document.write( "Question 142815: At 9 pm, Lucy started to gallop her horse at 20 miles per hour to catch Edmund, who had a three hour head start, and if Lucy was still 30 miles behind at 11 om, how fast was Edmund riding? \n" ); document.write( "

Algebra.Com's Answer #103906 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r
\n" ); document.write( "\n" ); document.write( "Let r=Edmund's rate of speed
\n" ); document.write( "Before Lucy started, Edmund had already travelled 3r miles. \r
\n" ); document.write( "\n" ); document.write( "In two hours, Lucy had travelled 20*2 or 40 miles
\n" ); document.write( "And in the same two hours, Edmund travelled another 2r miles\r
\n" ); document.write( "\n" ); document.write( "Now we are told that the distance Edmund travelled minus 30 mi equals the distance Lucy travelled, so:\r
\n" ); document.write( "\n" ); document.write( "5r-30=40 add 30 to each side
\n" ); document.write( "5r=70 divide each side by 5\r
\n" ); document.write( "\n" ); document.write( "r=14 mph----------------------------Edmund's rate of speed\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "3*14+2*14-30=40
\n" ); document.write( "42+28-30=40
\n" ); document.write( "70-30=40
\n" ); document.write( "40=40\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor \n" ); document.write( "

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