#### Answer

$x=\dfrac{3}{2}, y = 1 \text{ or } \left(\dfrac{3}{2}, 1\right)$

#### Work Step by Step

The system of equations is given by the following two equations:
$$\begin{cases}
2x+3y&=6\hspace{20pt} &(1)\\[3mm]
x-y&=\dfrac{1}{2} \hspace{20pt} &(2)
\end{cases}$$
Multiplying equation $(2)$ by $3$ so that the coefficients of y in the two equations are additive inverses gives:
$3x-3y=\dfrac{3}{2}\hspace{20pt} (1)$
$\text{Add the two equations to eliminate } y:$
\[
\begin{aligned} 2x+3y&=6 \\[2mm]
3x-3y&=\dfrac{3}{2} \ \\[3mm]
\hline 5x&=\dfrac{15}{2}\\[2mm]
\end{aligned}
\]
Divide both sides by $5$ to obtain:
$x=\dfrac{\dfrac{15}{2}}{5}$
$x=\dfrac{3}{2}$
$\text{Substituting in equation (2) for }x=\dfrac{3}{2}$ gives:
$\dfrac{3}{2}-y=\dfrac{1}{2}$
$y=\dfrac{3}{2}-\dfrac{1}{2}$
$y=1$
Therefore, the solution is:
$\boxed{x=\dfrac{3}{2} \hspace{20pt} y = 1}$