How do you solve $\dfrac{2}{3}x+1=\dfrac{5}{6}x+\dfrac{3}{4}$ ?

Hint: First take the l.c.m as ‘3’ on the left hand side of the equation and ‘12’ on the left hand side of the equation. Cancel out 12 from the denominator of both the sides by multiplying and dividing ‘4’ on the left hand side and making the denominator 12. Then separate the variable ‘x’ and constants on the different sides of the equation. Do the necessary calculations and solve for ‘x’ to get the required solution.

Complete step-by-step answer:
Solving the equation means we have to find the value of ‘x’ for which the equation gets satisfied.
Considering the given equation $\dfrac{2}{3}x+1=\dfrac{5}{6}x+\dfrac{3}{4}$
Taking the l.c.m. as‘3’ on the left hand side of the equation and ‘12’ on the left hand side of the equation, we get
$\dfrac{2x+3}{3}=\dfrac{10x+9}{12}$
Multiplying and dividing by ‘4’ on both the left hand side of the equation, we get
$\begin{align}
  & \Rightarrow \dfrac{4\left( 2x+3 \right)}{3\times 4}=\dfrac{10x+9}{12} \\
 & \Rightarrow \dfrac{8x+12}{12}=\dfrac{10x+9}{12} \\
\end{align}$
Cancelling out ‘12’ from the denominator of both the sides of the equation, we get
$\Rightarrow 8x+12=10x+9$
Now we have to separate the variable and constants.
Bringing all ‘x’ terms to left hand side and all constants to right hand side of the equation, we get
 $\begin{align}
  & \Rightarrow 8x-10x=9-12 \\
 & \Rightarrow -2x=-3 \\
\end{align}$
Dividing both sides by $-2$, we get
$\Rightarrow \dfrac{-2x}{-2}=\dfrac{-3}{-2}$
Cancelling out $-2$ both from the numerator and the denominator on the left hand side, we get
$\Rightarrow x=\dfrac{3}{2}$
This is the required solution of the given question.

Note: Taking l.c.m. of 9 on both sides of the equation should be the first approach to solve this question. Division part should be eliminated. Necessary calculations should be done to get the value of ‘x’ as per the requirement of the question.