Answered



Solve and graph the absolute value inequality: |2x + 1| ≤ 5.


A.number line with closed dots on negative 3 and 2 with shading going in the opposite directions.

Bnumber line with closed dots on negative 3 and 2 with shading in between.
Cnumber line with open dots on negative 3 and 2 with shading in between.

D.number line with closed dots on negative 2 and 2 with shading in between.

Verified Answer :

texaschic101

|2x + 1| < = 5

-5 < = 2x + 1 < = 5 ...subtract 1
-5 - 1 < = 2x + 1 - 1 < = 5 - 1
-6 < = 2x < = 4 ....divide by 2
-6/2 < = (2/2)x < = 4/2
-3 < = x < = 2

B. a number line with closed dots (because of the = sign) on -3 and 2 with shading in between

Answer:

B. number line with closed dots on negative 3 and 2 with shading in between.

Step-by-step explanation:

Solve and graph the absolute value inequality: |2x + 1| ≤ 5.

if |x|<=a  then x<=a  and x>=-a. consider two cases for absolute function

|2x + 1| ≤ 5

2x+1 <= 5              2x+1 >= -5  (solve both inequalities)

subtract 1 from both sides and then divide both sides by 2

2x <= 4                   2x>= -6

x<= 2                         x>= -3

combine both inequalities

-3<=x <= 2

x lies between -3 and 2

number line with closed dots on negative 3 and 2 with shading in between.

Related Questions