document.write( "Question 95190: A cargo plane flew to the maintenance facility and back. It took one hour less time to get there than it did to get back. The average speed on the trip there was 220 mph. The average speed on the way back was 200 mph. How many hours did the trip take? \n" ); document.write( "

Algebra.Com's Answer #69301 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
distance(d)=rate(r) times time(t) or d=rt; r=d/t and t=d/r
\n" ); document.write( "Let t=time coming back
\n" ); document.write( "Then t-1=time going
\n" ); document.write( "distance going =rate going times time going =220*(t-1)\r
\n" ); document.write( "\n" ); document.write( "distance coming back=rate coming back times time coming back=200*t\r
\n" ); document.write( "\n" ); document.write( "distance going=distance coming back so our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "220(t-1)=200t get rid of parens\r
\n" ); document.write( "\n" ); document.write( "220t-220=200t subtract 220t from both sides\r
\n" ); document.write( "\n" ); document.write( "220t-220t-220=200t-220t collect like terms\r
\n" ); document.write( "\n" ); document.write( "-20t=-220 divide both sides by -20\r
\n" ); document.write( "\n" ); document.write( "t=11 hours ----------------------------time coming back
\n" ); document.write( "t-1=11-1=10 hours ------------------time going\r
\n" ); document.write( "\n" ); document.write( "Total time of trip =11+10=21 hours\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "distance going =220*10=2200 mi\r
\n" ); document.write( "\n" ); document.write( "distance coming =200*11=2200mi\r
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