Solve the equation \[(3x+2)-5(6x-1)=2(x-8)-6(7x-4)\] .

Hint: First of all expand the given equation \[(3x+2)-5(6x-1)=2(x-8)-6(7x-4)\] . After expanding, make LHS of the equation in such a way that it only has terms of x. Similarly, we have to make RHS of the equation in such a way that it contains only constant terms. Now, solve it further.

Complete step-by-step answer:

According to the question, it is given that our equation is \[(3x+2)-5(6x-1)=2(x-8)-6(7x-4)\] ……………………..(1)
We can see that equation (1) is a linear equation in x where x is unknown. We have to find the value of x using the equation provided.
First of all, we have to simplify the equation \[(3x+2)-5(6x-1)=2(x-8)-6(7x-4)\] into a simpler form. To simplify, we have to expand it first.
Expanding equation (1), we get
\[(3x+2)-5(6x-1)=2(x-8)-6(7x-4)\]
\[\Rightarrow (3x+2)-30x+5=2x-16-42x+24)\]
\[\Rightarrow -27x+7=-40x+12\] ……………………..(2)
Now, we have to make LHS of equation (2) in such a way that it only has terms of x. Similarly, we have to make RHS of the equation in such a way that it contains only constant terms.
Shifting -40x of RHS to the LHS and +7 of LHS to RHS of the equation, we get
\[\Rightarrow -27x+7=-40x+12\]
\[\begin{align}
  & \Rightarrow -27x+40x=12-7 \\
 & \Rightarrow 13x=5 \\
\end{align}\]
Here, the above is in its simplest form and can be solved easily.
Now, dividing the above equation by 13 on both LHS and RHS, we get
\[\begin{align}
  & \Rightarrow \dfrac{13x}{13}=\dfrac{5}{13} \\
 & \Rightarrow x=\dfrac{5}{13} \\
\end{align}\]
Hence, the value of x is \[x=\dfrac{5}{13}\] .

Note: In this question, one can make a mistake in shifting +7 of LHS to the RHS and write +7 in RHS too, which is wrong. If we are moving some terms of LHS to RHS then its sign is changed and vice-versa. Here we are moving +7 of LHS to the RHS. So, we have to write -7 in the RHS of the equation. Similarly, we have to shift -40x of RHS to the LHS. So, we have to write +40x in the LHS of the equation.