How do you solve #|5+ x | > 7#?

2 Answers
smendyka
Jul 22, 2017

See a solution process below:

Explanation:

The absolute value function takes any negative or positive 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.

#-7 > 5 + x > 7#

#-color(red)(5) - 7 > -color(red)(5) + 5 + x > -color(red)(5) + 7#

#-12 > 0 + x > 2#

#-12 > x > 2#

Or

#x < -12# and #x > 2#

Or, in interval notation:

#(-oo, -12)# and #(2, +oo)#

Tony B
Jul 22, 2017

#-12>x>2#

I have given an in depth explanation. Under normal condition you would be expected to be more efficient in your approach. They way that smendyka has been.

Explanation:

#color(blue)("Determine the 'trigger values'")#

Just for a moment lets ignore the greater than and use equals instead,

#|5+x|=7#

This is the same as

#|+-7|=+7#

So for THIS to be true we have 2 condition

#5+x=-7" and "5+x=+7#

This gives rise to

#x=-12" and "x=+2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Translating the 'trigger values' to suit the question")#

#color(brown)("Consider the case "x=-12)#

Suppose #x=-11# this would give #|5-11| = |-6|=+6 ->" not ">7#
Suppose #x=-13# this would give #|5-13|=|-8|=+8->" is ">7#

So we need #x<-12#

#" "........................................#

#color(brown)("Consider the case "x=+2)#

Using the same sort of testing as above we end up with

So we need #x>+2#
#" ".........................................#

Thus we have:

#2 < x < -12#

same thing written differently

#-12>x>2#

The shaded area is the feasible domain / range

Tony B

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