Let's solve the given system of linear equations step by step:
[tex]\[
\begin{cases}
x - y = 1 \quad \text{(1)} \\
3x - 6z = 12 \quad \text{(2)} \\
3y + z = 9 \quad \text{(3)}
\end{cases}
\][/tex]
### Step 1: Simplify Equation (2)
First, we simplify Equation (2) by dividing every term by 3:
[tex]\[
3x - 6z = 12 \implies x - 2z = 4 \quad \text{(2')}
\][/tex]
Now we have:
[tex]\[
\begin{cases}
x - y = 1 \quad \text{(1)} \\
x - 2z = 4 \quad \text{(2')} \\
3y + z = 9 \quad \text{(3)}
\end{cases}
\][/tex]
### Step 2: Express [tex]\(x\)[/tex] from Equation (1)
From Equation (1), solve for [tex]\(x\)[/tex]:
[tex]\[
x = y + 1 \quad \text{(4)}
\][/tex]
### Step 3: Substitute [tex]\(x\)[/tex] into Equation (2')
Substitute [tex]\(x = y + 1\)[/tex] into Equation (2'):
[tex]\[
(y + 1) - 2z = 4
\][/tex]
Simplify this equation:
[tex]\[
y + 1 - 2z = 4 \implies y - 2z = 3 \quad \text{(5)}
\][/tex]
Now we have two equations with two variables:
[tex]\[
\begin{cases}
y - 2z = 3 \quad \text{(5)} \\
3y + z = 9 \quad \text{(3)}
\end{cases}
\][/tex]
### Step 4: Solve the system with [tex]\(y\)[/tex] and [tex]\(z\)[/tex]
First, solve Equation (5) for [tex]\(y\)[/tex]:
[tex]\[
y = 2z + 3 \quad \text{(6)}
\][/tex]
Substitute [tex]\(y = 2z + 3\)[/tex] into Equation (3):
[tex]\[
3(2z + 3) + z = 9
\][/tex]
Simplify and solve for [tex]\(z\)[/tex]:
[tex]\[
6z + 9 + z = 9 \implies 7z + 9 = 9 \implies 7z = 0 \implies z = 0
\][/tex]
### Step 5: Find [tex]\(y\)[/tex]
Substitute [tex]\(z = 0\)[/tex] back into Equation (6):
[tex]\[
y = 2(0) + 3 = 3
\][/tex]
### Step 6: Find [tex]\(x\)[/tex]
Substitute [tex]\(y = 3\)[/tex] back into Equation (4):
[tex]\[
x = 3 + 1 = 4
\][/tex]
### Final Solution
The solution to the system of equations is:
[tex]\[
(x, y, z) = (4, 3, 0)
\][/tex]
This is the step-by-step solution, where [tex]\(x = 4\)[/tex], [tex]\(y = 3\)[/tex], and [tex]\(z = 0\)[/tex].
