Question 588924
Working together, two pipes can fill one tank in 4 hours.
Rewrite this so it makes sense.
Working alone the smaller pipe would take 6 hours longer than the larger pipe to fill the tank
How many hours would it take the larger pipe the fill the tank, working alone? 
'
let x = time required by the larger pipe alone
then
(x+6) = time required by the smaller pipe alone
:
Let the completed job = 1 (a full tank)
:
Each pipe does a fraction of the job. The two fraction add up to 1
:
Large pipe = small pipe = a full tank
{{{4/x}}} + {{{4/((x+6))}}} = 1
multiply by x(x+6)
x(x+6)*{{{4/x}}} + x(x+6)*{{{4/((x+6))}}} = x(x+6)
cancel the denominators and you have
4(x+6) + 4x = x(x+6)
4x + 24 + 4x = x^2 + 6x
8x + 24 = x^2 + 6x
combine on the right
0 = x^2 + 6x - 8x - 24
a quadratic equation
x^2 -2x - 24 = 0
Factors to
(x-6)(x+4) = 0
the positive solution
x = 6 hrs, the large pipe alone
;
:
See if this checks out. (Small pipe will take 12 hrs alone)
{{{4/6}}} + {{{4/12}}} =
{{{2/3}}} + {{{1/3}}} = 1; confirms our solution