SOLVING:
[tex]\begin{cases}&2x+6y=6\quad\Longrightarrow\ Equation\ 1\\&3x-2y=20\quad\Longrightarrow\ Equation\ 2\end{cases}\\ \\ \\\textbf{Solve\ "x"\ in\ the\ second\ equation\ and\ replace\ in\ the\ first}\\\textbf{equation:}\\ \\\boxed{x=\dfrac{20+2y}{3}}\\ \\ \\2\left( \dfrac{20+2y}{3}\right)+6y=6\\ \\ \\2(20+2y)+3(6y)=3(6)\\ \\40+4y+18y=18\\ \\22y+40=18\\ \\22y=18-40\\ \\22y=-22\\ \\y= \dfrac{-22}{22}\\ \\y=-1\quad\to\boxed{\spadesuit\ Answer\ \spadesuit}[/tex]
[tex]\textbf{Replace\ "y"\ in\ the\ second\ equation:}\\ \\3x-2(-1)=20\\ \\3x+2=20\\ \\3x=20-2\\ \\3x=18\\ \\x= \dfrac{18}{3}\\ \\x=6\quad\to\boxed{\spadesuit\ Answer\ \spadesuit}\\ \\ \\\mathbb{ANSWERS:}\\ \\(x,y)\Longrightarrow\boxed{\boxed{B\ (6,\ -1)\ \checkmark}}\\ \\ \\\textbf{HOPE\ THIS\ HELPS...!!}[/tex]
