Problem 33 - 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} \frac{1}{x}+\frac{1}{y}-\frac{1}{z}=\frac{1}{4} \\ \frac{2}{x}-\frac{1}{y}+\frac{3}{z}=\frac{9}{4} \\ -\frac{1}{x}-\frac{2}{y}+\frac{4}{z}=1 \end{array}

Step-by-Step Solution

Answer

Solve the system using substitution and elimination, expressing in terms of \( z \).

Step 1: Substitute Variables

Let's substitute the variables as suggested: let \( t = \frac{1}{x} \), \( u = \frac{1}{y} \), and \( v = \frac{1}{z} \). This gives us the following system of equations: \[ t + u - v = \frac{1}{4} \] \[ 2t - u + 3v = \frac{9}{4} \] \[ -t - 2u + 4v = 1 \]

Step 2: Write in Matrix Form

Write the equations in matrix form: \[ \begin{pmatrix} 1 & 1 & -1 \ 2 & -1 & 3 \ -1 & -2 & 4 \end{pmatrix} \begin{pmatrix} t \ u \ v \end{pmatrix} = \begin{pmatrix} \frac{1}{4} \ \frac{9}{4} \ 1 \end{pmatrix} \]

Step 3: Use Gaussian Elimination

Use Gaussian Elimination to simplify the system. First, scale row 1 by multiplying by 4 to eliminate fractions. Then perform row operations until the matrix is in row-echelon form.

Step 4: Solve for One Variable

From the row-echelon form, solve for one of the variables in terms of another. Here, you can express all variables in terms of \( v \), for example.

Step 5: Determine Dependence

Analyze the row-echelon or reduced row-echelon form of the matrix. If you find a row of zeros, it indicates dependency among the equations. Use this to express \( t \), \( u \), and \( v \) in terms of one parameter, \( v \).

Step 6: Express the Solution Set

If the system is dependent, write the solution set as expressions for \( x \), \( y \), and \( z \) in terms of the parameter \( v \). Since \( v = \frac{1}{z} \), substitution can be used to find \( x \), \( y \), and \( z \).

Key Concepts

Gaussian Elimination

Imagine having a jumble of intertwining equations, each with multiple unknowns. To untangle them, one highly effective method is Gaussian Elimination. This technique helps simplify a system of linear equations. The main goal is to transform the system into a form where solving for the unknowns becomes straightforward.

When you first encounter the system of equations, the task is to systematically eliminate variables until each equation becomes easier. Begin by picking one equation and changing it such that one of the elements becomes 1, if it's not already.

From there, strategically subtract or add multiples of this equation to others in order to zero out coefficients in a certain variable for other equations. By repeating these steps, you simplify the entire system into an upper-triangular form, where every equation only has one new unknown to solve.

• Start by eliminating variables in the first column of lower rows.
• Then move to the second variable and repeat the process.

Once the system is in triangular form, you work backwards to find the solution starting from the last equation.

Matrix Form

Transforming a system of equations into a matrix makes the manipulation of the equations much simpler. Imagine each equation as a row of numbers, and the variables as columns. This transition is crucial for effectively applying Gaussian Elimination.

For example, the coefficients in each equation make up the entries of the matrix, turning a seemingly complex system into an organized grid. Consider the system written as: \[ \begin{pmatrix} 1 & 1 & -1 \ 2 & -1 & 3 \ -1 & -2 & 4 \ \end{pmatrix} \begin{pmatrix} t \ u \ v \ \end{pmatrix} = \begin{pmatrix} \frac{1}{4} \ \frac{9}{4} \ 1 \ \end{pmatrix} \]

- Each row here represents an equation from the original system.- Solve by converting this into a simpler form, which makes identifying solutions easier.
Matrix form manages the bulk of equations efficiently, allowing you to focus on the numbers, making the solution process systematic.

Dependent Equations

In the context of systems of equations, dependency indicates a special relationship among them. This means one equation can be figured out using a combination of others. Simply put, if you can express one of the equations as a sum or difference of the others, dependency is present. This often leads to infinite solutions.

When you find a row of zeros in the row-echelon form during Gaussian Elimination, it is a sign of such dependence. It ends up offering you a parameter for which you can express the other variables.

• The system does not have a unique solution.
• Use parameters to express each variable, like setting one variable in terms of another.

In our initial problem, solving involves expressing the solution set as equations in terms of a parameter. This will represent infinite sets of solutions. Here, the parameter was set as \( v \), giving solutions in terms of \( t \), \( u \), and resulting in respect to \( z \). These parameters give you a direction to find solutions between infinite possible sets.