The augmented matrix that corresponds to a system of linear equations in \( x_{1} \) and \( x_{2} \) is \( \left[\begin{array}{cc|c}1 & 2 & -7 \\ 1 & 3 & -12\end{array}\right] \) \( x_{1}= \) Ex: 42 \( x_{2}= \)

To solve the system represented by the augmented matrix, we can convert it into a more manageable form. The first step is to eliminate one of the variables. By subtracting the first row from the second, we get: \[ \left[\begin{array}{cc|c}1 & 2 & -7 \\ 0 & 1 & -5\end{array}\right] \] This indicates that \( x_{2} = -5 \). Now, plugging this value back into the first equation \( x_{1} + 2(-5) = -7 \), we find: \[ x_{1} = -7 + 10 = 3. \] Thus, \( x_{1} = 3 \) and \( x_{2} = -5 \). To verify your answers, plug \( x_{1} \) and \( x_{2} \) back into the original equations and ensure both are satisfied—this step is crucial to avoid embarrassing mistakes!