algebra
posted by ann .
what is x short leg for right triangle
when given long leg b is 1/2x +11
when given hypotenuse c is 2x + 1
i put (1/2x + 11) +(2x + 1) = x and came out with 20 but that did not work out when squared A + b = C.
I estimated x/a = 8, then b = 15 and c = 17 and that works 8^2 + 15^2 = 17^2 or 64 + 225 = 289 for A^2 +b^2 = c^2
could you please show me how to do it the right way?
thanks

You are not even using the correct formula.
it is A^2 + B^2 = C^2 , where C is the hypotenuse, so
x^2 + ((1/2)x + 11)^2 = (2x+1)^2
x^2 + (1/4)x^2 + 11x + 121 = 4x^2 + 4x + 1
times 4 to get rid of my fraction
4x^2 + x^2 + 44x + 484 = 16x^2 + 16x + 4
11x^2 + 28x + 480 = 0
11x^2  28x  480 = 0
(x8)(11x + 60) = 0
x = 8 or x = 60/11 , but x can't be negative
so x = 8
one side is 8
the other is (1/2)x+11 = 15
and the hypotenuse is 2x+1 = 17 
Reiny, thank you so much, I really needed that help.