Originally Posted by

**Soroban** Hello, someone21!

I'm not sure what your question is,

. . but I can solve the system for you . . .

We have: .$\displaystyle \left|\begin{array}{cc|c}1 & 2 & 1 \\ 2 & \text{-}3 & 4 \end{array}\right| $

. $\displaystyle \begin{array}{c} \\ R_2-2R_1\end{array} \left|\begin{array}{cc|c}1 & 2 & 1 \\ 0 & \text{-}7 & 2 \end{array}\right|$

. . . . $\displaystyle \begin{array}{c} \\ \text{-}\frac{1}{7}R_2 \end{array} \left|\begin{array}{cc|c}1 & 2 & 1 \\ 0 & 1 & \text{-}\frac{2}{7} \end{array}\right| $

. $\displaystyle \begin{array}{c} R_1-2R_2 \\ \\ \end{array}\left|\begin{array}{cc|c}1 & 0 & \frac{11}{7} \\ \\[-4mm] 0 & 1 & \text{-}\frac{2}{7}\end{array}\right| $

Solution: .$\displaystyle \boxed{x \:=\:\frac{11}{7}\quad y \:=\:\text{-}\frac{2}{7}}$