Question

Determinati nr x si y stiind ca : sunt invers proportionale cu 2 si 3 ,iar x la a 2 +y la a2 =208

Answer 1

2x=3y=n
x=n/2; y=n/3
x^2 + y^2=208
(n/2)^2 + (n/3)^2 = 208
n^2/4 + n^2/9 = 208
Aducem la numitor comun
n^2*9/36 + n^2*4/36 = 208
n^2*9+n^2*4/36 = 208
n^2*13/36 = 208
n^2*13 = 7488
n^2 = 576
n=24
x=12; y=8