document.write( "Question 199619: I had an algebra test this morning and I am still trying to figure out this problem. I'm sure it is not near as hard as I am making it. The problem read: Bob leaves the camp at 3:00 paddling downstream at 4mph (stillwater) with a 2mph current. Joe leaves the camp at 4:30 paddling downstream twice as fast as Bob (the water is flowing at the same rate). What time will Bob and Joe meet? \n" ); document.write( "
Algebra.Com's Answer #150021 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
Bob leaves the camp at 3:00 paddling downstream at 4mph (stillwater) with a 2mph current.
\n" ); document.write( "Joe leaves the camp at 4:30 paddling downstream twice as fast as Bob (the water
\n" ); document.write( "is flowing at the same rate).
\n" ); document.write( " What time will Bob and Joe meet?
\n" ); document.write( ":
\n" ); document.write( "Let t = the travel time of Bob
\n" ); document.write( "then
\n" ); document.write( "(t-1.5) = travel time of Joe (leaves 1.5 hrs after Bob)
\n" ); document.write( ";
\n" ); document.write( "Bob's speed with the current: 4 + 2 = 6 mph
\n" ); document.write( "Joe's speed with the current: 8 + 2 = 10 mph
\n" ); document.write( ":
\n" ); document.write( "When they meet they will have traveled the same distance. Write a dist equation:
\n" ); document.write( "Dist = speed * time
\n" ); document.write( ":
\n" ); document.write( "B's dist = J's dist
\n" ); document.write( "6t = 10(t-1.5)
\n" ); document.write( "6t = 10t - 15
\n" ); document.write( "15 = 10t - 6t
\n" ); document.write( "4t = 15
\n" ); document.write( "t = \"15%2F4\"
\n" ); document.write( "t = 3.75 hrs from 3:00 is 6:45 when they meet
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check solution by ensuring they did, in fact, travel the same distance.
\n" ); document.write( "6*3.75 = 22.5 mi
\n" ); document.write( "10*2.25 = 22.5 mi \n" ); document.write( "
\n" );