Answer :
Notice that the coefficient of y is the same in both equations:
[tex]\begin{gathered} 5x+2y=48 \\ 3x+2y=32 \end{gathered}[/tex]Then, use the elimination method to solve the system of equations. Subtract the second equation from the first one:
[tex]\begin{gathered} \Rightarrow(5x+2y)-(3x+2y)=48-32 \\ \\ \Rightarrow5x-3x+2y-2y=16 \\ \\ \Rightarrow2x=16 \\ \\ \Rightarrow x=\frac{16}{2} \\ \\ \therefore x=8 \end{gathered}[/tex]Replace x=8 in the first equation and solve for y:
[tex]\begin{gathered} 5x+2y=48 \\ \\ \Rightarrow5(8)+2y=48 \\ \\ \Rightarrow40+2y=48 \\ \\ \Rightarrow2y=48-40 \\ \\ \Rightarrow2y=8 \\ \\ \Rightarrow y=\frac{8}{2} \\ \\ \therefore y=4 \end{gathered}[/tex]Therefore, the solution to the system of equations is:
[tex]x=8\qquad\text{ and }\qquad y=4[/tex]