Solving the absolute value equation $2-3|x-1| = -4|x-1|+7$ $$2-3|x-1| = -4|x-1|+7$$
This is an example from my text book, and I do not understand how they got the answers.
Solution: (this is the solution in my textbook)
Isolate the absolute value of expression on one side
Add $4|x-1|$ to both sides $ \rightarrow 2+|x-1|=7$
Subtract $2$ from both sides $ \rightarrow |x-1|=5$
If the absolute value of an expression is equal to $5$, then the expression is equal to either $-5$ or $5$.
$$x-1= -5, \ x-1=5$$
$$\implies x= -4, \ \implies x= 6.$$
I don't understand where the $4|x-1|$ went or what happened to it.
 A: This is simply due to the fact that we can add or subtract anything to both sides of an equation and preserve the equality. Maybe this will help you understand it
\begin{align}
2-\color{red}{3|x-1|} &= \color{red}{-4|x-1|}+7 \\
2-\color{red}{3|x-1|} + \underbrace{\color{blue}{4|x-1|}}_{\text{Added}} &= \color{red}{-4|x-1|}+7+ \underbrace{\color{blue}{4|x-1|}}_{\text{Added}}\\
2+(\color{blue}{4}-\color{red}{3})|x-1| &= (\color{blue}{4}-\color{red}{4})|x-1| + 7\\
2+1\cdot|x-1|&= 0\cdot|x-1|+7\\
2+|x-1| &= 7\\
2-\color{blue}{2}+|x-1| &= 7-\color{blue}{2}\\
|x-1| &= 5.
\end{align}
I hope it makes sense now.
A: Let $y = |x-1|$.
Now your problem becomes:
$$2 - 3y = -4y + 7$$
Perhaps in this more recognizable guise, you can solve for $y$ to find:
$$y = 5$$
Returning to $y = |x-1|$, we now have: 
$$|x-1| = 5$$
Note that this will happen in two cases.


*

*If $x-1 = 5$, for then $|x-1| = |5| = 5$.

*If $x-1 = -5$, for then $|x-1| = |-5| = 5$.
In the former case, you find $x = 6$; in the latter, $x = -4$.
