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

$2\displaystyle \frac{2}{3}$ is 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: $2\displaystyle \frac{2}{3}=\frac{2\times 3+2}{3}=\frac{8}{3}$ Now, substitute into the equation: $LHS=\displaystyle \frac{8}{3}\div 6+\frac{1}{3}$ ...dividing means multiplying with the reciprocal... $=\displaystyle \frac{8}{3}\times\frac{1}{6}+\frac{1}{3}$ $=\displaystyle \frac{8}{18}+\frac{1}{3}$ ... the LCD is 18... $=\displaystyle \frac{8}{18}+\frac{1\times 6}{3\times 6}$ $=\displaystyle \frac{14\div 2}{18\div 2}$ $=\displaystyle \frac{7}{9}$ $RHS=\displaystyle \frac{8}{3}\div 2-\frac{5}{9}$ ...dividing means multiplying with the reciprocal... $=\displaystyle \frac{8}{3}\times\frac{1}{2}-\frac{5}{9}$ ...reduce the product by 2... $=\displaystyle \frac{4}{3}-\frac{5}{9}$ ... the LCD is 9... $=\displaystyle \frac{4\times 3}{3\times 3}-\frac{5}{9}$ $=\displaystyle \frac{12-5}{9}$ $=\displaystyle \frac{7}{9}$ $LHS$ = $RHS$ , so $2\displaystyle \frac{2}{3}$ is a solution.