Question 36067
using pythogrean theorem

hypotnuse^2 = side1^2 +side2^2

(x-2)^2 = (2x-11)^2 + (25-3x)^2

x^2 - 4x + 4 = 4x^2 -44x + 121 + 9x^2 -150x +625

12x^2 -190x + 742 = 0

using (-b +/- sqrt(b^2 -4*a*c))/(2*a)  where a =12 b = -190 c =742

we get x = 7 or x = 212/24 = 53/6 

if x =53/6 the the side 25-3x = 25 -53/2 = -3/2 which is not possible as the length of sides cannot be negative
hence x  =7
hypotnuse = 7-2 =5

side1 = 2*7-11 = 3
side2 = 25 - 3*7 =4