document.write( "Question 562303: Running at the average rate of 400 m/minute, Pam dashed to the end of the race course. She then walked back to her starting position at an average rate of 80 m/minute. How long was the course if it took her 6 minutes total to make the dash and the walk back. I cannot figure out the formula \n" ); document.write( "

Algebra.Com's Answer #364533 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"t\"$ = number of minutes for the run
\n" ); document.write( " $\"+6+-+t+\"$ = number of minutes for the walk
\n" ); document.write( "Let $\"+d+\"$ = length of course in meters
\n" ); document.write( "--------------
\n" ); document.write( "Equation for run:
\n" ); document.write( " $\"+d+=+400t+\"$
\n" ); document.write( "Equation for walk:
\n" ); document.write( " $\"+d+=+80%2A%28+6+-+t+%29+\"$
\n" ); document.write( "-----------------
\n" ); document.write( "By substitution:
\n" ); document.write( " $\"+400t+=+80%2A%28+6+-+t+%29+\"$
\n" ); document.write( " $\"+400t+=+480+-+80t+\"$
\n" ); document.write( " $\"+480t+=+480+\"$
\n" ); document.write( " $\"+t+=+1+\"$ min
\n" ); document.write( " $\"+6-+t+=+6+-+1+\"$
\n" ); document.write( " $\"+6+-+t+=+5+\"$ min
\n" ); document.write( "---------------
\n" ); document.write( " $\"+d+=+400t+\"$
\n" ); document.write( " $\"+d+=+400%2A1+\"$
\n" ); document.write( " $\"+d+=+400+\"$
\n" ); document.write( "check:
\n" ); document.write( " $\"+d+=+80%2A%28+6+-+t+%29+\"$
\n" ); document.write( " $\"+d+=+80%2A5+\"$
\n" ); document.write( " $\"+d+=+400+\"$
\n" ); document.write( "The length of the course was 400 m \n" ); document.write( "

\n" );