Problem 23 - 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{aligned} 8 x-3 y+6 z &=-2 \\ 4 x+9 y+4 z &=18 \\ 12 x-3 y+8 z &=-2 \end{aligned}

Step-by-Step Solution

Answer

The solution set is parameterized by \(v=c\) as \( ( t, u, v ) = \left( \frac{-11c - 6}{10c}, \frac{6 + 11c}{6c}, c \right) \).

Step 1: Introduction to Problem

We need to determine if the given system of equations is dependent, and if so, write the solution set in terms of the variable \(z\). We will use a substitution method with the hint provided, where \(t = \frac{1}{x}\), \(u = \frac{1}{y}\), and \(v = \frac{1}{z}\).

Step 1: Convert the Equations

Convert each equation into an equation in terms of \(t\), \(u\), and \(v\) using the substitutions: \(x = \frac{1}{t}\), \(y = \frac{1}{u}\), and \(z = \frac{1}{v}\). The original equations become: - \(8\left(\frac{1}{t}\right) - 3\left(\frac{1}{u}\right) + 6\left(\frac{1}{v}\right) = -2\)- \(4\left(\frac{1}{t}\right) + 9\left(\frac{1}{u}\right) + 4\left(\frac{1}{v}\right) = 18\)- \(12\left(\frac{1}{t}\right) - 3\left(\frac{1}{u}\right) + 8\left(\frac{1}{v}\right) = -2\)

Step 2: Clear Fractions

Clear the fractions by multiplying each equation by the common denominator to get equations in terms of \(t\), \(u\), and \(v\):

Step 4: 3: Simplified Form of Equations

Simplify each equation. Multiply through each equation by \(tuv\):- First equation: \(8uv - 3tv + 6tu = -2tuv\)- Second equation: \(4uv + 9tv + 4tu = 18tuv\)- Third equation: \(12uv - 3tv + 8tu = -2tuv\)

Step 4: Check for Dependent Equations

Check if the system of equations is dependent by trying to express one equation as a linear combination of the others. Notice that the first and third equations can be the same if multiplied by 1.5:

Step 5: Write Dependent Solutions

Since the system is dependent, express one variable in terms of others. From equations, express \(t, u\) in terms of \(v\). Let \(v = c\), a constant, so the solution set is expressed as:\[\left( t, u, c \right) = \left( \frac{-11c - 6}{10c}, \frac{6 + 11c}{6c}, c \right)\].

Step 7: Conclusion

The system of equations is dependent and the solution set can be parameterized by \(c\), which corresponds to \(v\) or \(\frac{1}{z}\).

Key Concepts

System of Equations

A system of equations involves multiple equations that have common variables. In this context, we are working with a system of three equations:

• \(8x - 3y + 6z = -2\)
• \(4x + 9y + 4z = 18\)
• \(12x - 3y + 8z = -2\)

A key objective in solving such systems is finding values for the variables \(x\), \(y\), and \(z\) that satisfy all equations simultaneously. This process often involves rearranging, substituting, or eliminating variables across the equations to find a common solution.
Systems like this can be classified as consistent or inconsistent, independent or dependent, based on whether they have one unique solution, infinitely many solutions, or no solution at all.

Solution Set

The solution set of a system of equations is the collection of all possible solutions that satisfy every equation in the system. It's represented as ordered triplets \((x, y, z)\).
For our problem, since the system is dependent (meaning the equations are not independent from each other), we arrive at an infinite number of solutions. This implies that the equations represent the same plane or are parallel with overlap in three-dimensional space.
Dependent systems typically mean there's no unique solution. Instead, solutions vary with parameters, usually represented by a free variable. In our case, the solution set is expressed in terms of \(z\).

Substitution Method

The substitution method is an algebraic technique used to solve a system of equations. It involves solving one of the equations for a single variable and then substituting this expression into the other equations.
In this problem, we employ the substitution method by using the conversions given: \(t = \frac{1}{x}\), \(u = \frac{1}{y}\), and \(v=\frac{1}{z}\). By expressing each variable in terms of these new ones, we transform the original equations into a clearer form. This allows us to analyze relationships between \(t\), \(u\), and \(v\) more easily, aiding in finding solutions that satisfy all transformed equations.By successfully substituting and simplifying, you eliminate one variable and solve for the others, helping to parameterize the solution.

Parametrization

Parametrization involves expressing the solution set of a dependent system using one or more free parameters. It is especially useful when dealing with infinitely many solutions, as it provides a concise way to describe them all.
For dependent systems like ours, we introduce a parameter \(c\) representing \(v\), which equals \(\frac{1}{z}\).
The solution can be generalized using expressions for \(t\) and \(u\) derived from one of the equations in terms of \(v\):

• \(t = \frac{-11c - 6}{10c}\)
• \(u = \frac{6 + 11c}{6c}\)
• And \(v = c\)

These parametrized expressions enable us to find any particular solution by assigning different values to the parameter \(c\). This encapsulates the entire solution space and is particularly handy in sharing the limitless options of solutions simply and effectively.