\(\displaystyle{x}_{{{1}}}-{s}_{{{3}}}+{3}{x}_{{{4}}}={9}\)

\(\displaystyle{x}_{{{2}}}+{2}{x}_{{{3}}}-{5}{x}_{{{4}}}={8}\)

\(\displaystyle{0}={0}\) The augmented matrix does not contain a row in which the only nonzero entry appears in the last column. Therefore, this system of equations must be consistent. Convert the augmented matrix into a system of equations. \(\displaystyle{x}_{{1}}={9}+{x}_{{3}}-{3}{x}_{{4}}\)

\(\displaystyle{x}_{{2}}={8}-{2}{x}_{{3}}+{5}{x}_{{4}}\)

\(\displaystyle{x}_{{3}},{\mathfrak{{e}}}{e}\)

\(\displaystyle{x}_{{4}},{\mathfrak{{e}}}{e}\) Solve for the leading entry for each individual equation. Determine the free variables, if any. \(\displaystyle{x}_{{1}}={9}+{s}-{3}{t}\)

\(\displaystyle{x}_{{2}}={8}-{2}{s}+{5}{t}\)

\(\displaystyle{x}_{{3}}={s}\)

\(\displaystyle{x}_{{4}}={t}\) Parameterize the free variables. \(x = \begin{bmatrix}9 \\ 8\\0\\0 \end{bmatrix} + s \begin{bmatrix}1 \\-2\\1\\0 \end{bmatrix} + t \begin{bmatrix}-3 \\ 5\\0\\1 \end{bmatrix}\) And write the solution in vector form.