Question 984537
{{{x}}}= speed of the freight train in miles per hour.
{{{x+56}}}= speed of the passenger train in miles per hour.
{{{1&3/4}}}{{{hours=1.75}}}{{{hours}}}
{{{45minutes(1hour/"60 minutes")=3/4}}}{{{hours=0.75hours}}}
A freight train going at {{{x}}}{{{mph}}} for {{{1.75hours}}}
covers a distance of {{{1.75*x}}}{{{miles}}} .
A passenger train going at {{{x+56}}}{{{mph}}} for {{{0.75hours}}}
covers a distance of {{{0.75*(x+56)}}}{{{miles}}} .
It is he same distance, so
{{{1.75*x=0.75*(x+56)}}}
{{{1.75x=0.75x+0.75*56}}}
{{{1.75x=0.75x+42}}}
{{{1.75x-0.75x=42}}}
{{{(1.75-0.75)x=42}}}
{{{(1)x=42}}}--->{{{x=42}}}--->{{{x+56=42+56=highlight(98)}}} .
The average speed of passenger train is {{{highlight(98mph)}}} .