Solve the following equation and check your result.
\[2y + \dfrac{5}{3} = \dfrac{{26}}{3} - y\]
\[A. y = \dfrac{9}{2}\]
\[B. y = \dfrac{7}{3}\]
\[C. y = \dfrac{7}{2}\]
\[D. y = \dfrac{1}{2}\]

Hint: Here we have a linear equation with a variable ‘y’. Here we need to solve for ‘y’. We can solve this using the transposition method. The common transposition method is to do the same thing (mathematically) to both sides of the equation, with the aim of bringing like terms together and isolating the variable (or the unknown quantity). That is, we group the ‘y’ terms on one side and constants on the other side of the equation.

Complete step-by-step solution:
Given, \[2y + \dfrac{5}{3} = \dfrac{{26}}{3} - y\].
We transpose \[\dfrac{5}{3}\] which is present in the left-hand side of the equation to the right-hand side of the equation by subtracting \[\dfrac{5}{3}\]on the right-hand side of the equation.
\[2y = \dfrac{{26}}{3} - y - \dfrac{5}{3}\]
We transpose ‘y’ to the right-hand side of the equation by adding ‘y’ on the right-hand side,
\[2y + y = \dfrac{{26}}{3} - \dfrac{5}{3}\]
Thus, we have separated the variable ‘y’ on one side of the equation and constant on the other side of the equation.
\[3y = \dfrac{{26}}{3} - \dfrac{5}{3}\]
Now taking the LCM and simplifying we have,
\[3y = \dfrac{{26 - 5}}{3}\]
\[\Rightarrow 3y = \dfrac{{21}}{3}\]
or
\[y = \dfrac{{21}}{3} \times \dfrac{1}{3}\]
\[\Rightarrow y = \dfrac{{21}}{9}\]
\[ \Rightarrow y = \dfrac{7}{3}\].
Hence the required answer is option (b).

Note: We can solve this by substituting the given options in the given problem. After simplification if we LHS is equal to RHS. Then our obtained result is correct. But it will be difficult to do and it consumes time.
In the above, we did the transpose of addition and subtraction. Similarly, if we have multiplication, we use division to transpose. If we have division, we use multiplication to transpose. Follow the same procedure for these kinds of problems.