Question

How do you solve and graph `abs(2x+3)<=11`?

Answer 1

`|Ax + B| <= k` is an interval when k > 0.

Explanation:

When solving an absolute value linear inequality of the form
`|Ax + B| SIGN k`
Where SIGN is one of `<, <=, >, or >=`,
first consider whether k > 0, k = 0, or k < 0.
k > 0 is the "normal case."

In such cases, `|Ax + B| > k` and `|Ax + B| >= k`
yield disjoint intervals as their solutions, and
In such cases, `|Ax + B| < k` and `|Ax + B| <= k`
yield single open or closed intervals.

The answer to your question will be a single closed interval.

`|2x + 3| <= 11` if and only if
`2x + 3 <= 11` AND `2x + 3 >= 11`.
You may solve these two inequalities separately for x.
Some textbooks will write an "and" situation as a "three-sided" inequality:
`-11<= 2x + 3 <= 11`

Subtract 3 and then divide by 2.

Leaving the steps to you, the solution is `-7<= x <= 4`. In interval form we write [-7, 4].

On a number line put two filled-in dots: one at -7 and the other at 4. Shade in between.