How do you solve and write the following in interval notation: #-1 + | 1/3 - x/2 | > -11/12#?

1 Answer
Aug 25, 2017

Answer:

See a solution process below:

Explanation:

First, add #color(red)(1)# to each side of the inequality to isolate the absolute value function while keeping the inequality balanced:

#color(red)(1) - 1 + abs(1/3 - x/2) > color(red)(1) - 11/12#

#0 + abs(1/3 - x/2) > (12/12 xx color(red)(1)) - 11/12#

#abs(1/3 - x/2) > 12/12 - 11/12#

#abs(1/3 - x/2) > 1/12#

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.

#-1/12 > 1/3 - x/2 > 1/12#

Next, subtract #color(red)(1/3)# from each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:

#-color(red)(1/3) + -1/12 > -color(red)(1/3) + 1/3 - x/2 > -color(red)(1/3) + 1/12#

#(4/4 xx -color(red)(1/3)) + -1/12 > 0 - x/2 > (4/4 xx -color(red)(1/3)) + 1/12#

#-4/12 + -1/12 > -x/2 > -4/12 + 1/12#

#-5/12 > -x/2 > -3/12#

#-5/12 > -x/2 > -1/4#

Now, multiply each segment by #color(blue)(-2)# to solve for #x# while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators:

#color(blue)(-2) xx -5/12 color(red)(<) color(blue)(-2) xx -x/2 color(red)(<) color(blue)(-2) xx -1/4#

#10/12 color(red)(<) -cancel(color(blue)(2)) xx -x/color(blue)(cancel(color(black)(2))) color(red)(<) 2/4#

#5/6 color(red)(<) x color(red)(<) 1/2#

Or

#x > 5/6# and #x < 1/2#

Or, in interval notation:

#(-oo, 1/2)" and "(5/6, +oo)#