If  $\frac{3}{x}+\frac{4}{y}=18,\frac{5}{x}+\frac{6}{y}=28$  , then $x+y=$

Given equations are

$\frac{3}{x}+\frac{4}{y}=18,\frac{5}{x}+\frac{6}{y}=28$

Let  $\frac{1}{x}=u,\frac{1}{y}=v$  replace in given equations

$3u+4v=18\to \left(1\right)$

$5u+6v=28\to \left(2\right)$

$\left(1\right)\times \left(5\right)-\left(2\right)\times \left(3\right)$

$\begin{array}{c}15u+20v=90\\ \underset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}{{}_{-}15u+{}_{-}18v={}_{-}84}\\ 2v=6\end{array}$

$v=\frac{6}{2}=3$

$\text{'}v=3\text{'}$  substitute in equation $\left(1\right)$

$3u+4\times 3=18\Rightarrow 3u+12=18$

$3u=18-12\Rightarrow 3u=6$

$u=\frac{6}{3}=2$

$u=2=\frac{1}{x}$

$\Rightarrow$  $x=\frac{1}{2}$

$v=3=\frac{1}{y}$

$\Rightarrow$  $y=\frac{1}{3}$

$x+y=\frac{1}{2}+\frac{1}{3}$

$=\frac{3+2}{6}=\frac{5}{6}$

$\therefore$   $x+y=\frac{5}{6}$