Step 1
Given set of Equations:
\[\left\{ \begin{array}{l}{x_1} + 5{x_2} = 7\\ - 2{x_1} - 7{x_2} = - 5\end{array} \right.\]We have to Solve the system of equations given above by using elementary row operations on the equations or on the augmented matrix. We will prefer the augmented matrix method.

Step 2
Convert the given system of Equations into augmented matrix form . \[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}1&5\\{ - 2}&{ - 7}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}7\\{ - 5}\end{array}} \right]\\\\A = \left[ {\begin{array}{*{20}{c}}1&5&7\\{ - 2}&{ - 7}&{ - 5}\end{array}} \right]\end{array}\]

Step 3: ${R_2} = {R_2} + 2{R_1}$
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&5&7\\{ - 2}&{ - 7}&{ - 5}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&5&7\\{ - 2 + 2\left( 1 \right)}&{ - 7 + 2\left( 5 \right)}&{ - 5 + 2\left( 7 \right)}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&5&7\\0&3&9\end{array}} \right]\end{array}\]

Step 4: ${R_2} = {R_2}/3 $
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&5&7\\0&3&9\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&5&7\\0&{3/3}&{9/3}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&5&7\\0&1&3\end{array}} \right]\end{array}\]

Step 5: Write the reduced form in system of equations
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&5&7\\0&1&3\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}1&5\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}7\\3\end{array}} \right]\\\\\left\{ \begin{array}{l}{x_1} + 5{x_2} = 7\\{x_2} = 3\end{array} \right.\end{array}\]

Step 6
From the last row, we have\[{x_2} = 3\]Now put the value of $x_3$ in first row
\[\begin{array}{l}{x_1} + 5{x_2} = 7\\{x_1} + 5\left( 3 \right) = 7\\{x_1} = 7 - 15\\{x_1} = - 8\end{array}\]

Step 4: ANSWERS
\[\left[ \begin{array}{l}{x_1} = - 8\\{x_2} = 3\end{array} \right]\]