How do you solve #abs(5 - 1/x)< 2#?

1 Answer
Jun 15, 2015

Answer:

First interpret the modulus to get: #-2 < 5 - 1/x < 2#
Then apply arithmetic operations to arrive at: #1/7 < x < 1/3#

Explanation:

#abs(5-1/x) < 2# is equivalent to #-2 < 5-1/x < 2#

Next subtract #5# from all parts of this inequality to get:

#-7 < -1/x < -3#

Next multiply all parts by #-1# and reverse the inequality signs to get:

#7 > 1/x > 3#

Since this forces #1/x# to be positive, that means #x > 0# too.

Multiply all parts by #x# to get:

#7x > 1 > 3x#. That is #7x > 1# and #3x < 1#.

Dividing the first of these by #7# we get #x > 1/7#

Diving the second by #3# we get #x < 1/3#

So #1/7 < x < 1/3#

In general, you can apply the following operations to an inequality and preserve its truth:

(1) Add or subtract the same value from all parts of the inequality.
(2) Multiply or divide all parts by the the same positive value.
(3) Multiply or divide all parts by the same negative value and reverse the inequalities (#<# becomes #>#, #>=# becomes #<=#, etc.)