Let's solve the system of equations step by step.
The given system is:
[tex]\[
\begin{aligned}
1) & \quad x - 2 = 3z \quad \text{(Equation 1)} \\
2) & \quad 3x = -4y - z - 4 \quad \text{(Equation 2)}
\end{aligned}
\][/tex]
### Step 1: Express [tex]\( x \)[/tex] in terms of [tex]\( z \)[/tex]
From Equation 1:
[tex]\[
x - 2 = 3z
\][/tex]
Adding 2 to both sides:
[tex]\[
x = 3z + 2 \quad \text{(Equation 3)}
\][/tex]
### Step 2: Substitute Equation 3 into Equation 2
Substitute [tex]\( x = 3z + 2 \)[/tex] into Equation 2:
[tex]\[
3(3z + 2) = -4y - z - 4
\][/tex]
Simplify the left side:
[tex]\[
9z + 6 = -4y - z - 4
\][/tex]
Add [tex]\( z \)[/tex] and 4 to both sides:
[tex]\[
9z + z + 6 + 4 = -4y
\][/tex]
[tex]\[
10z + 10 = -4y
\][/tex]
Divide both sides by -4 to solve for [tex]\( y \)[/tex]:
[tex]\[
y = -\frac{10z + 10}{4}
\][/tex]
Simplify the fraction:
[tex]\[
y = -\frac{10(z + 1)}{4}
\][/tex]
[tex]\[
y = -\frac{5}{2}(z + 1) \quad \text{(Equation 4)}
\][/tex]
### Step 3: Parameterize the free variable [tex]\( z \)[/tex]
Since [tex]\( z \)[/tex] is the free variable, we can use the parameter [tex]\( t \)[/tex] to represent [tex]\( z \)[/tex]:
Let [tex]\( z = t \)[/tex].
### Step 4: Express [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in terms of [tex]\( t \)[/tex]
Using [tex]\( z = t \)[/tex]:
From Equation 3:
[tex]\[
x = 3z + 2 = 3t + 2
\][/tex]
From Equation 4:
[tex]\[
y = -\frac{5}{2}(z + 1) = -\frac{5}{2}(t + 1)
\][/tex]
Thus, the parameterized solutions are:
[tex]\[
\begin{aligned}
x &= 3t + 2 \\
y &= -\frac{5}{2}(t + 1) \\
z &= t
\end{aligned}
\][/tex]
So, the final parameterized solution is:
[tex]\[
\begin{array}{l}
x = 3t + 2 \\
y = -\frac{5}{2}(t + 1) \\
z = t
\end{array}
\][/tex]
