Problem 19 - 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 \begin{array}{r} x-2 y+3 z=6 \\ 2 x-y+2 z=5 \end{array} \begin{aligned}

Step-by-Step Solution

Answer

The solution set can be explained as relations among \( t, u, v \) expressing terms of \( z \) where dependently resolved.

Step 1: Substitute Variables

First, substitute the variables based on the hint given. Let \( t = \frac{1}{x} \), \( u = \frac{1}{y} \), and \( v = \frac{1}{z} \). Rewrite the equations by substituting these new variable definitions.

Step 2: Reformulate the Equations

Rewrite the original equations using these substitutions: \( t - 2u + 3v = 6 \) and \( 2t - u + 2v = 5 \). These equations are the new system to solve in terms of \( t \), \( u \), and \( v \).

Step 3: Eliminate a Variable

To solve this system, try to eliminate one of the variables. Subtract the second equation \( 2t - u + 2v = 5 \) from the first equation \( t - 2u + 3v = 6 \). This results in a single equation referencing fewer variables.

Step 4: Solve for Remaining Variables

The subtraction yields \( -t - u + v = 1 \). Use this equation alongside one of the original new forms to express one variable in terms of the others, typically choosing the variable \( v \) given the problem's question.

Step 5: Express Solution in Terms of z

Since \( z \) is represented by \( v = \frac{1}{z} \), find its representation in terms of \( v \) after isolating \( v \) in the new equation(s). Substitute back to express the solution set.

Key Concepts

Solution Set

A solution set is the collection of all possible solutions that satisfy a given system of equations. In the exercise where we have the equations: \[x - 2y + 3z = 6 \ 2x - y + 2z = 5 \]we need to find values of the variables that make both equations true at the same time.
When equations are dependent, this means they essentially provide the same information; one can be derived from the other. In such cases, the solution set includes infinitely many solutions rather than a single pair or set of values.
To explicitly showcase the solution set, we express it in terms of a parameter, often a single variable, instead of providing a single value for each variable. In your exercise, you're asked to express the solution set in terms of \(z\), a process which involves parameterizing the system.

• This method is common when handling dependent equations in a system.
• Remember, the solution set may not be limited to simple pairs; it may instead describe a relationship of variables dependent on a parameter.

Substitution Method

The substitution method is a technique for solving systems of equations. It involves expressing variables in terms of others to simplify solving the equations. In the exercise, you applied the substitution method by first introducing new variables: \\( t = \frac{1}{x} \), \( u = \frac{1}{y} \), and \( v = \frac{1}{z} \).
Using these substitutions, the original system of equations transforms into:\[t - 2u + 3v = 6\] and \[2t - u + 2v = 5.\]This transformation simplifies the complexity of handling reciprocal relationships that the hint introduced and allows for easier manipulation to find expressions of one variable in terms of others.
As you eliminate one of the variables through subtraction or addition, you simplify the equations further, ultimately aiming to solve for one variable in terms of another, making it easier to isolate variables like \( v \) when seeking a solution in terms of \( z \).

• This method is especially useful when equations appear cumbersome or when handling complex transformations.
• In many scenarios, it's a highly efficient strategy to reduce a system to a more manageable form.

System of Equations

A system of equations is a set of equations with multiple variables. Solving a system means finding values for these variables that satisfy all equations simultaneously. In this exercise, you solved a system involving two equations and variables \(x\), \(y\), and \(z\).
This system initially posed:\[x - 2y + 3z = 6\] and \[2x - y + 2z = 5.\]Systems can either have:

• A unique solution, when each line (or plane, in higher dimensions) intersects at a single point.
• No solution, when the lines are parallel and never intersect.
• Infinitely many solutions, often occurring when the lines overlap, as in a dependent system.

In dependent systems such as this, the equations share proportional variables or are scalar multiples of each other.
They essentially communicate the same constraint, indicating a dependency, and typically, solving one in terms of another explicitly highlights this relationship. The problem then becomes finding a representative solutions set often described by a parameter.