system

As follows:

solving it this way is easier, i reckon.

$ \begin{cases} ab^2 = \frac{27}{5} \\  a^2b = 1080  \end{cases} $

lets work this -- $ ab^2=\frac{27}{5} $ ; take the $b^2$ to the other side and you get

$ a=\frac{27}{5b^2}   $

plugging that in the other equation you get \( (\frac{27}{5b^2})^2\times b=1080  \)

that is equal to $   \frac{729}{25b^3} \times b=1080 $

$ 729=27000b^3  $

$  b^3=\frac{27}{1000}$

$ b=\sqrt[3]{\frac{27}{1000}}   $

$  b=\frac{3}{10} $

back to our a equaltion:  $  ab^2 = \frac{27}{5} \ \ \ \Rightarrow \ \ \ \   a (\frac{3}{10})^2 = \frac{27}{5}  $ 

$a\frac{9}{100}=\frac{27}{5}$

$  \cancel{100}a\frac{9}{\cancel{100}}=\frac{27\times100}{5}  $

$  9a=540 $

$  a=60 $

now to our last condition $a+5b   \ \ \ \Rightarrow \ \ \  60+ 5(\frac{3}{10}) $

$  60+  \frac{5\times 3}{10} $

$  60+ \frac{3}{2}  $

$ \frac{60\times 2+3}{2} $

$  \frac{123}{2} $

:D