## Introductory Algebra for College Students (7th Edition)

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