If #2x- 5#, #x-1#, and #3x -8# are all integers and #x- 1# is the median of these integers, what is x?

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Answer 1 · 2016-12-06T00:50:52

#x=4#

Note first that for all three values to be integers, we must have #x# be an integer as well.

Because #x-1# is the median of the given values, we have

#2x-5 <= x-1 <= 3x-8#
or
#3x-8 <= x-1 <= 2x-5#.

We consider each case.

Case 1: #2x-5 <= x-1 <= 3x-8#

#2x-5 <= x-1#
#=> 2x-5-x+5 <= x-1-x+5#
#=> x <= 4#

#x-1 <= 3x-8#
#=> x-1-x+8 <= 3x-8-x+8#
#=> 7<=2x#
#=> 7/2<=x#

Taken together, we have #x in [7/2, 4]#. As #4# is the only integer in that range, the only solution in this case is #x=3#.

Case 2: #3x-8 <= x-1 <= 2x-5#

If we go through the same steps as above with the directions of the inequalities reverse, we get

#x>=4# and #x<= 7/2#. As #7/2<4#, there are no such values, meaning this case produces no solutions.

Having considered both cases, then, we have found the sole solution as #x=4#.