document.write( "Question 108864: AT 9:00 A.M. A TRUCK LEAVES THE TRUCK YARD AND
\n" ); document.write( "TRAVELS WEST AT A RATE OF 35 MI/HR.
\n" ); document.write( "TWO HOURS LATER, A SECOND TRUCK LEAVES ALONG THE SAME ROUTE
\n" ); document.write( "TRAVELING AT 70 MI/HR.
\n" ); document.write( "WHEN WILL THE SECOND TRUCK CATCH UP TO THE FIRST TRUCK?\r
\n" ); document.write( "\n" ); document.write( "THANKS
\n" ); document.write( "JMS
\n" ); document.write( "

Algebra.Com's Answer #79386 by checkley71(8403)\"\" \"About 
You can put this solution on YOUR website!
DISTANCE=RATE*TIME
\n" ); document.write( "SEEING AS THE DISTANCES ARE THE SAME WE HAVE:
\n" ); document.write( "35T=70(T-2)
\n" ); document.write( "35T=70T-140
\n" ); document.write( "35T-70T=-140
\n" ); document.write( "-35T=-140
\n" ); document.write( "T=-140/-35
\n" ); document.write( "T=4 HOURS THEY WILL MEET.
\n" ); document.write( "PROOF
\n" ); document.write( "35*4=70(4-2)
\n" ); document.write( "140=70*2
\n" ); document.write( "140=140
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