document.write( "Question 165248: a motorcycle breaks down and the rider must walk the rest of the way to work.The motorcyle was riden 45mph and the rider walks at the speed of 6mph. the distance from home to work is 25mi.And the total of the trip was 2hrs. How far did the motorcycle go before it broke down? \r
\n" ); document.write( "\n" ); document.write( "Question how do you set this system of equation up?
\n" ); document.write( "

Algebra.Com's Answer #121797 by checkley77(12844)\"\" \"About 
You can put this solution on YOUR website!
This rider/walker rode @ 45 mph. for x hours (45x) then walked @ 6 mph. for
\n" ); document.write( "(2-x) hours.
\n" ); document.write( "-----------------------------------------------------------
\n" ); document.write( "45x+6(2-x)=25
\n" ); document.write( "45x+12-6x=25
\n" ); document.write( "39x=25-12
\n" ); document.write( "39x=13
\n" ); document.write( "x=13/39
\n" ); document.write( "x=.333 hours for the 45 mph trip.
\n" ); document.write( "45*.333=15 miles is the motorcycle trip distance.
\n" ); document.write( "Proof:
\n" ); document.write( "45*.333+6(2-.333)=25
\n" ); document.write( "15+6(1.667)=25
\n" ); document.write( "15+10=25
\n" ); document.write( "25=25
\n" ); document.write( "
\n" );