## Algebra and Trigonometry 10th Edition

$\begin{bmatrix} 1 & 1 & 4 & |5\\ 2 & 1 & -1 & |9\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 1 & 4 & |5\\ 0 & -1 & -9 & |-1\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 0 & -5 & |4\\ 0 & -1 & -9 & |-1\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 0 & -5 & |4\\ 0 & 1 & 9 & |1\\ \end{bmatrix}$ The solution can be written as: x - 5z = 4 y + 9z = 1 The solution can then be written in terms of z: x = 5z + 4 y = -9z + 1 Then replace z with a in order to avoid using a system variable in the answer: x = 5a + 4 y = -9a + 1 z = a Where a is any real number.