Solve the following equation:
\[\dfrac{x-2}{4}+\dfrac{1}{3}=x-\dfrac{2x-1}{3}\]

Hint: We solve this problem by rearranging the terms in the given equation such that we get all the variable terms on one side and the remaining terms on the other side. Then, we find the value of \['x'\] which will be the solution of the given equation. After getting the solution of the given equation we check it by substituting the solution in the given equation to check whether our solution is correct or wrong.

Complete step-by-step solution
We are given that the equation as
\[\dfrac{x-2}{4}+\dfrac{1}{3}=x-\dfrac{2x-1}{3}\]
Now, let us separate the terms in the fractions in the above equation then we get
\[\Rightarrow \dfrac{x}{4}-\dfrac{1}{2}+\dfrac{1}{3}=x-\dfrac{2x}{3}+\dfrac{1}{3}\]
Now, let us rearrange the terms in such a way that all the variable terms gets one side and the remaining terms gets other side and adding the terms using the LCM then we get
\[\begin{align}
  & \Rightarrow \dfrac{x}{4}-x+\dfrac{2x}{3}=\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{3} \\
 & \Rightarrow \dfrac{3x-12x+8x}{12}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{-x}{12}=\dfrac{1}{2} \\
\end{align}\]
Now, by cross multiplying the terms from LHS to RHS we get
\[\Rightarrow x=-6\]
Therefore, the solution of given equation is \[x=-6\]

Note: Students may sometimes get the solutions of equations even though they do not satisfy the given equation. But, we need to give the solution that satisfies the given equation.
So, for finding the solution that satisfy the given equation we need to check whether the solution we got is correct or wrong by substituting the solution we got in the given equation as we did above that is
Now by substituting the value \[x=-6\] in the given equation we get
\[\begin{align}
  & \Rightarrow \dfrac{-6-2}{4}+\dfrac{1}{3}=-6-\dfrac{-12-1}{3} \\
 & \Rightarrow -2+\dfrac{1}{3}=-6+\dfrac{13}{3} \\
 & \Rightarrow \dfrac{-5}{3}=\dfrac{-5}{3} \\
\end{align}\]
Here, we can see that both LHS and RHS are equal to each other.
So, we can say that the solution we got that \[x=-6\] is correct.