document.write( "Question 1010852: when chris drives his car to work the trip takes 1/2hr when he rides the bus it takes 3/4hr the average rate of driving is 12mph more than his rate when riding the bus. Find the distance he travels to work (note:Driving distance=Distance on bus) \n" ); document.write( "

Algebra.Com's Answer #626332 by rothauserc(4718) $\"\"$ $\"About$

You can put this solution on YOUR website!
we have two (rate * time = distance) equations
\n" ); document.write( "***********************************************************
\n" ); document.write( "let r be the rate for the bus
\n" ); document.write( "r * (0.75) = distance
\n" ); document.write( "***********************************************************
\n" ); document.write( "r + 12 is the rate for the car
\n" ); document.write( "(r+12) * (0.50) = distance
\n" ); document.write( "**********************************************************
\n" ); document.write( "now set the two equations equal to each other since the car and the bus cover the same distance
\n" ); document.write( "0.75r = 0.50r + 6
\n" ); document.write( "0.25r = 6
\n" ); document.write( "r = 24 mph
\n" ); document.write( "************************************************************
\n" ); document.write( "you can use either equation to find the distance traveled
\n" ); document.write( "24 * (0.75) = 18 miles
\n" ); document.write( " \n" ); document.write( "

\n" );