How do you solve and write the following in interval notation: #x/2 + 3 ≤ 1#?

1 Answer
Sep 3, 2017

Answer:

See a solution process below:

Explanation:

This is a standard 2-Step problem:

Step 1: Isolate The Variable: Subtract #color(red)(3)# from each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#x/2 + 3 - color(red)(3) <= 1 - color(red)(3)#

#x/2 + 0 <= -2#

#x/2 <= -2#

Step 2: Multiply or Divide by the Appropriate Term to Solve for the Variable: Multiply each side of the inequality by #color(red)(2)# to solve for #x# while keeping the inequality balanced:

#color(red)(2) xx x/2 <= color(red)(2) xx -2#

#cancel(color(red)(2)) xx x/color(red)(cancel(color(black)(2))) <= -4#

#x <= -4#

Or, in interval notation:

#(-oo, -4]#