document.write( "Question 71972: A consultant traveled 4 hours to attend a meeting. The return trip took only 3 hours because the speed was 7 miles per hour faster. What was the consultant's speed each way? \n" ); document.write( "

Algebra.Com's Answer #51455 by Earlsdon(6294) $\"\"$ $\"About$

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Yuo can use the distance formula: d = rt where: d = distance, r = rate (speed), and t = time of trip. In this problem, the distance out is equal to the distance back, so $\"d%5B1%5D+=+d%5B2%5D\"$
\n" ); document.write( "For the outbound trip, t = 4 hrs, and speed = $\"r%5B1%5D\"$ so:
\n" ); document.write( " $\"d%5B1%5D+=+r%5B1%5D%284%29\"$
\n" ); document.write( "For the return trip, t = 3 hrs, and speed = $\"r%5B1%5D+%2B+7\"$ so:
\n" ); document.write( " $\"d%5B2%5D+=+%28r%5B1%5D%2B7%29%283%29\"$ but $\"d%5B1%5D+=+d%5B2%5D\"$ , so:
\n" ); document.write( " $\"r%5B1%5D%284%29+=+%28r%5B1%5D%2B7%29%283%29\"$ Simplify and solve for $\"r%5B1%5D\"$
\n" ); document.write( " $\"4r%5B1%5D+=+3r%5B1%5D%2B21\"$ Subtract $\"3r%5B1%5D\"$ from both sides.
\n" ); document.write( " $\"r%5B1%5D+=+21\"$ mph This is the outbound speed.
\n" ); document.write( " $\"r%5B2%5D+=+r%5B1%5D%2B7\"$
\n" ); document.write( " $\"r%5B2%5D+=+21%2B7\"$
\n" ); document.write( " $\"r%5B2%5D+=+28\"$ mph This is the return speed. \n" ); document.write( "

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