Let's solve the given system of linear equations step-by-step:
[tex]\[
\begin{aligned}
&1.\quad x + 2y - z = 3 \quad \text{(Equation 1)} \\
&2.\quad 2x + 4y - 2z = 6 \quad \text{(Equation 2)} \\
&3.\quad -x - 2y + 2 = 6 \quad \text{(Equation 3)} \\
\end{aligned}
\][/tex]
First, let's simplify Equation 3:
[tex]\[
-x - 2y + 2 = 6 \implies -x - 2y = 4 \implies x + 2y = -4 \quad \text{(Equation 4)}
\][/tex]
Next, let's observe Equations 1 and 2. Notice that Equation 2 is simply twice Equation 1:
[tex]\[
2(x + 2y - z) = 2 \cdot 3 \implies 2x + 4y - 2z = 6
\][/tex]
This means that Equation 2 does not provide any new information and is redundant.
Hence, we are left with two independent equations:
[tex]\[
\begin{aligned}
&1.\quad x + 2y - z = 3 \quad \text{(Equation 1)} \\
&4.\quad x + 2y = -4 \quad \text{(Equation 4)}
\end{aligned}
\][/tex]
Now, let's subtract Equation 4 from Equation 1 to eliminate [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[
(x + 2y - z) - (x + 2y) = 3 - (-4) \implies -z = 7 \implies z = -7
\][/tex]
We have found [tex]\(z = -7\)[/tex].
Next, let's substitute [tex]\(z = -7\)[/tex] back into Equation 4 to find [tex]\(x\)[/tex]:
[tex]\[
x + 2y = -4 \quad \text{(Equation 4)}
\][/tex]
Since [tex]\(x = -2y - 4\)[/tex], substituting [tex]\(x\)[/tex] we get:
Finally, we only have this relation [tex]\(x = -2y - 4\)[/tex] holding true.
Therefore, the solutions are given by:
[tex]\[
\begin{aligned}
&x = -2y - 4, \\
&z = -7
\end{aligned}
\][/tex]
Here, [tex]\(y\)[/tex] can be any real number, making [tex]\(x\)[/tex] and [tex]\(z\)[/tex] dependent on [tex]\(y\)[/tex].
