Ta có \[\left\{ {\begin{array}{*{20}{l}}{2x - y = 1}\\{{x^2} + 2xy - {y^2} = 7}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{y = 2x - 1}\\{{x^2} + 2x\left( {2x - 1} \right) - {{\left( {2x - 1} \right)}^2} = 7}\end{array}} \right.\]

\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{y = 2x - 1}\\{{x^2} + 4{x^2} - 2x - 4{x^2} + 4x - 1 - 7 = 0}\end{array}} \right.\]\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{y = 2x - 1}\\{{x^2} + 2x - 8 = 0}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{y = 2x - 1}\\{\left( {x + 4} \right)\left( {x - 2} \right) = 0}\end{array}} \right.\]

\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{y = 2x - 1}\\{\left[ {\begin{array}{*{20}{l}}{x + 4 = 0}\\{x - 2 = 0}\end{array}} \right.}\end{array}} \right. \Leftrightarrow \left\{ {\left\{ {\begin{array}{*{20}{l}}{y = 2x - 1}\\{\left[ {\begin{array}{*{20}{l}}{x = - 4}\\{x = 2}\end{array}} \right.}\end{array}} \right.} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{x = - 4}\\{y = - 9}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{l}}{x = 2}\\{y = 3}\end{array}} \right.}\end{array}} \right.\]. Chọn B.