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 29: 98


$1\displaystyle \frac{1}{3}$ is a solution.

Work Step by Step

For the given number to be a solution of the equation, it must hold true when we substitute the variable with the given number. Multiplication comes before addition, you can divide numerators and denominators by common factors before performing multiplication. Convert the given number to an improper fraction, and substitute into the equation. $1\displaystyle \frac{1}{3}=\frac{4}{3}$ LHS =$ \displaystyle \frac{2}{3}\cdot\frac{4}{3}+\frac{5}{6}\cdot\frac{4}{3}$ ... reduce,,, $= \displaystyle \frac{2}{3 }\cdot\frac{4}{3}+\frac{5}{6\div 2}\cdot\frac{4\div 2}{3}$ $= \displaystyle \frac{8}{9}+\frac{5\cdot 2}{3\cdot 3}$ $= \displaystyle \frac{8}{9}+\frac{10}{9}$ $=\displaystyle \frac{18}{9}$ $=\displaystyle \frac{18\div 9}{9\div 9}=2$ LHS = RHS, so $1\displaystyle \frac{1}{3}$ is a solution.
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