Linear Equations in Three Variables
For \( a, \; b, \;c, \;d \in R \; (a \neq 0, \; b\neq 0, \; c\neq 0) \), equations of the form \( ax + by + cz + d = 0 \) are called linear equations in three variables. A system consisting of at least two equations of this form is called a system of linear equations in three variables.
Finding the Solution Set:
\(1) \) If the number of variables is three and the number of equations is one, the solution set contains infinitely many elements.
Example:
There are infinitely many ordered triples such as \( (1, 1, 8), (2,1,6), \dots \) that satisfy the equation \( 2x – 3y + z = 7 \).
\(2) \) If the number of variables is three and the number of equations is two, the solution set contains infinitely many elements.
Example:
\[ x+ y – z = 1 \]
\[ 2x- 2y + z = 3 \]
If the substitution \( z = t \) is made in this system of equations,
\[ x+ y – t = 1 \Rightarrow x + y = 1 + t \]
\[ 2x- 2y + t = 3 \Rightarrow 2x-2y = 3- t \]
a system of linear equations in two variables expressed in terms of $t$ is obtained. From here,
\[x = \frac{t+5}{4} \quad \text{and} \quad y= \frac{3t-1}{4} \quad \text{are found. } \]
Where \( t \in R\), there are infinitely many ordered triples such as \( \large (\frac{t+5}{4}, \frac{3t-1}{4}, t ) \) that satisfy the system of equations.
\(3) \) If the number of variables is equal to the number of equations and is three, the solution set can contain a unique element (or can be the empty set, or contain infinitely many elements).
In this case, methods used for solving systems of linear equations in two variables can be applied to find the solution set.
Example:
\[\begin{aligned}
x+y+z =& 7\\
2x-y+z=& 4\\
3x+y+z=& 9\\
\end{aligned}\]
Let’s find the solution set of this system of equations. To do this, let’s add the first equation to the second equation, and the second equation to the third equation side by side.
\[\begin{aligned}
x+y+z =& 7\\
+ \quad 2x-y+z=& 4\\
\hline
3x+ 2z = & 11
\end{aligned}\]
\[\begin{aligned}
2x-y+z =& 4\\
+ \quad 3x+y+z=& 9\\
\hline
5x+ 2z = & 13
\end{aligned}\]
If we perform a simultaneous solution of the two resulting equations in two variables,
\[\begin{aligned}
5x+ 2z = & 13\\
– \quad 3x+ 2z = & 11\\
\hline
2x = & 2 \Rightarrow x = 1
\end{aligned}\]
from here \( x = 1, \; z = 4, \; y= 2 \) are found.
Question 21
\[
\begin{aligned}
\frac{1}{x} + \frac{1}{y} &= 5 \\
\frac{1}{y} + \frac{1}{z} &= 8 \\
\frac{1}{x} + \frac{1}{z} &= 7
\end{aligned}
\]
then what is the value of the product \( x \cdot y \cdot z \)?
\[
\text{A)} \frac{1}{6} \quad
\text{B) } 6 \quad
\text{C) } \frac{1}{5} \quad
\text{D) } 5 \quad
\text{E) } \frac{1}{30}
\]
Solution:
After multiplying both sides of the second equation by \(-1 \), if we add the three equations side by side:
\[
\begin{aligned}
\frac{1}{x}\; + \;\frac{1}{y} &= 5 \\
-\frac{1}{y} \;- \;\frac{1}{z} &= -8 \\
+ \quad \frac{1}{x} \;+ \;\frac{1}{z} &= 7\\
\hline\\
\frac{2}{x} = &4 \Rightarrow \frac{1}{x} = 2
\end{aligned}
\]
\[\frac{1}{x} = 2, \quad \frac{1}{y} = 3, \quad \frac{1}{z} = 5 \] are found. If these values are multiplied side by side:
\[ \frac{1}{x} \cdot \frac{1}{y} \cdot \frac{1}{z} = 2 \cdot 3 \cdot 5 = 30 \]
\[ x \cdot y \cdot z = \frac{1}{30} \]
\(\textbf{Answer: E} \)
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