Let the actual speed of the train be \\\\(x\\\\) km/hr and the actual time taken be \\\\(y\\\\) hours. Then,Distance covered \\\\(=(xy)km\\\\) ..(i) \\\\([\ herefore\\\\) Distance \\\\(=\\\\) Speed \\\\(\ imes\\\\) Time\\\\(]\\\\)If the speed is increased by \\\\(6\\\\) km/hr, then time of journey is reduced by \\\\(4\\\\) hours i.e., when speed is \\\\((x+6)\\\\)km/hr, time of journey is \\\\((y-4)\\\\) hours.\\\\(\ herefore\\\\) Distance covered \\\\(=(x+6)(y-4)\\\\)\\\\(\\Rightarrow xy=(x+6)(y-4)\\\\) [Using (i)]\\\\(\\Rightarrow -4x+6y-24=0\\\\)\\\\(\\Rightarrow -2x+3y-12=0\\\\) ..(ii)When the speed is reduced by \\\\(6\\\\) km/hr, then the time of journey is increased by \\\\(6\\\\) hours i.e., when speed is \\\\((x-6)\\\\) km/hr, time of journey is \\\\((y-6)\\\\) hours.\\\\(\ herefore\\\\) Distance covered \\\\(=(x-6)(y+6)\\\\)\\\\(\\Rightarrow xy=(x-6)(y+6)\\\\) [Using (i)]\\\\(\\Rightarrow 6x-6y-36=0\\\\)\\\\(\\Rightarrow x-y-6=0\\\\) (iii)Thus, we obtain the following system of equations:\\\\(-2x+3y-12=0\\\\)\\\\(x-y-6=0\\\\)By using cross-multiplication, we have,\\\\(\\dfrac{x}{3\ imes -6-(-1)\ imes -12}=\\dfrac{-y}{-2\ imes -6-1\ imes -12}=\\dfrac{1}{-2\ imes -1-1\ imes 3}\\\\)\\\\(\\Rightarrow \\dfrac{x}{-30}=\\dfrac{-y}{24}=\\dfrac{1}{-1}\\\\)\\\\(\\Rightarrow x=30\\\\) and \\\\(y=24\\\\)Putting the values of x and y in equation (i), we obtainDistance \\\\(=(30\ imes 24)\\\\)km \\\\(=720\\\\)km.Hence, the length of the journey is \\\\(720\\\\)km.