document.write( "Question 97149: I have tried what I thought to be the rate x time=distance formula for the problem below:\r
\n" ); document.write( "\n" ); document.write( "John and Bill left a bus terminal at the same time and travelled in opposite directions. John's bus was in heavy traffic and had to travel 20 mi/h slower than Bill's bus. After 3 hours, their buses were 270 miles apart. How fast was each bus going?\r
\n" ); document.write( "\n" ); document.write( "I tried doing:
\n" ); document.write( "x+x+20(3)=270
\n" ); document.write( "2x+60=270
\n" ); document.write( "-60=270-60
\n" ); document.write( "2x/2=210/2
\n" ); document.write( "x=105 mi/h
\n" ); document.write( "x+20=125 mi/h\r
\n" ); document.write( "\n" ); document.write( "This did not make much sense to me as to why a bus would be going 125 mi/h, and I just am not sure how to set up the correct formula. \n" ); document.write( "

Algebra.Com's Answer #70723 by checkley71(8403) $\"\"$ $\"About$

You can put this solution on YOUR website!
Distance=rate*time. Here you have to add both Bob's & John's distances.
\n" ); document.write( "270=3*b+3(b-20) [3*b is Bob's speed, 3(b-20)=John's speed]
\n" ); document.write( "270=3b+3b-60
\n" ); document.write( "6b=270+60
\n" ); document.write( "6b=330
\n" ); document.write( "b=330/6
\n" ); document.write( "b=55 mph.
\n" ); document.write( "proof
\n" ); document.write( "270=3*55+3(55-20)
\n" ); document.write( "270=165+3*35
\n" ); document.write( "270=165+105
\n" ); document.write( "270=270 \n" ); document.write( "

\n" );