x;y;z;t là số tiền của bốn bạn (t =1500)
\(\left\{{}\begin{matrix}x=\dfrac{y+z+t}{2}\left(1\right)\\y=\dfrac{x+z+t}{3}\left(2\right)\\z=\dfrac{x+y+t}{4}\left(3\right)\end{matrix}\right.\) cộng hai vế
\(x+y+z=\dfrac{y+z+t}{2}+\dfrac{x+z+t}{3}+\dfrac{x+y+t}{4}=\dfrac{\left(6y+6z+6t\right)+\left(4x+4z+4t\right)+\left(3x+3y+3t\right)}{12}=\dfrac{9\left(x+y+z\right)+13t-2x}{12}\)12(x+y+z) =9(x+y+z)+13t-2x
5x=13t-(y+z)
từ (1) => y+z =2x -t <=> 5x =13t -(2x-t)<=> 7x =14t => x =2t
y+z=4t -t =3t
x+y+z =3t+2t=6t
x+y+z+t =7t =7.1500 =10500