document.write( "Question 401141: During the first part of a trip a canoeist travels 48 miles at a certain speed. The canoeist travels 19 miles on the second part of the trip at a speed 5 mph slower. The total time for the trip is 3 hours. What was the speed on each part of the trip? \n" ); document.write( "

Algebra.Com's Answer #283986 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
Using the distance equals rate times time, d = r*t, we can write an equation for each part of the trip:
\n" ); document.write( " $\"d%5B1%5D+=+r%5B1%5D%2At%5B1%5D\"$
\n" ); document.write( " $\"d%5B2%5D+=+r%5B2%5D%2At%5B2%5D\"$
\n" ); document.write( "with the subscripts describing the part of the trip. Since we know the distances for each part we can substitute the numbers for those distaces:
\n" ); document.write( " $\"48+=+r%5B1%5D%2At%5B1%5D\"$
\n" ); document.write( " $\"19+=+r%5B2%5D%2At%5B2%5D\"$

\n" ); document.write( "At this point we have two equations and four variables. ( $\"r%5B1%5D\"$ and $\"r%5B2%5D\"$ count as separate variables. So do $\"t%5B1%5D\"$ and $\"t%5B2%5D\"$ .) So we need two more equations. We are told that the rate during the second part was 5 mph slower than the rate during the first part. So:
\n" ); document.write( " $\"r%5B2%5D+=+r%5B1%5D+-+5\"$
\n" ); document.write( "We are also told that the total time was 3 hours. So:
\n" ); document.write( " $\"t%5B1%5D+%2B+t%5B2%5D+=+3\"$
\n" ); document.write( "If we subtract $\"t%5B1%5D\"$ from each side we can solve for $\"t%5B2%5D\"$ :
\n" ); document.write( " $\"t%5B2%5D+=+3+-+t%5B1%5D\"$
\n" ); document.write( "Now we can use these last two equations and substitute for $\"r%5B2%5D\"$ and $\"t%5B2%5D\"$ in the $\"19+=+r%5B2%5D%2At%5B2%5D\"$ equation:
\n" ); document.write( " $\"19+=+%28r%5B1%5D-5%29%2A%283-t%5B1%5D%29\"$
\n" ); document.write( "which simplifies as follows:
\n" ); document.write( " $\"19+=+r%5B1%5D%2A3-r%5B1%5D%2At%5B1%5D-5%2A3-5%2At%5B1%5D\"$
\n" ); document.write( " $\"19+=+3r%5B1%5D-r%5B1%5D%2At%5B1%5D-15-5%2At%5B1%5D\"$
\n" ); document.write( "From $\"48+=+r%5B1%5D%2At%5B1%5D\"$ we know that $\"r%5B1%5D%2At%5B1%5D\"$ is 48. Replacing the $\"r%5B1%5D%2At%5B1%5D\"$ with 48 in the previous equation we get:
\n" ); document.write( " $\"19+=+3r%5B1%5D-48-15-5%2At%5B1%5D\"$
\n" ); document.write( "or
\n" ); document.write( " $\"19+=+3r%5B1%5D-63-5%2At%5B1%5D\"$
\n" ); document.write( "Adding 63 to each side we get:
\n" ); document.write( " $\"82+=+3r%5B1%5D-5%2At%5B1%5D\"$
\n" ); document.write( "Next we can solve for $\"t%5B1%5D\"$ in the equation $\"48+=+r%5B1%5D%2At%5B1%5D\"$ by dividing both sides by $\"r%5B1%5D\"$ :
\n" ); document.write( " $\"48%2Fr%5B1%5D+=+t%5B1%5D\"$
\n" ); document.write( "Now we can substitute for $\"t%5B1%5D\"$ in $\"82+=+3r%5B1%5D-5%2At%5B1%5D\"$ :
\n" ); document.write( " $\"82+=+3r%5B1%5D-5%2A%2848%2Fr%5B1%5D%29\"$
\n" ); document.write( "We finally have an equation with a single variable! We can now solve for $\"r%5B1%5D\"$ . Multiplying both sides by $\"r%5B1%5D\"$ (to eliminate the fraction) we get:
\n" ); document.write( " $\"82r%5B1%5D+=+3r%5B1%5D%5E2-240\"$
\n" ); document.write( "This is a quadratic equation so we want one side to be zero. Subtracting $\"82r%5B1%5D\"$ from each side we get:
\n" ); document.write( " $\"0+=+3r%5B1%5D%5E2-82r%5B1%5D-240\"$
\n" ); document.write( "There are just too many possible factors of 240 for me to want to try to factor this. So I will use the Quadratic Formula instead:
\n" ); document.write( " $\"r%5B1%5D+=+%28-%28-82%29+%2B-+sqrt%28%28-82%29%5E2-4%283%29%28-240%29%29%29%2F2%283%29\"$
\n" ); document.write( "which simplifies as follows:
\n" ); document.write( " $\"r%5B1%5D+=+%28-%28-82%29+%2B-+sqrt%286724-4%283%29%28-240%29%29%29%2F2%283%29\"$
\n" ); document.write( " $\"r%5B1%5D+=+%28-%28-82%29+%2B-+sqrt%286724%2B2880%29%29%2F2%283%29\"$
\n" ); document.write( " $\"r%5B1%5D+=+%28-%28-82%29+%2B-+sqrt%289604%29%29%2F2%283%29\"$
\n" ); document.write( " $\"r%5B1%5D+=+%2882+%2B-+sqrt%289604%29%29%2F6\"$
\n" ); document.write( " $\"r%5B1%5D+=+%2882+%2B-+98%29%2F6\"$
\n" ); document.write( "In long form this is:
\n" ); document.write( " $\"r%5B1%5D+=+%2882+%2B+98%29%2F6\"$ or $\"r%5B1%5D+=+%2882+-+98%29%2F6\"$
\n" ); document.write( "which simplify as:
\n" ); document.write( " $\"r%5B1%5D+=+%28180%29%2F6\"$ or $\"r%5B1%5D+=+%28-16%29%2F6\"$
\n" ); document.write( " $\"r%5B1%5D+=+30\"$ or $\"r%5B1%5D+=+%28-8%29%2F3\"$
\n" ); document.write( "(BTW, this means the equation would have factored into:
\n" ); document.write( " $\"0+=+%28r%5B1%5D-30%29%283r%5B1%5D%2B8%29\"$

\n" ); document.write( "The second solution for the rate is negative which makes no sense in the context of this word problem. So we will reject it.

\n" ); document.write( "So $\"r%5B1%5D\"$ , the rate during the first part of the trip, is 30 mph. The rate during the second part, $\"r%5B2%5D\"$ , is 5 mpg slower so it must be 30-5 or 25 mph. (Since the problem only asks for the rates we are finished. We could use these values to find $\"t%5B1%5D\"$ and $\"t%5B2%5D\"$ also.) \n" ); document.write( "

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