Question 789284
(a)
With just small pipe:
Rate of filling is ( 1 tank ) / ( x min )
Multiply by ( 1 min ) to get fraction 
of tank filled in {{{ 1 }}} min
{{{ ( 1/x)*1 = 1/x }}}
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With just large pipe:
{{{ ( 1/(x-8) )*1 = 1/( x-8 ) }}}
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(b)
{{{ 1/x + 1/(x-8) = 1/7.5 }}}
{{{ 1/x + 1/(x-8) = 2/15 }}}
Multiply both sides by {{{ x*(x-8)*15 }}}
{{{ 15*(x-8) + 15x = 2*x*(x-8) }}}
{{{ 15x - 120 + 15x = 2x^2 - 16x }}}
{{{ 2x^2 - 46x + 120 = 0 }}}
{{{ x^2 - 23x + 60 = 0 }}}
Use the quadratic formula
{{{ x = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = -23 }}}
{{{ c = 60 }}}
{{{ x = ( -(-23) +- sqrt( (-23)^2 - 4*1*60 )) / (2*1) }}}
{{{ x = ( 23 +- sqrt( 529 - 240 )) / 2 }}}
{{{ x = ( 23 +- sqrt( 289 )) / 2 }}}
{{{ x = ( 23 + 17 ) / 2 }}}
{{{ x = 40/2 }}}
{{{ x = 20 }}}
and, also,
{{{ x = 23 - 17 ) / 2 }}}
{{{ x = 6/2 }}}
{{{ x = 3 }}}
The answer can't be {{{ x=3 }}} because {{{ 8 }}} cannot be
subtracted from it to get a positive number
So, {{{ x = 20 }}} min
and
{{{ x - 8 = 20 - 8 }}}
{{{ x - 8 = 12 }}}
The large pipe fills the tank in 12 min