Question

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

Answer 1

The answer is:
`x=1` and
`x=-3`

The way to solve it is as follows.

Equal parts of an equation, left and right, can be multiplied by the same non-equal to zero multiplier getting an equivalent equation.
Let's multiply them by `-1`:
`|x+1|=2`

Now we have to remember the definition of the absolute value of a number.
If the number is positive or zero, its absolute value equals to itself:
if `Z>=0` then `|Z|=Z`.
If the number is negative, its absolute value equals to its opposite (or, not very scientifically, minus this number)
if `Z<0` then `|Z|=-Z`.

Applying this to a problem at hand:

CASE 1
Looking for solutions in the area defined by an inequality
`x+1>=0` (that is, `x>=-1`) then `|x+1|=x+1` and,
using our equation,
`x+1=2`, that is, `x=1`.
This value is within the area `x>=-1` and, therefore, is the first legitimate solution.

CASE 2
Looking for solutions in the area defined by an inequality
`x+1<0` (that is, `x<-1`) then `|x+1|=-(x+1)` and,
using our equation,
`-(x+1)=2`, that is, `x=-3`.
This value is within the area `x<-1` and, therefore, is the second legitimate solution.

We can demonstrate it graphically.
Our equation is equivalent to
`|x+1|-2=0`
Let's draw a graph of a function
`y=|x+1|-2`
graph{|x+1|-2 [-10, 10, -5, 5]}
As you see, it intersects the X-axis (that is, equals to zero) at points `x=-3` and `x=1`. This confirms our solutions.