To start, we want to see whether it will be easier to eliminate $x$ or $y\text{.}$ We see that the coefficients of $x$ in each equation are $2$ and $3\text{,}$ and the coefficients of $y$ are $-6$ and $2\text{.}$ Because $6$ is a multiple of $2$ and the coefficients already have opposite signs, the $y$ variable will be easier to eliminate. To eliminate the $y$ terms, we will multiply each side of the second equation by $3$ so that we will have $6y\text{.}$ We can call this process scaling the second equation by $3\text{.}$ $$\begin{gathered} 2x-6y=54\\ 3\left({3x+2y}\right)=3\left({-7}\right) \end{gathered}$$ Which gives us $$\begin{gathered} 2x-6y=54\\ 9x+6y=-21 \end{gathered}$$ We now have an equivalent system of equations where the $y$-terms can be eliminated: $$\begin{aligned} \begin{aligned}2x-6y\\{}+9x+6y\end{aligned}\amp=\begin{aligned}54\\{}+(-21)\end{aligned} \end{aligned}$$ So we have: $$\begin{aligned} 11x\amp=33\\ x\amp=3 \end{aligned}$$ To solve for $y\text{,}$ we will substitute $3$ for $x$ into either of the original equations or the new one. We will use the original first equation, $3x-4y=2\text{:}$ $$\begin{aligned} 2x-6y\amp=54\\ 2(\substitute{3})-6y\amp=54\\ 6-6y\amp=54\\ -6y\amp=48\\ y\amp=-8 \end{aligned}$$ Our solution is $x=3$ and $y=-8\text{,}$ or the point $(3,-8)\text{.}$ The checking step is left up to the reader.