Solve the following equation:
$\dfrac{{3\left( {2x - 5} \right)}}{4} - \dfrac{{5\left( {7 - 5x} \right)}}{6} = \dfrac{{7x}}{3}$

Hint: In this question we just have to take LCM and then we have to do cross-multiplication to simplify the equation. On the simplified equation we have to do some calculation to get to the value of x. This is a simple question of algebra.

Complete step-by-step solution:
In the given question, we have
$ \Rightarrow \dfrac{{3\left( {2x - 5} \right)}}{4} - \dfrac{{5\left( {7 - 5x} \right)}}{6} = \dfrac{{7x}}{3}$
Since here the variable is ‘x’ in the equation, we will find the value of ‘x’ using operations like multiplication, addition, subtraction and division.
On multiplying in denominator, we get
$ \Rightarrow \dfrac{{6x - 15}}{4} - \dfrac{{35 - 25x}}{6} = \dfrac{{7x}}{3}$
Now, taking LCM of $4$ and $6$
$ \Rightarrow \dfrac{{6\left( {6x - 15} \right) - 4\left( {35 - 25x} \right)}}{{24}} = \dfrac{{7x}}{3}$
Calculate products in numerator of LHS
$ \Rightarrow \dfrac{{36x - 90 - 140 + 100x}}{{24}} = \dfrac{{7x}}{3}$
On cross-multiplication, we get
$ \Rightarrow 3\left( {36x - 90 - 140 + 100x} \right) = 24\left( {7x} \right)$
On simplification, we get
$ \Rightarrow 3\left( {136x - 230} \right) = 24\left( {7x} \right)$
On multiplication in LHS and RHS, we get
$ \Rightarrow 408x - 690 = 168x$
Shift constant to right side and variable to left side
$ \Rightarrow 408x - 168x = 690$
On simplification, we get
$ \Rightarrow 240x = 690$
On division, we get
$ \Rightarrow x = \dfrac{{690}}{{240}}$
Divide numerator and denominator by $10$
$ \Rightarrow x = \dfrac{{69}}{{24}}$
Divide numerator and denominator by $3$
$ \Rightarrow x = \dfrac{{23}}{8}$
Hence, the value of x is $\dfrac{{23}}{8}$.

Note: Students are likely to make mistakes in the calculations part where they shift values from one side of the equation to another, so always keep in mind that the sign changes from positive to negative and vice versa when shifting constants or variables to the other side. Also, take the LCM carefully while solving the equation and multiply with the correct number in the numerator. Always add the constant part with the constant and the variable part with the variable.