Considering the coefficients of $x$ ($5$ and $3$) and the coefficients of $y$ ($2$ and $-5$), we see that we cannot eliminate the $x$ or the $y$ variable by scaling a single equation. We will need to scale both. The $y$-terms already have opposite signs, so we choose to eliminate $y\text{.}$ The least common multiple of $2$ and $5$ is $10\text{.}$ We can scale the first equation by $5$ and the second equation by $2$ so that the equations have terms $10y$ and $-10y\text{,}$ which will cancel when added. $$\begin{gathered} 5\left({5x+2y}\right)=5\left({16}\right)\\ 2\left({3x-5y}\right)=2\left({22}\right) \end{gathered}$$ Which gives us $$\begin{gathered} 25x+10y=80\\ 6x-10y=44 \end{gathered}$$ At this point we can add the corresponding sides from the two equations and solve for $x\text{:}$ $$\begin{aligned} \begin{aligned}25x+10y\\{}6x-10y\end{aligned}\amp=\begin{aligned}80\\{}+44\end{aligned} \end{aligned}$$ So we have: $$\begin{gathered} 31x=124\\ x=4 \end{gathered}$$ To solve for $y\text{,}$ we’ll replace $x$ with $4$ in $5x+2y=16\text{:}$ $$\begin{aligned} 5x+2y\amp=16\\ 5(\substitute{4})+2y\amp=16\\ 20+2y\amp=16\\ 2y\amp=-4\\ y\amp=-2 \end{aligned}$$ So the system’s solution is $(4,-2)$ and the solution set is $\{(4,-2)\}\text{.}$ The check is left up to the reader.