You know x+y!

$\begin{cases} x+y = \dfrac {16}5 & ...(1) \\ \dfrac 1x + \dfrac 1y = \dfrac {16}3 & ...(2) \end{cases}$

$\begin{aligned} (2): \quad \frac 1x + \frac 1y & = \frac {16}3 \\ \frac {\color{#3D99F6}{x+y}}{xy} & = \frac {16}3 & \small \color{#3D99F6}{(1): \ x+y = \frac {16}5} \\ \frac {\color{#3D99F6}{16}}{\color{#3D99F6}{5}xy} & = \frac {16}3 \\ \implies xy & = \frac 35 \quad ...(3) \end{aligned}$

$\begin{aligned} (1)\times x: \quad x^2+\color{#3D99F6}{xy} & = \frac {16}5x & \small \color{#3D99F6}{(3): \ xy = \frac 35} \\ x^2+\color{#3D99F6}{\frac 35} & = \frac {16}5x \\ 5x^2 -16x+3 & = 0 \\ (5x-1)(x-3) & = 0 \\ \implies x,y & = \frac 15, 3 \end{aligned}$

$\implies 4x + 10y - 5xy = \begin{cases} 4 \times \dfrac 15 + 10 \times 3 - 5 \times \dfrac 35 = \color{#D61F06}{27.8} & \color{#D61F06} {\text{Not integer, rejected.}} \\ 4 \times 3 + 10 \times \frac 15 - 5 \times \dfrac 35 = \boxed{\color{#3D99F6}{11}} & \color{#3D99F6} {\text{Integer, accepted.}} \end{cases}$