Problem 29 - 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}{c} x-y+z=-6 \\ 4 x+y+z=7 \end{array}

Step-by-Step Solution

Answer

The solution set is \( x = \frac{1-2z}{5}, \ y = \frac{1+3z}{5}, \ z = z \).

Step 1: Substitute Variables

Given the hint, substitute the variables such that \( t=\frac{1}{x}, u=\frac{1}{y}, v=\frac{1}{z} \). However, in our main equations, we can use substitution to express them with \( x, y, \) and \( z \) as it is given that we need to solve for them.

Step 2: Set Up System of Equations

Based on the system provided: \( x - y + z = -6 \) and \( 4x + y + z = 7 \). These are the equations we need to work with.

Step 3: Eliminate \(y\)

To eliminate \(y\) from the equations, add the two equations. This results in: \( (x-y+z) + (4x+y+z) = -6 + 7 \). This simplifies to \( 5x + 2z = 1 \).

Step 4: Solve for a Variable

From the equation \( 5x + 2z = 1 \), express one of the variables in terms of the other - for example, solving for \( x \) gives \( x = \frac{1 - 2z}{5} \).

Step 5: Express Solution Set in Terms of \(z\)

Since we are given the equations are dependent, express the other variables in terms of \( z \). Use substitution: from Step 4, we have \( x = \frac{1 - 2z}{5} \). Plugging this into one of the original equations helps find \( y \). For example, substitute \( x \) in the first original equation \( x - y + z = -6 \) to solve for \( y \).

Step 6: Solve for \(y\)

For simplification, from Step 5, use \( x = \frac{1-2z}{5} \) in \( x - y + z = -6 \): \( \frac{1-2z}{5} - y + z = -6 \). Solve for \( y \) yielding: \( y = \frac{1-2z}{5} + z + 6 \).

Step 7: Final Solution Set in Terms of \(z\)

Thus, the solution can be expressed as: \( x = \frac{1-2z}{5}, \, y = \frac{1+3z}{5}, \, \text{and variable } z \text{ is free} \). Hence, the solution set is in terms of \( z \).

Key Concepts

Solution Set

A solution set is a collection of values that satisfy a given system of equations. When dealing with dependent equations, the solution set typically exhibits dependency on one or more parameters. In scenarios like this, instead of finding a unique solution, we express the solution set in terms of a free variable, such as \( z \) in the given exercise. This means that instead of pinpointing exact numbers for \( x \) and \( y \), we articulate them as expressions containing \( z \). This approach is especially valuable as it highlights the infinite nature of solutions when equations don't intersect at a single point, but rather along a line or plane in space.

System of Equations

A system of equations consists of two or more equations with the same set of unknowns. Solving a system means finding values for the unknowns that satisfy each equation simultaneously.
In the given problem, we have the equations:\( x - y + z = -6 \) and \( 4x + y + z = 7 \). The challenge often lies in how to manage and manipulate these equations to identify the relationships amongst variables.
This system may be solved using several methods, such as substitution or elimination, each assisting in finding solutions by reducing complexity step by step. Understanding how these methods function can unlock solutions to even more complicated systems.

Substitution Method

The substitution method is a powerful tool for solving systems of equations. This approach involves expressing one variable in terms of another and substituting this expression into the remaining equations.
For the exercise, we first aimed to simplify the equations by eliminating \( y \) using addition, resulting in a clearer equation: \( 5x + 2z = 1 \). This simplified equation allowed further manipulation by isolating \( x \) in terms of \( z \). This step paved the way to express \( y \) similarly in terms of \( z \).
The substitution method is particularly effective when equations have parameters or constraints that make direct solutions challenging. By systematically reducing equations, it transforms a multi-variable problem into single-variable expressions, making it easier to handle.