document.write( "Question 1141846: Two people leave from two towns that are 195 miles apart at the same time
\n" ); document.write( "and travel along the same road toward each other. The first person drives 5
\n" ); document.write( "miles slower than the second person. If they meet in 3 hours, at what rate
\n" ); document.write( "of speed did each travel? \n" ); document.write( "
Algebra.Com's Answer #762478 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
An alternate way to work such problems is to use rate of approach when two\r\n" );
document.write( "things are approaching each other, or rate of separation if they are separating.\r\n" );
document.write( "\r\n" );
document.write( "Let r = the rate of the second person. \r\n" );
document.write( "\r\n" );
document.write( "
The first person drives 5 miles slower than the second person.
\r\n" );
document.write( "\r\n" );
document.write( "So the rate of the first person is r-5.\r\n" );
document.write( "\r\n" );
document.write( "Their approach rate is the sum of their rates or r+r-5 or 2r-5 miles per hour.\r\n" );
document.write( "\r\n" );
document.write( "
If they meet in 3 hours,
\r\n" );
document.write( "\r\n" );
document.write( "Distance = Rate × Time\r\n" );
document.write( "     195 = (2r-5)×(3)\r\n" );
document.write( "     195 = 3(2r-5)\r\n" );
document.write( "     195 = 6r-15\r\n" );
document.write( "     210 = 6r\r\n" );
document.write( "      35 = r\r\n" );
document.write( "\r\n" );
document.write( "So the faster one went r=35 mph and the slower one went r-5=30 mph.\r\n" );
document.write( "\r\n" );
document.write( "Checking:  The faster one went 35 mph for 3 hours and so went 105 miles.\r\n" );
document.write( "The slower one went 30 mph for 3 hours and so went 90 miles. And sure\r\n" );
document.write( "enough the total distance they covered was 105+90 = 195 miles.\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );