document.write( "Question 521501: A cyclist leaves Toronto at 10am riding West. At noon, a car starts along the same road, driving 3 times as fast as the bike. When will the car catch up with the bike? \n" ); document.write( "
Algebra.Com's Answer #346499 by richwmiller(17219) You can put this solution on YOUR website! r*t=d \n" ); document.write( "We don't know time and we don't know distance. \n" ); document.write( "We know the car is 3 times as fast the bike. \n" ); document.write( "We also know the distances are the same for the bike and the car. \n" ); document.write( "We also know \n" ); document.write( "r*(t+2)=d \n" ); document.write( "3r*t=d \n" ); document.write( "r*(t+2)=3*r*t \n" ); document.write( "rt+2r=3rt \n" ); document.write( "2r=2rt \n" ); document.write( "1=t \n" ); document.write( "They will meet in one hour.\r \n" ); document.write( "\n" ); document.write( "Also \n" ); document.write( "r*(t+2)=3*r*t \n" ); document.write( "divide by r \n" ); document.write( "t+2=3t \n" ); document.write( "2=2t \n" ); document.write( "1=t \n" ); document.write( "They will meet in one hour\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |