[tex]Using \ Substitution \ Method; \\ \\ Let \ 3x+y=8 \ be \ the \
first \ equation. \\ Let \ 3x-3y=12 \ be \ second \ equation.\\ \\
3x+y=8 \\ \\ y=-3x+8 \\ \\ Now \ I'll \ substitute \ "y" \ to \ the \
second \ equation \ and \ solve \ for \ "x". \\ \\ 3x-3(-3x+8)=12 \\
\\ 3x+9x+8=12 \\ \\ 3x+9x=12-8 \\ \\ \frac{12x}{12}= \frac{4}{12} \\
\\ \boxed{x= \frac{1}{3}} \\ \\ y=3\times \frac{1}{3}+8 \\ \\
\boxed{y=7} \\ \\ Thus,the \ solution \ is \ \boxed{(x,y)=(
\frac{1}{3},7)} [/tex]
