document.write( "Question 137639: On a round trip of 700km in each direction,Mr.anderson averaged 7km/h more returning than he did going.If the return trip took 5hours less than the trip going, find the time he spent traveling in each direction \n" ); document.write( "
Algebra.Com's Answer #100739 by ankor@dixie-net.com(22740)\"\" \"About 
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On a round trip of 700km in each direction,Mr.anderson averaged 7km/h more returning than he did going.If the return trip took 5hours less than the trip going, find the time he spent traveling in each direction
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\n" ); document.write( "From the text we can assume the trip out was 350 km and return trip was 350 km
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\n" ); document.write( "Let s = speed out
\n" ); document.write( "then
\n" ); document.write( "(s+7) = return speed
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\n" ); document.write( "Write a time equation Time = Distance/speed
\n" ); document.write( "It says it took 5 hrs less time to return
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\n" ); document.write( "\"350%2F%28%28s%2B7%29%29\" - 5 = \"350%2Fs\"
\n" ); document.write( "Multiply equation by s(x+7) to get rid of the denominators:
\n" ); document.write( "s(s+7)*\"350%2F%28%28s%2B7%29%29\" - 5(s(s+7)) = s(s+7)*\"350%2Fs\"
\n" ); document.write( "Leaving us with:
\n" ); document.write( "350s + 5(s^2 + 7s) = 350(s+7)
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\n" ); document.write( "350s + 5s^2 + 35s = 350s + 2450
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\n" ); document.write( "Subtract 350s from both sides and arrange as a quadratic equation on the left
\n" ); document.write( "5s^2 + 35s - 2450 = 0
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\n" ); document.write( "Simplify, divide by 5
\n" ); document.write( "s^2 + 7s - 490 = 0
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\n" ); document.write( "We have to use the quadratic formula:
\n" ); document.write( "\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
\n" ); document.write( "a = 1; b = 7; c = -490; solve for s;
\n" ); document.write( "\"s+=+%28-7+%2B-+sqrt%287%5E2-+4+%2A+1%2A+-490+%29%29%2F%282%2A1%29+\"
\n" ); document.write( "\"s+=+%28-7+%2B-+sqrt%2849+%2B+1960%29%29%2F%282%29+\"
\n" ); document.write( "Do the math here and you should get a positive solution:
\n" ); document.write( "s = 18.91 km/h is outbound speed
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\n" ); document.write( "18.91 + 7 = 25.91 km/hr return speed
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\n" ); document.write( "Find the time required for each direction:
\n" ); document.write( "\"350%2F18.91\" = 18.51 hrs outbound
\n" ); document.write( "and
\n" ); document.write( "\"350%2F25.91\" = 13.51 hrs return; note the 5 hr difference confirms our solutions\r
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