Basic Manipulation

$\begin{aligned}(\color{#20A900}{\sqrt{3})^{x+y}=}&\ \color{#20A900}9\Rightarrow(\color{#20A900}1)\\\color{#624F41}{(\sqrt{2})^{x-y}=}&\ \color{#624F41}{32}\Rightarrow(\color{#624F41}2)\end{aligned}$

From equation $(\color{#20A900}1)$ that is,

$\begin{aligned}\color{#20A900}{(\sqrt{3})^{x+y}=}&\ \color{#20A900}9\\(3^{\frac{1}{2}})^{x+y}=&\ 3^2\\3^{\frac{x+y}{2}}=&\ 3^2\\\frac{x+y}{2}=&\ 2\\\color{#D61F06}{x+y=}&\ \color{#D61F06}4\Rightarrow(\color{#D61F06}3)\end{aligned}$

From equation $(\color{#624F41}2)$ that is,

$\begin{aligned}\color{#624F41}{(\sqrt{2})^{x-y}=}&\ \color{#624F41}{32}\\(2^{\frac{1}{2}})^{x-y}=&\ 2^5\\2^{\frac{x-y}{2}}=&\ 2^5\\\frac{x-y}{2}=&\ 5\\\color{#3D99F6}{x-y=}&\ \color{#3D99F6}{10}\Rightarrow(\color{#3D99F6}4)\end{aligned}$

Equation $(\color{#D61F06}3)$ $+$ Equation $(\color{#3D99F6}4)$ that is,

$\begin{aligned}2x=14\Rightarrow\color{#69047E}{x=7}\end{aligned}$

Equation $(\color{#D61F06}3)$ $-$ Equation $(\color{#3D99F6}4)$ that is,

$\begin{aligned}2y=-6\Rightarrow\color{#302B94}{y=-3}\end{aligned}$

Thus, $2\color{#69047E}x+\color{#302B94}y=2(\color{#69047E}7)+(\color{#302B94}{-3})=14-3=\boxed{11}$

x + y = 4 ,x-y= 10 ,x= 7 ,y=-3 ,2(7)+(-3) = 11