2}{3}$ ...dividing means multiplying with the reciprocal... $=\displaystyle \frac{13}{3}\times\frac{1}{6}+\frac{2}{3}$ $=\displaystyle \frac{13}{18}+\frac{2}{3}$ ... the LCD is 18... $=\displaystyle \frac{13}{18}+\frac{2\times 6}{3\times 6}$ $=\displaystyle \frac{13+12}{18}$ $=\displaystyle \frac{25}{18}$ $RHS=\displaystyle \frac{13}{3}\div 2+\frac{2}{3}$ ...dividing means multiplying with the reciprocal... $=\displaystyle \frac{13}{3}\times\frac{1}{2}+\frac{2}{3}$ $=\displaystyle \frac{13}{6}+\frac{2}{3}$ ... the LCD is $6$... $=\displaystyle \frac{13}{6}+\frac{2\times 2}{3\times 2}$ $=\displaystyle \frac{13+4}{6}$ $=\displaystyle \frac{17}{6}$ $LHS\neq RHS$ , so $4\displaystyle \frac{1}{3}$ is not a solution.
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