document.write( "Question 695774:
\n" ); document.write( "On Monday, Roger drove to work in 38 minutes. On Tuesday, he averaged 10 miles per hour more, and it took him 5 minutes less to get to work. How far (in miles) does he travel to work?
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Algebra.Com's Answer #428774 by ankor@dixie-net.com(22740)\"\" \"About 
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On Monday, Roger drove to work in 38 minutes.
\n" ); document.write( " On Tuesday, he averaged 10 miles per hour more, and it took him 5 minutes less to get to work.
\n" ); document.write( " How far (in miles) does he travel to work?
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\n" ); document.write( "Find his speed first
\n" ); document.write( ":
\n" ); document.write( "The dist is the same each day, write a distance equation dist = time * speed
\n" ); document.write( "we have to change the minutes to hours
\n" ); document.write( ":
\n" ); document.write( "\"38%2F60\"s = \"33%2F60\"*(s+10)
\n" ); document.write( "multiply both sides by 60
\n" ); document.write( "38s = 33(s + 10)
\n" ); document.write( "38s - 33s = 330
\n" ); document.write( "5s = 330
\n" ); document.write( "s = 330/5
\n" ); document.write( "s = 66 mph on Monday
\n" ); document.write( "and
\n" ); document.write( "76 mph on Tuesday
\n" ); document.write( ":
\n" ); document.write( "Find the distance
\n" ); document.write( "\"38%2F60\"(66) = 41.8 miles
\n" ); document.write( "check that the distance is the same on Tuesday
\n" ); document.write( "\"33%2F60\"(76) = 41.8 miles
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