Answer
$2\displaystyle \frac{2}{3}$ is 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:
$2\displaystyle \frac{2}{3}=\frac{2\times 3+2}{3}=\frac{8}{3}$
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.
