m = Moe's rate
a = Andrew's rate
we know both of them together can do the whole thing in 6 hours, so in 1 hour they have done 1/6 of the job.
since Moe can do the job by himself in "m" hours, in 1 hour he has done 1/m of it then, and whilst Andrew can do it in "a" hours, in 1 hour he has done 1/a of it.
now, let's keep in mind that Andrew is slower, it took Andrew 9 hours longer than it took Moe, so if Moe took "m" hours, Andrew took "m + 9", so a = m + 9, and therefore Andrew in 1 hour has done 1/(m+9).
[tex]\bf \stackrel{Moe}{\cfrac{1}{m}}+\stackrel{Andrew}{\cfrac{1}{m+9}}=\stackrel{whole~job}{\cfrac{1}{6}}\impliedby \textit{let's use the LCD of m(m+9)}
\\\\\\
\cfrac{m+9+m}{m(m+9)}=\cfrac{1}{6}\implies \cfrac{2m+9}{m(m+9)}=\cfrac{1}{6}\implies 12m+54=m(m+9)
\\\\\\
12m+54=m^2+9m\implies 0=m^2-3m-54\implies 0=(m-9)(m+6)
\\\\\\
m=
\begin{cases}
\boxed{9}\\
-6
\end{cases}[/tex]
since it's a speed rate, it cannot be -6.
so, Moe can do the job by himself in 9 hours, well, Andrew can do it in, 9+9.
