\(y=\frac{1}{2}x \\
2x+3y=28 \\ \\
\hbox{substitute } \frac{1}{2}x \hbox{ for y in the second equation:} \\
2x+3 (\frac{1}{2}x)=28 \\
2x+\frac{3}{2}x=28 \ \ \ \ \ |\times 2 \\
4x+3x=56 \\
7x=56 \ \ \ \ \ \ \ \ \ \ \ \ |\div 7 \\
x=8 \\ \\
y=\frac{1}{2}x=\frac{1}{2} \times 8=4 \\ \\
\boxed{(x,y)=(8,4)} \Leftarrow \hbox{answer B}\)
