document.write( "Question 823617: Art leaves his house at 10 o'clock traveling at 50 mph. Jennifer leaves her house at 10:30 traveling at 60 mph. At what time will Jennifer catch up to Art? \n" ); document.write( "

Algebra.Com's Answer #495700 by algebrahouse.com(1659) $\"\"$ $\"About$

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Distance = rate x time\r
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\n" ); document.write( "\n" ); document.write( "Art's distance
\n" ); document.write( "D = rt {distance = rate x time}
\n" ); document.write( "rate = 50
\n" ); document.write( "time = t
\n" ); document.write( "d = 50t\r
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\n" ); document.write( "\n" ); document.write( "Jennifer's distance
\n" ); document.write( "D = rt {distance = rate x time}
\n" ); document.write( "rate = 60
\n" ); document.write( "time = t - 0.5 {left 1/2 hour later than Art}
\n" ); document.write( "d = 60(t - 0.5) = 60t - 30 {used distributive property}\r
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\n" ); document.write( "\n" ); document.write( "50t = 60t - 30 {when Jennifer catches up with Art, the distances will be equal}
\n" ); document.write( "-10t = -30 {subtracted 60t from each side}
\n" ); document.write( "t = 3 {divided each side by -10}\r
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\n" ); document.write( "\n" ); document.write( "In 3 hours, Jennifer will catch up with Art
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