document.write( "Question 1059932: A train traveling at the rate of 90 miles per hour (mi/hr) leaved New York City. Two hours later, another train traveling at the rate of 120 mi/hr also leaves New York City on a parallel track. How long will it take the faster train to catch up to the slower train? \n" ); document.write( "

Algebra.Com's Answer #674970 by algebrahouse.com(1659) $\"\"$ $\"About$

You can put this solution on YOUR website!
distance = rate x time
\n" ); document.write( "d = rt\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "One Train:
\n" ); document.write( "rate = 90
\n" ); document.write( "time = t
\n" ); document.write( "d = 90t {distance = rate x time}\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Other Train
\n" ); document.write( "rate = 120
\n" ); document.write( "time = t - 2 {left 2 hours later}
\n" ); document.write( "d = 120(t - 2) {distance = rate x time}\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "When the faster train catches up with the slower train, their distances will be equal.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "90t = 120(t - 2) {set distances equal to each other}
\n" ); document.write( "90t = 120t - 240 {used distributive property}
\n" ); document.write( "-30t = -240 {subtracted 120t from each side}
\n" ); document.write( "t = 8 {divided each side by -30}\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "t - 2 corresponds to the time of the faster train
\n" ); document.write( "= 8 - 2 {substituted 8, in for t, into (t - 2)
\n" ); document.write( "= 6 {subtracted}\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "It will take the faster train 6 hours to catch up with the slower train.
\n" ); document.write( "
For more help from me, visit: www.algebrahouse.com

\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );