How do you solve and write the following in interval notation: #| 4x | > -12#?

Step 1. Separate the absolute value
#abs(4)abs(x)> -12#

Step 2. Convert #abs(4)=4#
#4abs(x) > -12#

Step 3. Divide both sides by #4#
#abs(x) > -3#

Step 4. Add #3# to both sides
#abs(x)+3>0#

The trick to solving this is knowing what the absolute value function does. The absolute value essentially changes any negative number to a positive number. Let's plug in values to see what we get:

• If #x=-4#, then #abs(-4)+3=4+3=7>0# is TRUE
• If #x=-2#, then #abs(-2)+3=2=3+3=5>0# is TRUE
• If #x=-0#, then #abs(-0)+3=0+3=3>0# is TRUE

As you can see, any number less than or equal to zero is going to be true. And any number greater than zero is also true. So this express is true for all possible values for #x#!

#color(blue)("Preamble")#

The double vertical lines indicate something called absolute values.

This means that what is inside them is always considered as ending up being positive.

Consider the following example:

Making the inequality an equality such that we had

#|4x|=12#

Then what we really have is

#|+-12|=12#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering the question")#

We have #|4x| > -12 color(red)(" There is a trap in this")#

#|4x|# is always positive no matter what value we assign to #x#.

If #|4x|# is always positive then it is always more than -12

Consequently #x# belongs to the set of all 'Real numbers' written as:

#{x:x in RR}#

Or if you like #x in (-oo,+oo)#