document.write( "Question 974726: Sandy left Burbank driving 79 miles per hour. Mary, in order to catch up, drove 88 miles per hour. Mary caught up in 8 hours. How long was Sandy driving before Mary caught up?\r
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Algebra.Com's Answer #596577 by josgarithmetic(39620)\"\" \"About 
You can put this solution on YOUR website!
Sandy left first and Mary left later; but traveled less time than Sandy. Think about that
\n" ); document.write( "a while,... and then assign t as the travel time for Sandy BEFORE Mary began her travel.\r
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\n" ); document.write( "\n" ); document.write( "Each Sandy and Mary travel the same distance, d, upon catch-up.\r
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\n" ); document.write( "\n" ); document.write( "______________________speed____________time__________distance
\n" ); document.write( "Sandy_________________79_______________t+8____________d
\n" ); document.write( "Mary__________________88________________8_____________d\r
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\n" ); document.write( "\n" ); document.write( "\"system%2879%28t%2B8%29=d%2C88%2A8=d%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"d=704\"
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\n" ); document.write( "\"79t%2B78%2A8=704\"
\n" ); document.write( "\"79t=704-78%2A8\"
\n" ); document.write( "\"79t=80\"
\n" ); document.write( "\"t=80%2F79\" which IS close-enough to 1 hour. Slightly longer.
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\n" ); document.write( "1 HOUR 46 SECONDS
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