document.write( "Question 320359: At first two different pipes running together will fill a tank in 20/3 minutes. The rate that water runs through each of the pipes is then adjusted. If one pipe, running alone, takes 1 minute less to fill the tank at its new rate, and the other pipe, running alone, takes 2 minutes more to fill the tank at its new rate, then the two running together will fill the tank in 7 minutes. Find in what time the tank will be filled by each pipe running alone at the new rates. \n" ); document.write( "
Algebra.Com's Answer #229563 by ankor@dixie-net.com(22740)\"\" \"About 
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At first two different pipes running together will fill a tank in 20/3 minutes.
\n" ); document.write( " The rate that water runs through each of the pipes is then adjusted.
\n" ); document.write( " If one pipe, running alone, takes 1 minute less to fill the tank at its new
\n" ); document.write( "rate,
\n" ); document.write( " and the other pipe, running alone, takes 2 minutes more to fill the tank
\n" ); document.write( " at its new rate,
\n" ); document.write( " then the two running together will fill the tank in 7 minutes.
\n" ); document.write( ":
\n" ); document.write( " Find in what time the tank will be filled by each pipe running alone at the new rates.
\n" ); document.write( ":
\n" ); document.write( "let a = first pipe original time alone (in minutes)
\n" ); document.write( "let b = second pipe original time alone
\n" ); document.write( "then
\n" ); document.write( "(a-1) = first pipe adjusted rate time alone
\n" ); document.write( "(b+2) = 2nd pipe adjusted rate time alone
\n" ); document.write( ":
\n" ); document.write( "Let the completed job = 1
\n" ); document.write( ":
\n" ); document.write( "Original, working together equation
\n" ); document.write( "\"%2820%2F3%29%2Fa\" + \"%2820%2F3%29%2Fb\" = 1
\n" ); document.write( "We convert this to
\n" ); document.write( "\"20%2F%283a%29\" + \"20%2F%283b%29\" = 1
\n" ); document.write( "multiply by 3ab, results
\n" ); document.write( "20b + 20a = 3ab
\n" ); document.write( "20b - 3ab = -20a
\n" ); document.write( "b(20-3a) = -20a
\n" ); document.write( "b = \"-%2820a%29%2F%2820-3a%29\"; use this form for substitution
\n" ); document.write( ":
\n" ); document.write( "Adjusted, working together equation
\n" ); document.write( "\"7%2F%28a-1%29\" + \"7%2F%28b%2B2%29\" = 1
\n" ); document.write( "multiply by (a-1)(b+2), results
\n" ); document.write( "7(b+2) + 7(a-1) = (a-1)(b+2)
\n" ); document.write( "7b + 14 + 7a - 7 = ab + 2a - b - 2
\n" ); document.write( "Combine
\n" ); document.write( "7b + b + 7a - 2a + 14 - 7 + 2 = ab
\n" ); document.write( "8b + 5a + 9 = ab
\n" ); document.write( "Multiply by 3,
\n" ); document.write( "24b + 15a + 27 = 3ab
\n" ); document.write( "The two equations = 3ab, therefore
\n" ); document.write( "24b + 15a + 27 = 20b + 20a
\n" ); document.write( "24b - 20b + 15a - 20a = -27
\n" ); document.write( "4b - 5a = -27
\n" ); document.write( "replace b with \"-%2820a%29%2F%2820-3a%29\"
\n" ); document.write( "4(\"-%2820a%29%2F%2820-3a%29\") - 5a = -27
\n" ); document.write( "\"-%2880a%29%2F%2820-3a%29\" - 5a = -27
\n" ); document.write( "Get rid of all those negatives, multiply by -1
\n" ); document.write( "\"%2880a%29%2F%2820-3a%29\" + 5a = 27
\n" ); document.write( "Multiply by (20-3a), results
\n" ); document.write( "80a + 5a(20-3a) = 27(20-3a)
\n" ); document.write( "80a + 100a - 15a^2 = 540 - 81a
\n" ); document.write( "180a - 15a^2 = 540 - 81a
\n" ); document.write( "Combine on the right
\n" ); document.write( "0 = 15a^2 - 81a - 180a + 540
\n" ); document.write( "A quadratic equation
\n" ); document.write( "15a^2 - 261a + 540 = 0
\n" ); document.write( "Simplify, divide by 3
\n" ); document.write( "5a^2 - 87a + 180 = 0
\n" ); document.write( "Factor this
\n" ); document.write( "(5a-12)(a-15) = 0
\n" ); document.write( "Reasonable solution here
\n" ); document.write( "a = 15 hrs pipe a originally
\n" ); document.write( "then
\n" ); document.write( "15 - 1 = 14 hrs pipe a alone at the new rate
\n" ); document.write( ":
\n" ); document.write( "Find b original value
\n" ); document.write( "b = \"-%2820a%29%2F%2820-3a%29\"
\n" ); document.write( "b = \"-%2820%2815%29%29%2F%2820-3%2815%29%29\"
\n" ); document.write( "b = \"-%28300%29%2F%2820-45%29\"
\n" ); document.write( "b = \"%28-300%29%2F%28-25%29\"
\n" ); document.write( "b = +12 hrs pipe b originally
\n" ); document.write( "then
\n" ); document.write( "12 + 2 = 14 hrs pipe b alone at the new rate
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check solution in the adjusted equation
\n" ); document.write( "\"7%2F14\" + \"7%2F14\" = 1
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Seems like there should be an easier way to do this, but I see no one else has
\n" ); document.write( "worked this problem in 12 hrs, so here it is.\r
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