Answer
x = 5a + 4
y = -9a + 1
z = a
Where a is any real number.
Work Step by Step
$\begin{bmatrix}
1 & 1 & 4 & |5\\
2 & 1 & -1 & |9\\
\end{bmatrix}$ ~ $\begin{bmatrix}
1 & 1 & 4 & |5\\
0 & -1 & -9 & |-1\\
\end{bmatrix}$ ~ $\begin{bmatrix}
1 & 0 & -5 & |4\\
0 & -1 & -9 & |-1\\
\end{bmatrix}$ ~ $\begin{bmatrix}
1 & 0 & -5 & |4\\
0 & 1 & 9 & |1\\
\end{bmatrix}$
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.
