document.write( "Question 500880: can someone please help me?\r
\n" ); document.write( "\n" ); document.write( "A porsche and VW bug that were 450 miles apart traveled toward eachother. They met after 5 hours. If one car traveled twice as fast as the other, what were their speed? \n" ); document.write( "
Algebra.Com's Answer #338317 by geetha_rama(94)\"\" \"About 
You can put this solution on YOUR website!
Let x be the location of Porshe car and y be the lcoation of VW
\n" ); document.write( "distance between x and y = 450 miles
\n" ); document.write( "speed of porshe = a
\n" ); document.write( "speed of VW = b\r
\n" ); document.write( "\n" ); document.write( "b = 2a -> equation 1(speed of 1 car is twice the other) \r
\n" ); document.write( "\n" ); document.write( "After 5 hours, the cars meet.
\n" ); document.write( "i.e., both the cars together would have covered the 450 miles in between them
\n" ); document.write( "distance = spped * time
\n" ); document.write( "and hence 5a + 5b = 450
\n" ); document.write( "dividing by 5,
\n" ); document.write( " => a+b = 90
\n" ); document.write( " 3a = 90 (from equation 1)
\n" ); document.write( "a = 30
\n" ); document.write( "b = 2a = 60\r
\n" ); document.write( "\n" ); document.write( "Hence the speed of the cars are 30 miles/hour and 60 miles/hour\r
\n" ); document.write( "\n" ); document.write( "Check: 5a + 5b = 450 => 5*30 + 5* 60 = 150 + 300 = 450
\n" ); document.write( " \n" ); document.write( "
\n" );