Problem 12 - A Graphical Approach to College Algebra

Question

If the equations are dependent, write the solution set in terms of the variable \(z\). (Hint: In Exercises 33-36, let \(t=\frac{1}{x}, u=\frac{1}{y},\) and \(v=\frac{1}{z} .\) Solve for \(t, u,\) and \(v,\) and then find $x, $$\begin{aligned} 3 x-2 y-8 z &=1 \\ 9 x-6 y-24 z &=-2 \\ x-y+z &=1 \end{aligned}$$

Step-by-Step Solution

Answer

The equations are dependent, express solution in terms of \(z\) as: \( x = \frac{1}{t}, y = \frac{1}{u}, z = \frac{1}{v} \), where \( v = 1 - t + u \).

Step 1: Identify the Hint and Variables

Based on the hint given, replace the variables using the substitutions: \[ t = \frac{1}{x}, \quad u = \frac{1}{y}, \quad v = \frac{1}{z} \]This implies that \( x = \frac{1}{t}, y = \frac{1}{u}, z = \frac{1}{v} \).

Step 2: Rewrite the Equations using t, u, v

Substitute the expressions for \(x, y, \) and \(z\) into the equations.1. \( 3 \times \frac{1}{t} - 2 \times \frac{1}{u} - 8 \times \frac{1}{v} = 1 \) becomes \[ 3t - 2u - 8v = 1 \] 2. \( 9 \times \frac{1}{t} - 6 \times \frac{1}{u} - 24 \times \frac{1}{v} = -2 \) becomes \[ 9t - 6u - 24v = -2 \] 3. \( \frac{1}{t} - \frac{1}{u} + \frac{1}{v} = 1 \) becomes \[ t - u + v = 1 \].

Step 3: Identify and Check Dependency

Check if the equations are linear combinations of each other by comparing the coefficients.Notice that the second equation is 3 times the first equation, indicating dependency:- First equation: \( 3t - 2u - 8v = 1 \)- Second equation: \( 9t - 6u - 24v = -2 \) is actually \( 3 \times ( 3t - 2u - 8v = 1 ) \).

Step 4: Solve for Remaining Variables

Use the independent equations to find one variable in terms of another.Combine equations:- \( 3t - 2u - 8v = 1 \)- \( t - u + v = 1 \).Substitute \( v \) in terms of \( t \): From the second independent equation \(t - u + v = 1\),\[ v = 1 - t + u \].

Step 5: Express the Solution Set in Terms of z

Since the equations are dependent, express the variables in terms of \(z\) or \(v\).Substitute \( v = \frac{1}{z} \) back into expressions:- \( v = 1 - t + u \)Then substitute back to find \( x, y \) in terms of \( v \):- \( x = \frac{1}{t} \)- \( y = \frac{1}{u} \)- \( z = \frac{1}{v} \)Hence, the solution set is parameterized by \(z\).

Key Concepts

Linear Systems

A linear system consists of two or more linear equations involving the same set of variables. These systems can represent multiple lines in a plane, and the solutions are the points where these lines intersect. When dealing with such systems, there are typically three types of relationships:

• Independent: Equations intersect at a single point, giving one unique solution.
• Inconsistent: Equations do not intersect at any point, resulting in no solution.
• Dependent: Equations represent the same line, thus having infinitely many solutions.

In this exercise, we are examining a dependent system because one equation is a multiple of another. This identifies that there is a dependency among the equations, leading to infinitely many solutions that can be parameterized.

Substitutions in Algebra

Substitution is a crucial technique in algebra, especially when working with linear systems, as it can simplify equations significantly. The given exercise utilized substitution by introducing new variables:

• \[ t = \frac{1}{x}, \, u = \frac{1}{y}, \, v = \frac{1}{z} \]

This substitution transformed the original complex system of equations into a simpler form, allowing for easier manipulation and solving. Such substitutions can simplify solving by effectively reducing complexity and revealing underlying relationships between variables. In this case, it helped to recognize the dependency between equations, thus pinpointing the solution set more effectively.

Solution Sets

In the context of linear systems, a solution set consists of all possible solutions that satisfy all equations within the system. For dependent systems, like the one in this exercise, the solutions form an infinite set. These are not isolated points but rather expressed as parameterized variables known in terms of another, as shown with:

• Equations like \( t - u + v = 1 \) can be solved for a variable in terms of others.
• This leads to expressing \( v \) (or \( z \), due to substitution) as a free parameter \( z \).

This method reveals all possible solutions by parameterizing in terms of one variable. Consequently, this shows how the equations define a line or plane of solutions, rather than a single point.