How do you solve `2-3abs(x-1)=-4abs(x-1)+7`?

(1) `x>=1` so `x-1` is non-negative
The equality turns into:
`2-3(x-1)=-4(x-1)+7->`
`2-3x+3=-4x+4+7->`
`-3x+4x=4+7-2-3->`
`x=6`
(and this is `>=1`) -- always check this!

(2) `x<1` so `x-1` is negative. The absolute bars will turn the sign around and the equality turns into:
`2-3(1-x)=-4(1-x)+7->`
`2-3+3x=-4+4x+7->`
`3x-4x=-4+7-2+3->`
`-x=4 -> x=-4`
(and this is `<1`)

`2-3abs(x-1)=-4abs(x-1)+7`

Observe that the expression inside the absolute value signs is the same in both absolute values. It is `x-1`. This approach will first find out what `abs(x-1)` is equal to, and then, find out what `x` must be.

Until you get some experience with working with expressions as if they were variables, it will help to actually do a substitution.
Our first goal is to find `abs(x-1)`. Since this is an absolute value, let's call it `a` for the time being:

Let `a = abs(x-1)`.

Substituting, we have:

`2-3a=-4a+7` Find `a`. (add `4a` and subtract `2` on both sides)

`-3a+4a=7-2` (simplify)

`a = 5`

Good. We've finished step 1. Now we still need to find `x`. We'll use `a=abs(x-1)` to write:

`abs(x-1) = 5`

The two numbers whose absolute value is `5` are `-5` and `5`,, so

`x-1=-5` `color(white)"xx"` or `color(white)"xx"` `x-1 = 5` `color(white)"xx"` so

`x=-4` `color(white)"xx"` or `color(white)"xx"` `x = 6`

Less details looks like this:

`2-3abs(x-1)=-4abs(x-1)+7`

`-3abs(x-1) +4abs(x-1) = 7-2`

`abs(x-1) =5`

`x-1 = -5` or `x-1=5`

`x=-4 " or " 6`

Solutions: `-4`, and `6`