Let $x$ be the weight of a bag of apples and $y$ the weight of a bag of oranges. We’re told that $3x+2y=32$ and $4x+3y=44$:
$$\left\{\begin{align*}
&3x+2y=32\\
&4x+3y=44\;.
\end{align*}\right.\tag{1}$$
We want to know what $2x+y$ is.
One way to answer the question is to solve $(1)$ for $x$ and $y$ and substitute into $2x+y$. Multiply the top equation of $(1)$ by $3$ and the bottom by $2$, so as to get equations with the same coefficient on $y$:
$$\left\{\begin{align*}
&9x+6y=96\\
&8x+6y=88\;.
\end{align*}\right.\tag{2}$$
If you now subtract the bottom equation in $(2)$ from the top you find that $x=8$. Substitute that value of $x$ into any of the equations in $(1)$ or $(2)$ to find $y$; I’ll use the top equation in $(1)$, since it has the smallest coefficients. From it I find that $3\cdot8+2y=32$, $24+2y=32$, $2y=8$, and $y=4$. Thus, $2x+y=2\cdot8+4=20$.
If you happen to notice that $2(3x+2y)-(4x+3y)=2x+y$, you can take advantage of a shortcut (which I see Marvis has already pointed out), but if not, solving the system is guaranteed to work, and fairly mechanically, too.