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

Dec 6, 2016

$x = 4$

#### Explanation:

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

$2 x - 5 \le x - 1 \le 3 x - 8$
or
$3 x - 8 \le x - 1 \le 2 x - 5$.

We consider each case.

Case 1: $2 x - 5 \le x - 1 \le 3 x - 8$

$2 x - 5 \le x - 1$
$\implies 2 x - 5 - x + 5 \le x - 1 - x + 5$
$\implies x \le 4$

$x - 1 \le 3 x - 8$
$\implies x - 1 - x + 8 \le 3 x - 8 - x + 8$
$\implies 7 \le 2 x$
$\implies \frac{7}{2} \le x$

Taken together, we have $x \in \left[\frac{7}{2} , 4\right]$. As $4$ is the only integer in that range, the only solution in this case is $x = 3$.

Case 2: $3 x - 8 \le x - 1 \le 2 x - 5$

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

$x \ge 4$ and $x \le \frac{7}{2}$. As $\frac{7}{2} < 4$, there are no such values, meaning this case produces no solutions.

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