Problem 32 - 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} 10 x+2 y-3 z &=0 \\ 5 x+4 y+6 z &=-1 \\ 6 y+3 z &=2 \end{aligned}

Step-by-Step Solution

Answer

The solution set is \( \left( \frac{3z - 2}{30}, \frac{2 - 3z}{6}, z \right) \)."

Step 1: Rewrite Variables

Introduce new variables according to the hint. Let \( t = \frac{1}{x} \), \( u = \frac{1}{y} \), and \( v = \frac{1}{z} \). With this substitution, the given equations in terms of \( x, y, z \) don't directly transform into linear equations for \( t, u, v \) due to the nonlinear nature of the substitutions. We'll focus first on solving the original system for dependencies.

Step 2: Analyze and Simplify System

Inspect the given equations: \( 10x + 2y - 3z = 0 \), \( 5x + 4y + 6z = -1 \), and \( 6y + 3z = 2 \). Solve for a possible dependency. Notice that the first and second equations can be simplified and compared. Multiply the first equation by 2 to see dependency with the second equation.

Step 3: Multiply and Compare

Multiply the entire first equation by 2: \( 20x + 4y - 6z = 0 \). Now compare it with the second equation: \( 5x + 4y + 6z = -1 \). Notice they are not direct multiples as the constants do not line up after comparison, revealing no dependency yet. Instead, consider the system involving all three equations.

Step 4: Substitute Third Equation

Use the third equation \( 6y + 3z = 2 \) to express \( y \) in terms of \( z \): \( y = \frac{2-3z}{6} \). Now substitute this into the first and second equations to reduce them further in terms of \( z \).

Step 5: Substitute in First Equation

Understanding dependency, substitute \( y = \frac{2-3z}{6} \) into the first equation: \( 10x + 2\times\frac{2-3z}{6} - 3z = 0 \). Simplify: \( 10x + \frac{2}{3} - \frac{z}{3} - 3z = 0 \). Thus: \( 10x - \frac{3z}{3} - 3z = -\frac{2}{3} \), simplifying to \( 10x - dz = -\frac{2}{3} \) as dependent terms.

Step 6: Solve for x in Terms of z

Extract and solve for \( x \) from the previous equation: \( 10x = \frac{3z - 2}{3} \), leading to \( x = \frac{3z - 2}{30} \). This expression of \( x \) in terms of \( z \) provides a parameterized solution context.

Step 7: Implication and Solution Set

Conclusively, given the changes, note the parametric dependency options: \( x = \frac{3z - 2}{30} \), \( y = \frac{2 - 3z}{6} \), along with \( z \). Specifically, express solutions using \( z \) as a parameter: \( x = \frac{3z - 2}{30}, y = \frac{2 - 3z}{6} \).

Step 8: Summarize

The equations are dependent, producing a solution set parameterized by \( z \): \( \left( \frac{3z - 2}{30}, \frac{2 - 3z}{6}, z \right) \).

Key Concepts

Parameterization

Parameterization is a method in mathematics that allows us to express variables in terms of one or more parameters. Here, the concept is used to express the solution of a system of equations in terms of the variable \( z \). Instead of a single point, a parameterized solution describes a family of solutions based on a free variable - the parameter. When solving dependent systems of equations, this technique helps in finding all possible solutions linked by a certain variable.

• Parameters as Free Variables: In our context, \( z \) acts as a parameter, meaning it can take any value, and correspondingly, \( x \) and \( y \) adjust according to the relationships established in the equations.
• Benefits: With parameterization, a complex relationship can be defined in simpler terms, allowing us to see all possible solutions or outcomes, rather than finding just one specific solution.

Linear Equations

Linear equations are equations of the first order, which include variables raised to the power of one. The provided system involves three linear equations. They are: \[\begin{aligned}10x + 2y - 3z &= 0 \5x + 4y + 6z &= -1 \6y + 3z &= 2\end{aligned}\]These equations can represent planes in a three-dimensional space, and the solutions are where these planes intersect.

• Characteristics: Each equation can be graphically represented as a plane in 3D space, and their solutions reflect where these planes might intersect or overlap.
• System Types: Our particular system is dependent, meaning the solutions can be expressed on a shared line or plane instead of a single point, requiring parameterization.

Solution Sets

A solution set is the collection of all possible solutions of a system of equations. For dependent systems, the solution set typically depends on one or more parameters. In our exercise, the solution set for the system is derived by expressing all variables in terms of the parameter \( z \):\[(x, y, z) = \left(\frac{3z - 2}{30}, \frac{2 - 3z}{6}, z\right)\]

• Understanding Dependencies: Solution sets in dependent systems reveal connections among variables, indicating that a change in one directly affects the others.
• Flexibility: This parametric form allows us to input any value for \( z \) and find corresponding \( x \) and \( y \) values to satisfy the system.

Algebraic Substitution

Algebraic substitution is a foundational technique in mathematics used to simplify complex systems and equations by replacing one variable with another equivalent expression, typically to solve for one of the variables.In the exercise:- Substitution was used as a strategy to reduce the number of variables by replacing variables with expressions found from other equations.- For instance, from the equation \( 6y + 3z = 2 \), we derived \( y = \frac{2 - 3z}{6} \), which was substituted into other equations to find simpler forms depending on \( z \).- This helps isolate a particular variable or reveal dependencies among variables.

• Simplifying Equations: Substitution is often employed to convert a system into one with fewer variables, like expressing \( x \) in terms of \( z \) here.
• Step-by-step Reduction: As seen, each substitution simplifies system complexity, progressively unraveling the dependencies within.