How do you solve #1<abs(s-2)<3#?

2 Answers
Jun 17, 2015

Answer:

It's complicated.

Explanation:

First let's split this into two parts.

#1 < |s-2|# and #|s-2|<3#

Here's where things get confusing. Absolute value means that the number could either be positive or negative. If it's negative, the sign is flipped.

#1 < s-2# or #-1 > s-2# and #s-2<3# or #s-2> -3#

#3 < s# or #1>s# and #s < 5# or #s> -1#

Wait, look at that 2nd part. If s can be less than 5 (a bigger number), or more than -1 (a smaller number), then s can be anything (at least in one conditional). So this can be rephrased as:

#3< s# or #1>s# and #s = #all real numbers

The "and" means that 's' can be anything where these 2 conditionals overlap. It's pointless to ask if something overlaps with all real numbers, so we can get rid of that, leaving us with the final answer:

#3< s# or #1>s#

Jun 17, 2015

Answer:

There is no one sentence "How".
The solution set is: #(-1,1) uu (3,5)#

Explanation:

Algebraic solution

#abs (s-2) = s-2 " or " (-s-2)#. So solve the two inequalities:

#1 < s-2 < 3 # which gives us #3 < s < 5#

#1 < -s+2 < 3# which yields #-1 < -s < 1#. Multiply through by #-1# (change the direction of the inequalities) to get #1 > s >-1#

Geometric solution

#abs (a-b)# is the distance between #a# and #b#.

So #1 < abs (s-2) <3# means that #s# is more than #1# and less than #3# away from #2#.

On the number line, find #2#.
Look on the right for the points (numbers) more than #1#, but less that #3# away from 2. These are the numbers between 3 and 5 (not including 3 and 5.)
Look on the left for the points (numbers) more than #1#, but less that #3# away from 2. These are the numbers between 1 and -1 (not including 1 and -1.)

So we need s to be between #-1# and #1# OR between #3# and #5#.