Solve for x:
\[3x-\dfrac{{x - 2}}{3} = 4-\dfrac{{x{\text{ }}-{\text{ }}1}}{4}{\text{ }}\]

Hint: Linear equation of one variable contains one variable and constants. Variables are kept on one side and then simplified in order to solve the question. It should be done step by step to avoid mistakes. In an equation we could add, subtract, multiply or divide the same value on both sides, it does not change the value or result of the equation.

Complete step-by-step answer:
We are given a linear equation in one variable and we have to find the value of x.
The given equation is:
\[3x-\dfrac{{x - 2}}{3} = 4-\dfrac{{x{\text{ }}-{\text{ }}1}}{4}{\text{ }}\]
Taking LCM both sides we get,
\[ \Rightarrow \dfrac{{3 \times 3x-\left( {x - 2} \right)}}{3} = \dfrac{{4 \times 4-\left( {x-1} \right){\text{ }}}}{4}\]
Simplifying Numerator both sides
\[ \Rightarrow \dfrac{{9x-\left( {x - 2} \right)}}{3} = \dfrac{{16-\left( {x{\text{ }}-1} \right)}}{4}\]
Again opening bracket and simplifying (as sign of numbers under bracket will change in case of subtraction)
\[ \Rightarrow \dfrac{{9x-x + 2}}{3} = \dfrac{{16-x + 1}}{4}\]
\[ \Rightarrow \dfrac{{8x + 2}}{3} = \dfrac{{17-x}}{4}\]
For further we have to cross multiply to get the equation in one line and so we get,
\[ \Rightarrow \left( {8x + 2} \right) \times 4{\text{ }} = {\text{ }}\left( {17 - x} \right) \times 3\]
Multiplying on opening the bracket we get,
\[ \Rightarrow 32x + 8{\text{ }} = {\text{ 51}} - 3x\]
Now we have to keep variable in one side and constant in other side, we do it by changing the sign of the variable or constant whose side we are going to change
So,
\[ \Rightarrow 32x + 3x{\text{ }} = {\text{ 51}} - 8\]
Simplifying this we get
\[ \Rightarrow 35x{\text{ }} = {\text{ 43}}\]
Now dividing 35 both sides of the equation to get the value of \[x\]
\[ \Rightarrow x = \dfrac{{{\text{43}}}}{{35}}\]
Hence the value of \[x\] is \[\dfrac{{43}}{{35}}\]
So, the correct answer is “ \[\dfrac{{43}}{{35}}\]”.

Note: BODMAS rule is must for any arithmetic operation. The number of equations needed to solve must be equal to the number of variables to be obtained. Here it is a linear equation in one variable so one equation is sufficient to solve the given equation. Similarly two equations are needed to solve linear equations containing two variables.