Problem 27 - 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} 3 x+2 y-z &=-1 \\ 3 y+z &=12 \\ x-3 z &=-3 \end{aligned}

Step-by-Step Solution

Answer

The solution set is \((x, y, z) = (-3, 4, 0)\).

Step 1: Substitute Variables

Given the equations:1. \(3x + 2y - z = -1\)2. \(3y + z = 12\)3. \(x - 3z = -3\)Use the hint: let \(t = \frac{1}{x}, u = \frac{1}{y}, v = \frac{1}{z}\). However, since these new variables are not used directly in the remaining equations, we will focus on solving the system with the original variables first.

Step 2: Express x and y in Terms of z

Use equation 3 to find \(x\) in terms of \(z\):\[x = 3z - 3\]Substitute \(x\) into equation 1:\[3(3z - 3) + 2y - z = -1\]\[9z - 9 + 2y - z = -1\]\[8z + 2y = 8\]\[2y = 8 - 8z\]\[y = 4 - 4z\]

Step 3: Solve for y in Terms of z

To check consistency, substitute \(y\) from Step 2 into equation 2:\[3(4 - 4z) + z = 12\]\[12 - 12z + z = 12\]\[-11z + 12 = 12\]\[-11z = 0\]\[z = 0\]Therefore, \(y\) is \(y = 4\) and \(x = -3\).

Step 4: Verify the Solution

With \(x = -3, y = 4, z = 0\), check in original equations:For equation 1:\[3(-3) + 2(4) - 0 = -9 + 8 = -1\] which is correct.For equation 2:\[3(4) + 0 = 12\] which is correct.For equation 3:\[-3 - 3(0) = -3\] which is correct.Thus, the solution is consistent and correct.

Key Concepts

Solution Set

In mathematical terms, a solution set is the collection of all possible solutions to a given system of equations. This is particularly important for understanding dependent equations, where the solutions often form a continuous set rather than discrete points.
In the provided system of equations, we determined a unique solution set by expressing variables in terms of another.
In this particular situation, where equations are dependent, they share the same geometric plane.

• The solution must satisfy all the original equations for the specific set of values found.
• If equations are dependent, they will not contradict each other since they describe the same line or plane.

Here, we expressed the solution set in terms of the variable \( z \) to ensure all elements within the set can satisfy each original equation consistently.

Systems of Equations

A system of equations is a collection of two or more equations with a common set of variables. The main goal is to find values for the variables that can simultaneously satisfy all the equations in the system.
In the exercise above, we had a system with three equations and three variables: \( x \), \( y \), and \( z \). Each equation provides a relationship between the variables.
There are different types of systems, such as:

• Independent Systems: These have a single solution where all equations intersect at one point.
• Dependent Systems: These have infinitely many solutions, often because the equations represent the same line or plane.
• Inconsistent Systems: These have no solution since the equations represent parallel lines or planes that never intersect.

In the given problem, the system was dependent, leading to a shared solution that was consistent across each equation.

Variables Substitution

Variables substitution is a method used to simplify solving systems of equations. By substituting one variable for another or a combination of other expressions, it becomes easier to isolate and solve for specific variables.
In our exercise, it helped express \( x \) and \( y \) in terms of \( z \).

• The third equation was rearranged to express \( x \) in terms of \( z \) as \( x = 3z - 3 \).
• By substituting \( x = 3z - 3 \) into the first equation, we could express \( y \) in terms of \( z \).

This substitution allows each variable to be easily determined once \( z \) is found, simplifying the solution process.
The ultimate advantage of using variable substitution comes during verification, as it ensures that the found values satisfy all original equations consistently. In this instance, the process was simplified by verifying if all relationships held with the derived values, ensuring the correctness of the solution set.