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

Apr 3, 2017

$| A x + B | \le k$ is an interval when k > 0.

#### Explanation:

When solving an absolute value linear inequality of the form
$| A x + B | S I G N k$
Where SIGN is one of $< , \le , > , \mathmr{and} \ge$,
first consider whether k > 0, k = 0, or k < 0.
k > 0 is the "normal case."

In such cases, $| A x + B | > k$ and $| A x + B | \ge k$
yield disjoint intervals as their solutions, and
In such cases, $| A x + B | < k$ and $| A x + B | \le k$
yield single open or closed intervals.

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

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

Subtract 3 and then divide by 2.

Leaving the steps to you, the solution is $- 7 \le x \le 4$. 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.