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**Hint**: In order to solve this type of linear equation in one variable, first send all the constants to the right hand side and variables to the left hand side of the equation with the help of algebraic operations and then divide both the sides with the coefficient of the variable to get the desired solution for the equation.

**:**

__Complete step-by-step answer__To solve the given equation $ 3x - \dfrac{1}{3} = 5 $ , we will first send all the variables to the left hand side (L.H.S.) of the equation and constants to the right hand side of the equation (R.H.S.), so we can see that in the given equation $ 3x - \dfrac{1}{3} = 5 $ , we have to send only one constant from the left hand side to the right hand side, for this we will add both sides $ \dfrac{1}{3} $ , we will get

$

\Rightarrow 3x - \dfrac{1}{3} = 5 \\

\Rightarrow 3x - \dfrac{1}{3} + \dfrac{1}{3} = 5 + \dfrac{1}{3} \\

\Rightarrow 3x = 5 + \dfrac{1}{3} \;

$

Now taking L.C.M. in order to add \[5\;{\text{and}}\;\dfrac{1}{3}\]

$

\Rightarrow 3x = 5 + \dfrac{1}{3} \\

\Rightarrow 3x = \dfrac{{5 \times 3 + 1}}{3} \\

\Rightarrow 3x = \dfrac{{16}}{3} \;

$

Dividing both sides with the coefficient of $ x $ that is $ 3 $ to get the value for $ x $

$

\Rightarrow 3x = \dfrac{{16}}{3} \\

\Rightarrow \dfrac{{3x}}{3} = \dfrac{{16}}{{3 \times 3}} \\

\Rightarrow x = \dfrac{{16}}{9} \;

$

Therefore $ x = \dfrac{{16}}{9} $ is the required solution for the equation $ 3x - \dfrac{1}{3} = 5 $

**So, the correct answer is “ $ x = \dfrac{{16}}{9} $ ”.**

**Note**: The final result is in improper fraction, which means the numerical value of the numerator is greater than the numerical value of the denominator. So either convert the result into mixed fraction or write it in decimal form with the help of long division method.

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