To solve the system of linear equations given in terms of the nonzero constants [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\begin{aligned}
&1) \quad a x - b y - 2 c z = 15, \\
&2) \quad a x + b y + c z = 0, \\
&3) \quad 7 a x - b y + c z = 6,
\end{aligned}
\][/tex]
we will find the values of [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex].
Step-by-Step Solution:
### Equation Setup
First, we label the equations for reference:
[tex]\[
\begin{aligned}
& (1) \quad a x - b y - 2 c z = 15, \\
& (2) \quad a x + b y + c z = 0, \\
& (3) \quad 7 a x - b y + c z = 6.
\end{aligned}
\][/tex]
### Step 1: Combine Equations
Combine equations to eliminate variables and isolate one.
Combine (1) and (2):
Adding (1) and (2):
[tex]\[
(a x - b y - 2 c z) + (a x + b y + c z) = 15 + 0,
\][/tex]
[tex]\[
2 a x - c z = 15.
\][/tex]
This gives us:
[tex]\[
2 a x - c z = 15 \quad \text{(4)}.
\][/tex]
Combine (2) and (3):
Adding (2) and (3):
[tex]\[
(a x + b y + c z) + (7 a x - b y + c z) = 0 + 6,
\][/tex]
[tex]\[
8 a x + 2 c z = 6.
\][/tex]
This gives us:
[tex]\[
4 a x + c z = 3 \quad \text{(5)}.
\][/tex]
### Step 2: Solve for [tex]\( x \)[/tex] and [tex]\( z \)[/tex]
Next, we have two new equations (4) and (5):
[tex]\[
\begin{aligned}
& (4) \quad 2 a x - c z = 15, \\
& (5) \quad 4 a x + c z = 3.
\end{aligned}
\][/tex]
Add equations (4) and (5):
[tex]\[
(2 a x - c z) + (4 a x + c z) = 15 + 3,
\][/tex]
[tex]\[
6 a x = 18,
\][/tex]
[tex]\[
x = \frac{18}{6 a} = \frac{3}{a}.
\][/tex]
Solve for [tex]\( z \)[/tex] using the value of [tex]\( x \)[/tex] in (5):
[tex]\[
4 a \left(\frac{3}{a}\right) + c z = 3,
\][/tex]
[tex]\[
4 \cdot 3 + c z = 3,
\][/tex]
[tex]\[
12 + c z = 3,
\][/tex]
[tex]\[
c z = 3 - 12 = -9,
\][/tex]
[tex]\[
z = \frac{-9}{c}.
\][/tex]
### Step 3: Solve for [tex]\( y \)[/tex]
Substitute [tex]\( x = \frac{3}{a} \)[/tex] and [tex]\( z = \frac{-9}{c} \)[/tex] into equation (2):
[tex]\[
a \left(\frac{3}{a}\right) + b y + c \left(\frac{-9}{c}\right) = 0,
\][/tex]
[tex]\[
3 + b y - 9 = 0,
\][/tex]
[tex]\[
b y = 6,
\][/tex]
[tex]\[
y = \frac{6}{b}.
\][/tex]
### Solution Set
Thus, the solution set for the system is:
[tex]\[
\left( x, y, z \right) = \left( \frac{3}{a}, \frac{6}{b}, \frac{-9}{c} \right).
\][/tex]
So, the solution set is [tex]\(\left\{ \left( \frac{3}{a}, \frac{6}{b}, \frac{-9}{c} \right) \right\}\)[/tex].
