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

1 Answer
Apr 3, 2017

Answer:

#|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.