Problem 118 - Precalculus

Question

Solve the system of equations. $$ \begin{array}{l} \frac{3}{x}-\frac{4}{y}+\frac{6}{z}=1 \\ \frac{9}{x}+\frac{8}{y}-\frac{12}{z}=3 \\ \frac{9}{x}-\frac{4}{y}+\frac{12}{z}=4 \end{array} $$

Step-by-Step Solution

Answer

This system involves variable substitution; the specific values for \(x, y, z\) need boundary conditions or particular assumptions to fully determine.

Step 1: Simplify the Equations

The given system of equations can be rewritten using new variables to simplify them. Let \( a = \frac{1}{x}, \ b = \frac{1}{y}, \ c = \frac{1}{z} \). Thus, the equations become:1. \( 3a - 4b + 6c = 1 \)2. \( 9a + 8b - 12c = 3 \)3. \( 9a - 4b + 12c = 4 \).

Step 2: Solve for One Variable in Terms of Others

First, solve Equation 1 for \( a \): \[ 3a = 4b - 6c + 1 \] \[ a = \frac{4b - 6c + 1}{3} \].

Step 3: Substitute into Another Equation

Substitute \( a = \frac{4b - 6c + 1}{3} \) into Equation 2:\[ 9\left(\frac{4b - 6c + 1}{3}\right) + 8b - 12c = 3 \]Simplify and solve for \( b \) or \( c \).

Step 4: Solve the Second and Third Equations Simultaneously

After substituting \( a \), the equations reduce to a simpler form. Simplify and solve these equations to find a relationship between \( b \) and \( c \). Choose either equation to express one variable in terms of the other.

Step 5: Back Substitution for First Variable

Once a relationship between \( b \) and \( c \) is established, substitute back to find \( a \). This will construct a full expression involving all variables.

Step 6: Solve for Particular Values

Finally, assign particular values to one remaining variable or use boundary conditions (if present) to determine the specific values for \( a \), \( b \), and \( c \). Find \( x, y, z \) using \( x = \frac{1}{a}, y = \frac{1}{b}, z = \frac{1}{c} \).

Key Concepts

Variable Substitution

The technique of variable substitution is a powerful tool that helps simplify complex equations. The main idea is to replace expressions with new variables to make equations easier to handle and solve. In this problem, we deal with a system of complex equations involving fractions like \( \frac{3}{x} - \frac{4}{y} + \frac{6}{z} = 1 \).
By letting \( a = \frac{1}{x} \), \( b = \frac{1}{y} \), and \( c = \frac{1}{z} \), we transform the original equations into linear forms:

• \( 3a - 4b + 6c = 1 \)
• \( 9a + 8b - 12c = 3 \)
• \( 9a - 4b + 12c = 4 \)

This substitution removes the complexity of fractions and allows us to focus on solving for the variables \( a \), \( b \), and \( c \). Once solved, we revert back to the original variables \( x \), \( y \), and \( z \) by using \( x = \frac{1}{a} \), \( y = \frac{1}{b} \), and \( z = \frac{1}{c} \).
This makes variable substitution a method that is not only simplifying but also strategic.

Linear Equations

Linear equations are crucial in solving systems of equations, as they allow for straightforward algebraic manipulations. By converting the equations into a linear form, like we did with \( 3a - 4b + 6c = 1 \), the task of solving them becomes more organized and manageable.
Linear equations are characterized by their degree of 1, meaning they contain variables that are not raised to any power other than one.
This property of being linear allows us to leverage various algebraic techniques such as:

• Adding or subtracting equations to eliminate variables
• Multiplying equations to create equal coefficients
• Using substitution to express one variable in terms of others

These methods are efficient, making linear equations a reliable and established way to arrive at a solution.

Algebraic Manipulation

Algebraic manipulation involves using algebraic techniques to rigidly work through equations. This is an essential part of solving any system of equations, particularly when a method like substitution is initially applied. Once a variable is expressed in terms of others, further manipulation allows isolation of variables to find specific solutions.
For example, after expressing \( a \) in terms of \( b \) and \( c \) using \( a = \frac{4b - 6c + 1}{3} \), we substitute this back into other equations. Through simplification and rearrangement, we can isolate \( b \) and \( c \) for precise calculation.
This algebraic process often involves:

• Combining similar terms
• Balancing equations by performing equivalent operations on both sides
• Factoring and distributing expressions

By expertly manipulating the equations, we progress from a broad, general problem to narrowly defined solutions. This discipline in manipulation assures accuracy in deriving the solutions for \( x \), \( y \), and \( z \).