How do you solve #abs(32x-16)>32#?

1 Answer
Nov 10, 2017

Answer:

See a solution process below:

Explanation:

The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

#-32 > 32x - 16 > 32#

First, add #color(red)(16)# to each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:

#-32 + color(red)(16) > 32x - 16 + color(red)(16) > 32 + color(red)(16)#

#-16 > 32x - 0 > 48#

#-16 > 32x > 48#

Now, divide each segment by #color(red)(32)# to solve for #x# while keeping the system balanced:

#-16/color(red)(32) > (32x)/color(red)(32) > 48/color(red)(32)#

#-16/color(red)(16 xx 2) > (color(red)(cancel(color(black)(32)))x)/cancel(color(red)(32)) > (16 xx 3)/color(red)(16 xx 2)#

#(-color(red)(cancel(color(black)(16)))1)/color(red)(color(black)(cancel(color(red)(16))) xx 2) > (color(red)(cancel(color(black)(32)))x)/cancel(color(red)(32)) > (color(red)(cancel(color(black)(16))) xx 3)/color(red)(color(black)(cancel(color(red)(16))) xx 2)#

#-1/2 > x > 3/2#

Or

#x < -1/2# and #x > 3/2#

Or, in interval notation

#(oo, -1/2)# and #(3/2, +oo)#