It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?(A) z(y – x)/x + y(B) z(x – y)/x + y(C) z(x + y)/y – x(D) xy(x – y)/x + y(E) xy(y – x)/x + y

Answer:[tex]\Delta d =\frac{y-x}{y}*d_{Hs}\[/tex]Step-by-step explanation:Fist the velocity of high-speed train will be given by:[tex]V_{Hs} =\frac{z}{x}[/tex]And the velocity of the regular will be given by:[tex]V_{R} =\frac{z}{y}[/tex]The position equations of each movement will be given by:[tex]d_{Hs} =\frac{z}{x} t[/tex][tex]d_{R} =z-\frac{z}{y} t[/tex]We can get the encounter point isolating t in the first equation and reepalcing it in the second one:First isolating:[tex]d_{Hs}\frac{x}{z} = t[/tex]Second reeplacing t:[tex]d_{R} =z-\frac{z}{y}*d_{Hs}\frac{x}{z}[/tex][tex]d_{R}=z-\frac{x}{y}*d_{Hs}\[/tex]In the  moment of the encounter the high-speed train have gone round  [tex]d_{Hs}\[/tex] and the regular one [tex]\frac{x}{y}*d_{Hs}\[/tex]Then the difference  between the distance traveled the high-speed train and the regular one is:[tex]\Delta d =d_{Hs}-\frac{x}{y}*d_{Hs}\[/tex][tex]\Delta d =\frac{y-x}{y}*d_{Hs}\[/tex]