How do you graph and solve #|4x+8|>=20#?

1 Answer
Jun 30, 2016

Answer:

The solution is a union of two intervals:
#x<=-7# or #x in (-oo,-7]#
UNION with
#x >= 3# or #x in [3,+oo)#

Explanation:

The value of #|4x+8|# is defined as follows:
if #4x+8 >= 0# then #|4x+8| = 4x+8#
otherwise (that is, id #4x+8 < 0#), #|4x+8| = -(4x+8)#.

Let's consider a point #x=-2#, where term #4x+8# changes from negative (to the left of #x=-2#) to positive (to the right of #x=-2#).

From definition of absolute value mentioned above,
if #x < -2#, 4x+8 < 0 and, therefore, #|4x+8| = -(4x+8)# and our initial inequality looks like this:
#-(4x+8) >= 20# or
#-4x-8 >= 20# or
#-8-20 >= 4x# or
#4x <= -28# or
#x <= -7#
Since #x<=-7# lies inside the interval #x < -2# that we consider, all #x<=-7# are solutions.

From definition of absolute value mentioned above,
if #x >= -2#, 4x+8 >= 0 and, therefore, #|4x+8| = 4x+8# and our initial inequality looks like this:
#4x+8 >= 20# or
#4x >= 12# or
#x >= 3#
Since #x>=3# lies inside the interval #x >= -2# that we consider, all #x>=3# are solutions.

So, we have two separate intervals that represent the solutions to this inequality:
#x <= -7# and #x >= 3#.

Graphically, it looks like this:
graph{|4x+8| [-24, 24, -23.12, 23.12]}
It can be observed that this graph is above the line #y=20# when #x<=-7# or #x >= 3#.