document.write( "Question 187200: This is from my text book. If you could show me how to set it up to solve that would be great!! Problem: Donna is late for a sales meeting after traveling from one town to another at a speed of 32mph. If she had traveled 4mph faster, she could have made the trip in 1/2 hr less time. How far apart are the towns? \n" ); document.write( "

Algebra.Com's Answer #140325 by jojo14344(1513) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "
\n" ); document.write( "Just to show also of one way of looking the problem:
\n" ); document.write( "Based on our working eqn $\"system%28Speed=Distance%2Ftime%29\"$
\n" ); document.write( "We get, $\"Distance=Speed%2Atime\"$
\n" ); document.write( "We'll equate the 2 conditions: Normal Speed= Faster Speed
\n" ); document.write( " $\"S%5B1%5Dt=S%5B2%5D%28t-0.5hr%29\"$
\n" ); document.write( " $\"32%2At=highlight%2836%29%28t-0.5%29\"$ ---> $\"S%5B2%5D\"$ = 4 mph faster
\n" ); document.write( " $\"32t=36t-18\"$
\n" ); document.write( " $\"18=36t-32t\"$ -----> $\"18=4t\"$ ----> $\"cross%2818%294.5%2Fcross%284%29=cross%284%29t%2Fcross%284%29\"$
\n" ); document.write( " $\"t=4.5hrs\"$
\n" ); document.write( "Therefore, going back to Normal Speed:
\n" ); document.write( " $\"D=S%5B1%5D%2At=32%2A4.5=highlight%28144miles%29\"$ , Distance bet. towns
\n" ); document.write( "Going back at Faster Speed:
\n" ); document.write( " $\"D=S%5B2%5D%28t-0.5%29=36%284.5-0.5%29=36%284%29=highlight%28144+miles%29\"$ , Distance bet. towns
\n" ); document.write( "Thank you,
\n" ); document.write( "Jojo \n" ); document.write( "

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