Problem 72 - Intermediate Algebra

Question

Solve each system. $$ \left\\{\begin{array}{l} \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3 \\ \frac{2}{x}+\frac{1}{y}-\frac{1}{z}=0 \\ \frac{1}{x}-\frac{2}{y}+\frac{4}{z}=21 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{3} \), \( y = -\frac{1}{3} \), \( z = \frac{1}{3} \).

Step 1: Substitution for Simplicity

Let us simplify the given system of equations by using substitution. Introduce new variables to replace the fractions: \( a = \frac{1}{x} \), \( b = \frac{1}{y} \), and \( c = \frac{1}{z} \). This gives us the following system of equations: \( a + b + c = 3 \), \( 2a + b - c = 0 \), and \( a - 2b + 4c = 21 \).

Step 2: Solving for One Variable

Let's begin by solving one of the equations for a variable. Using the first equation \( a + b + c = 3 \), we can express \( a \) in terms of \( b \) and \( c \): \( a = 3 - b - c \).

Step 3: Substitution into Other Equations

Substitute \( a = 3 - b - c \) into the other two equations:1. Substituting into \( 2a + b - c = 0 \) gives \( 2(3 - b - c) + b - c = 0 \), simplifying to \( 6 - 2b - 2c + b - c = 0 \), which simplifies further to \( 6 - b - 3c = 0 \).2. Substituting into \( a - 2b + 4c = 21 \) gives \( (3 - b - c) - 2b + 4c = 21 \), simplifying to \( 3 - b - c - 2b + 4c = 21 \), which reduces to \( 3 - 3b + 3c = 21 \).

Step 4: Simplification and Solving New Equations

We now have two equations in \( b \) and \( c \): 1. \( 6 - b - 3c = 0 \) 2. \( 3 - 3b + 3c = 21 \)Let's solve the first equation for \( b \): \( b = 6 - 3c \).Substitute \( b = 6 - 3c \) into the second equation: \( 3 - 3(6 - 3c) + 3c = 21 \). Simplifying gives \( 3 - 18 + 9c + 3c = 21 \) or \( 12c = 36 \).

Step 5: Solving for c and b

From \( 12c = 36 \), solve for \( c \):\( c = \frac{36}{12} = 3 \). Now, use \( c = 3 \) in the expression for \( b \):\( b = 6 - 3 \cdot 3 = 6 - 9 = -3 \).

Step 6: Solving for a using b and c

Substitute \( b = -3 \) and \( c = 3 \) back into \( a = 3 - b - c \):\( a = 3 - (-3) - 3 = 3 + 3 - 3 = 3 \).

Step 7: Solving for original variables x, y, z

We have \( a = \frac{1}{x} = 3 \), \( b = \frac{1}{y} = -3 \), and \( c = \frac{1}{z} = 3 \). Solving these gives:\( x = \frac{1}{3} \), \( y = -\frac{1}{3} \), \( z = \frac{1}{3} \).

Key Concepts

Substitution Method

The substitution method is a powerful tool for solving systems of equations. This method involves expressing one variable in terms of the others and substituting this expression into the other equations. In this way, you reduce the number of variables in each equation, gradually simplifying the system until each variable is isolated and its value discovered.

In our problem, we used substitution at multiple steps. First, by setting the fractions as new variables:

• Let \( a = \frac{1}{x} \), \( b = \frac{1}{y} \), and \( c = \frac{1}{z} \).
• This change simplifies each equation by eliminating the fractions.

Then, one equation is solved for a variable. We picked the first equation to solve for \( a \):

• \( a = 3 - b - c \)
• Next, substituting \( a \) in the other equations reduces these equations to only involve two variables: \( b \) and \( c \).

This step-by-step reduction, using substitution, is key in managing and solving complex systems like the one discussed here.

Algebraic Fractions

Algebraic fractions arise when variables are in the denominator of the fractions within your equations. They require careful handling to avoid complications in calculations. In our original system, each equation featured algebraic fractions like \( \frac{1}{x} \), \( \frac{1}{y} \), and \( \frac{1}{z} \).

To manage these fractions effectively:

• Start by introducing new variables in place of the fractions. This simplifies manipulation and avoids potential pitfalls of direct fraction operations.
• For example, convert \( \frac{1}{x} = a \), \( \frac{1}{y} = b \), and \( \frac{1}{z} = c \).

By making these substitutions, the equations become linear, facilitating straightforward application of substitution and elimination methods.

Finally, after solving for the new variables, simply reverse the substitutions to find the values of the original variables \( x \), \( y \), and \( z \). This reverse operation involves multiplying both sides of each equation by the variable (e.g., \( x = \frac{1}{a} \)) to solve for it directly.

Variables Elimination

The elimination method in systems of equations involves removing one variable entirely by combining two equations. This technique is particularly useful when the system is simplified down to two variables after initial substitutions.

In our exercise, variable elimination took place after the substitution step:

• After substitution, we had two key equations involving \( b \) and \( c \):
  \( 6 - b - 3c = 0 \) and \( 3 - 3b + 3c = 21 \).
• By expressing \( b = 6 - 3c \) and substituting into the second equation, an equation solely in terms of \( c \) was obtained.

This reduction to a single variable allows for straightforward solving. After determining \( c \), it can be back-substituted to find \( b \) and subsequently \( a \). By using elimination strategically, one variable is canceled out, making it easier to solve step-by-step for all the unknowns.