document.write( "Question 100770: On Wednesday, Roman drove to work in 45 minutes. On Thursday he averaged 12miles per hour more, and it took him 9 minutes less to get to work. Hoew far does he travel to work? \n" ); document.write( "
Algebra.Com's Answer #73380 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
distance(d)=rate(r) times time(t) or d=rt;r=d/t and t=d/r\r
\n" ); document.write( "\n" ); document.write( "Let d=distance to work
\n" ); document.write( "And let r=his rate on Wednesday
\n" ); document.write( "Then r+12=his rate on Thursday\r
\n" ); document.write( "\n" ); document.write( "distance to work on Wed=rt=r*45 or 45r
\n" ); document.write( "distance to work on Thursday=rt=(r+12)*(45-9)=36(r+12)\r
\n" ); document.write( "\n" ); document.write( "distance to work on Wed=distance to work on Thu. So our eq to solve is:\r
\n" ); document.write( "\n" ); document.write( "45r=36(r+12) get rid of parens\r
\n" ); document.write( "\n" ); document.write( "45r=36r+432 subtract 36 from both sides
\n" ); document.write( "45r-36r=432 collect like terms\r
\n" ); document.write( "\n" ); document.write( "9r=432 divide both sides by 9\r
\n" ); document.write( "\n" ); document.write( "r=48mph--------------------rate on Wed
\n" ); document.write( "r+12=48+12=60mph---------------rate on Thu\r
\n" ); document.write( "\n" ); document.write( "d=rt=48*(3/4)=36 mi (Wed)(note:45 min=3/4 hr)--------------------Ans
\n" ); document.write( "d=rt=60*(3/5)=36 mi (thu)(note: 36 min=3/5hr)\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps--ptaylor\r
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