document.write( "Question 972881: Praying that someone out there can help me assist my daughter with this problem.\r
\n" ); document.write( "\n" ); document.write( "Two girls went on a run. Cary began running 40 seconds earlier than Kristie.
\n" ); document.write( " *Cary ran at a speed of 10 feet per second
\n" ); document.write( " *Kristie ran at a speed of 12 feet per second
\n" ); document.write( "For how many seconds had Cary been running at the moment when the two girls had ran exactly the same distance.\r
\n" ); document.write( "\n" ); document.write( "This problem falls under \"Linear equations and proportions.\"
\n" ); document.write( "Any feedback would be greatly appreciated.
\n" ); document.write( "Thanks in advance. \n" ); document.write( "
Algebra.Com's Answer #595121 by macston(5194)\"\" \"About 
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\n" ); document.write( "distance=rate x time
\n" ); document.write( "Since distance is the same:
\n" ); document.write( "Carey's rate x Carey's time=Kristie's rate x (Carey's time - 40 seconds)
\n" ); document.write( "Let T=Carey's time
\n" ); document.write( "(10ft/s)(T)=(12ft/s)(T-40s)
\n" ); document.write( "(10ft/s)T=(12ft/s)T-480ft Subtract (12ft/s)T from each side.
\n" ); document.write( "(-2ft/s)T=-480ft Divide each side by (-2ft/sec)
\n" ); document.write( "T=240s Carey's time when both girls had run the same distance was 240 seconds.
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