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Sagot :
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].
[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].
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