Answer :

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[tex]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}[/tex]

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