document.write( "Question 1048117: A tour bus averaged 50 miles per hour between two cities on the first leg of a trip and 45 miles per hour on the return trip. The return trip took 1/2 hour longer. Find the distance between the two cities. \n" ); document.write( "

Algebra.Com's Answer #663689 by josgarithmetic(39626) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "                    RATE             TIME          DISTANCE\r\n" );
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document.write( "ONE WAY             50                t             d\r\n" );
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document.write( "OTHER WAY           45               t+1/2          d\r\n" );
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\n" ); document.write( "\n" ); document.write( "Generalizing the situation using only variables:
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document.write( "g for the rate GOING and r for the rate RETURNING;\r\n" );
document.write( "Unknown variables are t and d.\r\n" );
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document.write( "                    RATE             TIME          DISTANCE\r\n" );
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document.write( "ONE WAY             g                t              d\r\n" );
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document.write( "OTHER WAY           r               t+h             d\r\n" );
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\n" ); document.write( "\n" ); document.write( "The question means, find d.
\n" ); document.write( " $\"system%28gt=d%2Cr%28t%2Bh%29=d%29\"$
\n" ); document.write( "-\r
\n" ); document.write( "\n" ); document.write( " $\"t=d%2Fg\"$
\n" ); document.write( "-
\n" ); document.write( " $\"r%28d%2Fg%2Bh%29=d\"$
\n" ); document.write( " $\"rd%2Fg%2Brh=d\"$
\n" ); document.write( " $\"rd%2Fg-d%2Brh=0\"$
\n" ); document.write( " $\"rd%2Fg-d=-rh\"$
\n" ); document.write( " $\"rd-gd=-rgh\"$
\n" ); document.write( " $\"d%28r-g%29=-rgh\"$
\n" ); document.write( " $\"d%28g-r%29=rgh\"$
\n" ); document.write( " $\"highlight%28d=rgh%2F%28g-r%29%29\"$ \n" ); document.write( "

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