How do you solve \[\dfrac{2}{3}n-\dfrac{2}{3}=\dfrac{n}{6}+\dfrac{4}{3}\] ?

Hint: These types of problems are linear in nature and are easy to solve. These are easy representations of linear equations which can be solved by analytical methods. First of all we need to understand the distributive property. The distributive property for a set of integers, say, \[a\] , \[b\] , \[c\] is defined as,
\[a\left( b+c \right)=a\cdot b+a\cdot c\] . Now using this property, we can very easily solve the above given problem.

Complete step by step answer:
Now starting off with the solution to the problem,
\[\begin{align}
  & \dfrac{2}{3}n-\dfrac{2}{3}=\dfrac{n}{6}+\dfrac{4}{3} \\
 & \Rightarrow \dfrac{2}{3}n-\dfrac{n}{6}=\dfrac{2}{3}+\dfrac{4}{3} \\
\end{align}\]
Now, taking \[n\] as common from the left hand side, we get,
\[n\left( \dfrac{2}{3}-\dfrac{1}{6} \right)=\dfrac{2}{3}+\dfrac{4}{3}\]
We now subtract the like terms of the equation on the left hand side and add the like terms on the right hand side we get,
\[n\left( \dfrac{3}{6} \right)=\dfrac{6}{3}\]
Now, cross multiplying both sides, we get,
\[n=\dfrac{6\times 6}{3\times 3}\]
We get from above that,
\[n=\dfrac{36}{9}\]
Now performing the necessary calculations we get,
\[n=4\]
Thus, we get the value of \[n\] from the above problem as \[4\] . Thus our answer is \[n=4\] .
We can easily cross check the answer by putting the value of \[n=4\] in the original given equation and observe that both the right hand side and left hand side come out to be equal. From this we can endure that we have solved the question correctly.

Note:
In context to the above problem, we must remember the theory of linear equations or else we won’t be able to solve the problem. This sum can also be solved with the aid of graphs. In such a case we assume the left hand side of the equation to be a function as well as the right hand side. Now after plotting the graph, we need to find the intersecting point for the required answer.