Answer
$4\displaystyle \frac{1}{3}$ 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.
