Solve the following equation and also check your result in each case :
\[\dfrac{{6x + 1}}{2} + 1 = \dfrac{{7x - 3}}{3}\]

Hint: We have to find the value of \[x\] from the given expression \[\dfrac{{6x + 1}}{2} + 1 = \dfrac{{7x - 3}}{3}\] and then check the solution . We solve this question using the concept of solving linear equations . First we would simplify the terms of the left hand side by taking the L.C.M. and then cross multiplying both sides , we would obtain a relation in terms of \[x\] . On further solving the expression we get the value of \[x\]. To check the solution , we will put the value of \[x\] in the given expression .

Complete step-by-step solution:
Given :
\[\dfrac{{6x + 1}}{2} + 1 = \dfrac{{7x - 3}}{3}\]
On taking L.C.M. , we get the expression as :
\[\dfrac{{6x + 1 + 2}}{2} = \dfrac{{7x - 3}}{3}\]
Further simplifying , we get
\[\dfrac{{6x + 3}}{2} = \dfrac{{7x - 3}}{3}\]
Now cross multiplying the terms of the expression , we get
\[3 \times \left( {6x + 3} \right) = 2 \times \left( {7x - 3} \right)\]
On further multiplication of the terms , we get
\[18x + 9 = 14x - 6\]
On solving , we get
\[18x - 14x = - 6 - 9\]
\[4x = - 15\]
We get the value of x as :
\[x = \dfrac{{ - 15}}{4}\]
Now , for the verification of the value of \[x\] .
Putting the value of \[x\] in the given expression \[\dfrac{{6x + 1}}{2} + 1 = \dfrac{{7x - 3}}{3}\] , we get solution for left hand side as:
\[\dfrac{{6x + 1}}{2} + 1 = \dfrac{{6 \times \dfrac{{ - 15}}{4} + 1}}{2} + 1\]
$\dfrac{6x+1}{2} + 1 = \dfrac{\dfrac{-45}{2}+1}2 + 1$
Further taking the L.C.M. , we get
\[\dfrac{{6x + 1}}{2} + 1 = \dfrac{{\dfrac{{ - 45}}{2} + 1}}{2} + 1\]
\[\dfrac{{6x + 1}}{2} + 1 = \dfrac{{ - 43}}{4} + 1\]
Further , we get
\[\dfrac{{6x + 1}}{2} + 1 = \dfrac{{ - 43 + 4}}{4}\]
\[\dfrac{{6x + 1}}{2} + 1 = \dfrac{{ - 39}}{4}\]
Hence , the value of the left hand side is \[\dfrac{{ - 39}}{4}\] .
Now solution for right hand side as :
\[\dfrac{{7x - 3}}{3} = \dfrac{{7 \times \dfrac{{ - 15}}{4} - 3}}{3}\]
Further taking the L.C.M. , we get
\[\dfrac{{7x - 3}}{3} = \dfrac{{ - 105 - 12}}{{12}}\]
\[\dfrac{{7x - 3}}{3} = \dfrac{{ - 117}}{{12}}\]
Further on cancelling the terms , we get
\[\dfrac{{7x - 3}}{3} = \dfrac{{ - 39}}{4}\]
Hence , the value of the right hand side is \[\dfrac{{ - 39}}{4}\] .
As the value of the left hand side and the right hand side are equal , we conclude that the solution for the value of \[x\] for the given expression is correct .
Hence , the value of \[x\] for the given expression \[\dfrac{{6x + 1}}{2} + 1 = \dfrac{{7x - 3}}{3}\] is \[\dfrac{{ - 15}}{4}\].

Note: After solving the question we can verify that we have calculated the correct value of \[x\] by putting the obtained value of \[x\] back into the expression and if we obtain both the left hand side and the right hand side equal then we have calculated the correct value .
We could have skipped the cross - multiplication part by taking the terms of the right hand side to the left and then taking the L.C.M. and then solving the expression we can get the value of \[x\] .