Solve the following equation and check your results:
$2y+\dfrac{5}{3}=\dfrac{26}{3}-y$

Hint: Solve the expression and find the value of y. To verify it, take the LHS and RHS of the given expression separately and put the value of y. Now, prove that LHS is equal to RHS.

Complete step-by-step answer:
We have been given an expression with only one variable ‘y’. We need to solve the expression and find the value of y.
We have been given the expression as,
$2y+\dfrac{5}{3}=\dfrac{26}{3}-y$ …………………………………..(1)
Now, let us rearrange the above expression and simplify it,
$2y+\dfrac{5}{3}=\dfrac{26}{3}-y$
$\Rightarrow 2y+y=\dfrac{26}{3}-\dfrac{5}{3}$
$3y=\dfrac{26-5}{3}\Rightarrow 3y=\dfrac{21}{3}$
$\therefore 3y=7$ i.e. $y=\dfrac{7}{3}$
Thus, we got the value of y as $\dfrac{7}{3}$ .
Now we need to check our result. Let us put $y=\dfrac{7}{3}$ in equation (1). First let us take the LHS of the equation and put $y=\dfrac{7}{3}$ .
$LHS=2y+\dfrac{5}{3}=2\times \dfrac{7}{3}+\dfrac{5}{3}$
$=\dfrac{14}{3}+\dfrac{5}{3}=\dfrac{19}{3}$
Thus, we got LHS $=\dfrac{19}{3}$ .
Now, let us take the RHS of the expression and put $y=\dfrac{7}{3}$
RHS $=\dfrac{26}{3}-y=\dfrac{26}{3}-\dfrac{7}{3}=\dfrac{19}{3}$
Therefore, we got RHS $=\dfrac{19}{3}$ .
Thus, we have checked our result and we got that $y=\dfrac{7}{3}$ is the ideal solution for the expression.

Note: You can also prove the LHS is equal to RHS in a few steps, rather than proving it both separately. You will get that LHS $=$ RHS $=\dfrac{19}{3}$ . Be careful while solving and carry out the arithmetic operation without mistake.