How do you graph #1/2x+8<=3#?

1 Answer
Oct 26, 2017

Answer:

See a solution process below:

Explanation:

First, solve the inequality for #x#:

Subtract #color(red)(8)# from each side of the inequality to isolate the #x# term while keeping the inequality balanced:

#1/2x + 8 - color(red)(8) <= 3 - color(red)(8)#

#1/2x + 0 <= -5#

#1/2x <= -5#

Multiply each side of the inequality by #color(red)(2)# to solve for #x# while keeping the inequality balanced:

#color(red)(2) xx 1/2x <= color(red)(2) xx -5#

#cancel(color(red)(2)) xx 1/color(red)(cancel(color(black)(2)))x <= -10#

#x <= -10#

To graph this we will draw a vertical line at #-10# on the horizontal axis.

The line will be a solid line because the inequality operator contains an "or equal to" clause.

We will shade to the left side of the line because the inequality operator also contains a "less than" clause:

graph{x <= -10 [-20, 20, -10, 10]}