document.write( "Question 118934: a motorcycle breaks down and the rider has to walk the rest of the way to work.the motorcycle was traveling at 45mi/h,and the rider walks at a speed of 6mi/h.the distance from home to work is 25 miles,and the total time for the trip was 2 hrs.how far did the motorcycle go before it broke down? \n" ); document.write( "

Algebra.Com's Answer #87027 by mayank(15) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let the motorcycle ride for x hrs and the rider walk for y hrs.\r
\n" ); document.write( "\n" ); document.write( "Then as we know total distance travelled is 25 miles and rider is traveling at 6 mi/hr and motorcycle at 45mil/hr we can have equation as
\n" ); document.write( "45x+6y=25----(i)\r
\n" ); document.write( "\n" ); document.write( "now as total duration of trip is 2 hrs, x+y=2---(ii)\r
\n" ); document.write( "\n" ); document.write( "Thus we get two equations and we need to find two variables.\r
\n" ); document.write( "\n" ); document.write( "multiplying equation (ii) by 6 we get 6x+6y=12---(iii)\r
\n" ); document.write( "\n" ); document.write( "subtracting equation iii from equation (i) we get 39x=13 or x=1/3 \r
\n" ); document.write( "\n" ); document.write( "Substituting value of x in equation (ii) we get y=5/3.\r
\n" ); document.write( "\n" ); document.write( "Thus total miles covered by motorcycle is 45x= 45*(1/3)=45/3=15 miles.\r
\n" ); document.write( "\n" ); document.write( "Thus ans is 15 miles. \n" ); document.write( "

\n" );