KL is a right triangle; JK = 3x – 6, KL = 2x + 1, JL = 20.

Answer 1

IF 3x-6 is the hypotenuse, then

9x^2 -36x +36 = 400 + 4x^2 +4x +1

5x^2 -40x +35 =0

x^2 -8x +7 =0

(x-7)(x-1) = 0

x = 1 or 7, but 1 makes the hypotenuse negative, so x = only 7

hypotenuse = 21-6 = 15

one side =20, so 3x-6 can't be the hypotenuse as 20>15

let 20 = the hypotenuse

then

400 = (3x-6)^2 + (2x+1)^2 = 9x^2 -36x+36 +4x^2 +4x+1 = 13x^2 -32x +37

13x^2 -32x -363 = 0

x = 32/26 + or - (1/26)sqr(32^2 +52(363)) =about 1.2 + 5.4 = 6.6

then 3x-6 = 3(6.6)-6 = 13.8

and the other side = 2x+1 = 2(6.6)+1 = 13.2+1 = 14.2

last alternative would be to try 2x+1 as the hypotenuse

then

4x^2 + 4x + 1 = 400+ 9x^2 -36x + 36

5x^2 -40x +35 =0

x^2 -8x +7 = 0

x =1 or 7, but either way 2x+1<20 so it can't be the hypotenuse

Answer seems to be it's about a 20-14.2-13.8 right triangle with 20 as the hypotenuse