Solution:
Given the system;
[tex]\begin{gathered} 4x+6y=32.............equation1 \\ \\ 3x-6y=3..............equation2 \end{gathered}[/tex]
Add equation1 to equation2, we have;
[tex]\begin{gathered} 4x+3x+6y-6y=32+3 \\ \\ 7x=35 \\ \\ \frac{7x}{7}=\frac{35}{7} \\ \\ x=5 \end{gathered}[/tex]
Substitute x = 5 in equation2;
[tex]\begin{gathered} 3x-6y=3 \\ \\ 3(5)-6y=3 \\ \\ 15-6y=3 \\ \\ 6y=15-3 \\ \\ 6y=12 \\ \\ y=2 \end{gathered}[/tex]
ANSWER:
[tex]x=5,y=2[/tex]
