Question 1197446

We let the time taken when ....
the larger hose alone is used be{{{ (x - 6)}}}
and the smaller hose as {{{x}}}.

This would make their rates  and  respectively.

Take note that:

{{{1/rate = time }}}

so,

{{{1/rate}}}  of both hoses = {{{4}}}

rate of both hoses = {{{1/4}}}


{{{1/(x-6)+1/x=1/4}}}


{{{(x+(x-6))/x(x-6)=1/4}}}


{{{(2x-6)/(x^2-6x)=1/4}}}


{{{4(2x-6)=1(x^2-6x)}}}


{{{8x-24=x^2-6x}}}


{{{0=x^2-6x-8x+24}}}


{{{0=x^2-14x+24}}}


{{{0=x^2-2x-12x+24}}}


{{{0=(x^2-2x)-(12x-24)}}}


{{{0=x(x-2)-12(x-2)}}}


{{{0=(x-12)(x-2)}}}


This would mean that {{{x = 12}}} or {{{x=2}}} but {{{x }}}cannot be {{{2}}} since that would make the time taken for the larger hose negative.

Therefore it will take the smaller hose {{{12 }}}hours to fill the swimming pool alone.