Solve $5x + \left( {3x - \dfrac{{4x + 6}}{5}} \right) = \dfrac{{\left( {5x + 3} \right)}}{6} - 2$

Hint: Given equation is a linear equation because the highest exponent of x is 1 and there are no x and x multiplications in the given equation which can increase the exponent of x so the given equation is linear. Multiply and divide with required integers on the left hand side and right hand side to find the value of x.

Complete step-by-step answer:
We are given an equation in one variable x and the equation is linear.
The equation is $5x + \left( {3x - \dfrac{{4x + 6}}{5}} \right) = \dfrac{{\left( {5x + 3} \right)}}{6} - 2$ and we have to solve the equation and find the value of x.
$5x + \left( {3x - \dfrac{{4x + 6}}{5}} \right) = \dfrac{{\left( {5x + 3} \right)}}{6} - 2$
Multiply and divide with 5 on the left hand side; multiply and divide with 6 on the right side.
$
  5x + \left( {3x - \dfrac{{4x + 6}}{5}} \right) = \dfrac{{\left( {5x + 3} \right)}}{6} - 2 \\
  \dfrac{5}{5}\left[ {5x + \left( {3x - \dfrac{{4x + 6}}{5}} \right)} \right] = \dfrac{6}{6}\left[ {\dfrac{{\left( {5x + 3} \right)}}{6} - 2} \right] \\
  \dfrac{{25x + \left( {15x - 4x + 6} \right)}}{5} = \dfrac{{\left( {5x + 3} \right) - 12}}{6} \\
 $
Send denominator 5 which is in the LHS to the RHS and send denominator 6 which is in the RHS to the LHS.
$
  6\left( {25x + 15x - 4x + 6} \right) = 5\left( {5x + 3 - 12} \right) \\
  150x + 90x - 24x + 36 = 25x + 15 - 60 \\
 $
Send all the terms containing x to the left hand side and all the constant terms to the right hand side.
$
  150x + 90x - 24x - 25x = 15 - 60 - 36 \\
  240x - 49x = 15 - 96 \\
  191x = - 81 \\
 $
Divide -81 by 191 to get the value of x.
$x = \dfrac{{ - 81}}{{191}}$
Therefore, the value of x is $\dfrac{{ - 81}}{{191}}$

Note: A linear equation is an equation of a straight line having maximum one variable. The power of the variable will be 1. To solve any equation in one variable, put all the variable terms on the left hand side and all the numerical values on the right hand side for an easy approach.