Question

How do you graph and solve `abs(x+1)<=11`?

Answer 1

Solution:
`-12 <= x <= 10`

Explanation:

Start from a definition of the absolute value.
For non-negative real `x` the value of `abs(x)` is `x`.
For negative real `x` the value of `abs(x)` is `-x`.

In short:
if `x>=0`, `|x| = x`
if `x<0`, `|x| = -x`

Using this definition, let's divide the domain of the `|x+1|` in two intervals:
1. `x+1 >= 0`, where `|x+1| = x+1`
2. `x+1 < 0`, where `|x+1| = -(x+1)`

Case 1. Looking for solutions in interval `x+1 >= 0` (or, equivalently, `x >= -1`)
Since in this interval `|x+1| = x+1`, our inequality looks like
`x+1 <= 11` and the solution is `x <= 10`.
Together with a boundary of the interval, we have come to a solution:
`-1 <= x <= 10`

Case 2. Looking for solutions in interval `x+1 < 0` (or, equivalently, `x < -1`)
Since in this interval `|x+1| = -(x+1)`, our inequality looks like
`-(x+1) <= 11` and the solution is `x >= -12`.
Together with a boundary of the interval, we have come to a solution:
`-12 <= x < -1`

So, two intervals are the solution to an original inequality:
`-1 <= x <= 10` and
`-12 <= x < -1`

These can be combined into
`-12 <= x <= 10`

Here is a graph of `y=|x+1|`

graph{|x+1| [-20, 20, -15, 15]}
It illustrates the solution