document.write( "Question 558650: John drove to a distant city in 5 hours. When he returned, there was less traffic, and the trip took only 3 hours. If John averaged 26 Mph faster in the return trip, how fast did he drive each way? \n" ); document.write( "

Algebra.Com's Answer #363108 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"+d+\"$ = distance to city
\n" ); document.write( "Let $\"+s+\"$ = his speed going to the city
\n" ); document.write( "given:
\n" ); document.write( "Going to city:
\n" ); document.write( " $\"+d+=+s%2A5+\"$
\n" ); document.write( "Coming back:
\n" ); document.write( " $\"+d+=+%28+s+%2B+26+%29%2A3+\"$
\n" ); document.write( "By substitution:
\n" ); document.write( " $\"+5s+=+%28+s+%2B+26+%29%2A3+\"$
\n" ); document.write( " $\"+5s+=+3s+%2B+78+\"$
\n" ); document.write( " $\"+2s+=+78+\"$
\n" ); document.write( " $\"+s+=+39+\"$
\n" ); document.write( " $\"+s+%2B+26+=+65+\"$
\n" ); document.write( "He drove 39 mi/hr to the city and 65 mi/hr coming back \n" ); document.write( "

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