Question 378651
First we must find which side must be the hypotenuse. Clearly, 3x - 13 cannot be the hypotenuse as 3x - 4 is larger. Therefore either 2x - 1 or 3x - 4 is larger. Furthermore, since all the sides are positive, 3x - 13 > 0 --> x > 13/3. We can see that for all values of x larger than 13/3, 3x - 4 > 2x - 1, so 3x - 4 is the hypotenuse.

Now we can solve for x using the Pythagorean theorem. We have

{{{(2x - 1)^2 + (3x - 13)^2 = (3x - 4)^2}}}

{{{4x^2 - 4x + 1 + 9x^2 - 78x + 169 = 9x^2 - 24x + 16}}}

{{{13x^2 - 82x + 170 = 9x^2 - 24x + 16}}}

{{{4x^2 - 58x + 154 = 0}}}

{{{2x^2 - 29x + 77 = 0}}}

{{{(2x - 7)(x - 11) = 0}}}

x = 7/2 or x = 11

However, if x = 7/2, then 3x - 13 is negative -- therefore the only value for x is x = 11.