Question 945325
(FOR THE EASIER SOLUTION METHOD, SEE THE ONE BELOW!)


Let x be time for "another hose" to fill the tank.
A hose, {{{1/6}}}
Another hose, {{{1/x}}}
Two combined hoses, {{{1/6+1/x=(x+6)/(6x)}}}



This seems unnatural to analyze:
the "Another hose needs 3 hours more to fill the pool then the 2 hoses combined. "


Best idea seems to try to compare the denominators for {{{1/x}}} and {{{(x+6)/(6x)}}} but this second expression must be used as a complex fraction with numerator of 1.


Combined hoses, {{{1/((x6)/(x+6))}}}, meaning the time for combined hoses to fill the tank is {{{6x/(x+6)}}};
the description says that {{{highlight(x=3+(6x)/(x+6))}}}........



Reread and study the problem description and question; recheck my (incomplete) solution analysis , and think carefully about all parts.  When you can make sense of it, then you can continue.  Understanding this description and solution process here is far far far more important than getting the final result.




<b>EASIER SOLUTION METHOD</b>


x, time for another hose to fill tank
y, time for combined hoses to fill tank
-
RATES
A hose, {{{1/6}}}
Another hose, {{{1/x}}}
Combined hoses, {{{1/y}}}
-
According to the given description, {{{highlight_green(x=3+y)}}}.


Combined hoses rate,
{{{1/6+1/x=1/y}}}
{{{x/(6x)+6/(6x)=1/y}}}
{{{(x+6)/(6x)=1/y}}}


Using the given description between x and y times, find {{{highlight_green(y=x-3)}}};
Substitute this into the combined hoses rate...


{{{highlight_green((x+6)/(6x)=1/(x-3))}}}

{{{(x+6)(x-3)=6x}}}
{{{x^2+3x-18=6x}}}
{{{x^2-3x-18=0}}}
which is factorable;
{{{(x+3)(x-6)=0}}}


The solution which makes meaningful sense is {{{highlight(x=6)}}}.
The "another hose" by itself will fill the tank in 6 hours.