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 $x=4$ .

If 2x- 52x- 5, x-1x-1, and 3x -83x -8 are all integers and x- 1x- 1 is the median of these integers, what is x?