A train covered a certain distance …

Suppose the actual speed of the train be x km/hr and the actual time taken be y hours. Then,
Distance covered =(xy) km ...........(i) [{tex}\therefore{/tex} Distance = Speed {tex}\times{/tex}Time]
If the speed is increased by {tex}6km/hr{/tex}, then time of journey is reduced by 4 hours i.e., when speed is {tex}(x + 6)km/hr{/tex}, time of journey is {tex}(y-4)hours.{/tex}
{tex}\therefore{/tex} Distance covered = {tex}(x + 6) (y - 4){/tex}
{tex}\Rightarrow{/tex}{tex}xy = (x + 6) (y - 4){/tex} [Using (i)]
{tex}\Rightarrow{/tex}{tex}-4x + 6y -24=0{/tex}
{tex}\Rightarrow{/tex}{tex}-2x + 3y -12=0{/tex} ..........(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 {tex}(x-6)km/hr{/tex}, time of journey is {tex}(y - 6)hours.{/tex}
{tex}\therefore{/tex} Distance covered ={tex}(x -6) (y + 6){/tex}
{tex}\Rightarrow{/tex} {tex}xy = (x - 6) (y + 6){/tex} [Using (i)]
{tex}\Rightarrow{/tex} {tex}6x - 6y -36 =0{/tex}
{tex}\Rightarrow{/tex} {tex}x - y -6=0 {/tex}...........(iii)
Thus, we obtain the following system of equations:
{tex}-2x + 3y -12 =0{/tex}
{tex}x - y - 6=0{/tex}
By using cross-multiplication, we have
{tex}\frac { x } { 3 \times - 6 - ( - 1 ) \times - 12 } = \frac { - y } { - 2 \times - 6 - 1 \times - 12 } = \frac { 1 } { - 2 \times - 1 - 1 \times 3 }{/tex}
{tex}\Rightarrow \quad \frac { x } { - 30 } = \frac { - y } { 24 } = \frac { 1 } { - 1 }{/tex}
{tex}\Rightarrow{/tex} {tex}x =30\ and\ y=24.{/tex}
Putting the values of x and y in equation (i), we obtain
Distance= {tex}(30\times24)km =720km.{/tex}
Hence, the length of the journey is {tex}720 km{/tex}.