Problem 68 - A Graphical Approach to Precalculus with Limits

Question

Solve each system. Write solutions in terms of \(z\) if necessary. $$\begin{aligned} 3 x+5 y-z &=0 \\ 4 x-y+2 z &=1 \\ -6 x-10 y+2 z &=0 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{5 - 9z}{23}, y = \frac{50z - 15}{115}, z = z\).

Step 1: Analyze the System of Equations

We are given a system of three equations: \(3x + 5y - z = 0\), \(4x - y + 2z = 1\), and \(-6x - 10y + 2z = 0\). To solve it, we want to find values for \(x\), \(y\), and \(z\). We'll aim to express solutions in terms of one of the variables, ideally \(z\), to simplify the system.

Step 2: Eliminate a Variable

First, notice that the third equation is a multiple of the first equation (multiplied by -2). So, it doesn't provide new information. We'll focus on the first two equations. We can eliminate \(y\) by multiplying the second equation by 5 and adding to the first equation: \(5(4x - y + 2z) = 5\). This gives \(20x - 5y + 10z = 5\). Adding to the first equation, we have \(3x + 5y - z + 20x - 5y + 10z = 0 + 5\), simplifying to \(23x + 9z = 5\).

Step 3: Solve for One Variable in Terms of Another

From the resulting equation \(23x + 9z = 5\), we can express \(x\) in terms of \(z\): \(x = \frac{5 - 9z}{23}\).

Step 4: Substitute and Solve for Another Variable

Substitute \(x = \frac{5 - 9z}{23}\) into the first equation \(3x + 5y - z = 0\):\(3\left(\frac{5 - 9z}{23}\right) + 5y - z = 0\). This simplifies to \(\frac{15 - 27z}{23} + 5y - z = 0\). To eliminate fractions, multiply by 23: \(15 - 27z + 115y - 23z = 0\). Solving for \(y\), \(115y = 50z - 15\), giving \(y = \frac{50z - 15}{115}\).

Step 5: Express All Variables in Terms of \(z\)

Having expressed \(x\) and \(y\) in terms of \(z\), we have \[x = \frac{5 - 9z}{23}, \, y = \frac{50z - 15}{115}, \, z = z\]. This is the set of solutions where \(x\), \(y\), and \(z\) depend linearly on the parameter \(z\).

Key Concepts

Systems of Equations

The idea of solving a system of equations is to find values for variables that satisfy all given equations simultaneously. In this problem, we have a system of three linear equations with three variables: \(x\), \(y\), and \(z\). A linear equation in this context is an equation of the form \(ax + by + cz = d\), where \(a\), \(b\), \(c\), and \(d\) are constants.

When dealing with multiple equations, their solution can either be:

• Consistent and independent, meaning there is exactly one set of values for \(x\), \(y\), and \(z\) that satisfy all equations.
• Consistent and dependent, implying the equations are multiple representations of the same line or plane in space, having infinitely many solutions.
• Inconsistent, where no single set of values will satisfy all equations.

In this exercise, the system reveals infinite solutions dependent on a parameter, indicating it's consistent and dependent. By determining the relationship between the variables, particularly expressing \(x\) and \(y\) in terms of \(z\), we find a general solution.

Variable Elimination

Variable elimination is a strategic approach to solving systems of equations by removing one variable at a time, simplifying the problem. In our exercise, this method is used to reduce the complexity of three equations. We do this by:

Transforming or manipulating the equations to remove one of the variables. The ultimate goal is to produce a simpler system that can be solved more easily.

• First, notice dependencies such as the third equation being merely a multiple of the first. This insight immediately reduces the problem's complexity by allowing us to focus on a simplified, independent system.
• Then, focus on eliminating one variable. The given solution shows how we eliminated \(y\) by combining multiples of the first two equations. This step effectively reduced the system from three equations to two equations by canceling \(y\) out.

Variable elimination streamlines the process of finding parametric solutions by clearing the path to express one variable in terms of another. As evidenced in the exercise, it results in an equation solely outlining \(x\) as a function of \(z\).

Parametric Solutions

In some cases, such as with dependent systems, we express solutions parametrically, meaning in terms of a parameter. This approach unifies solutions under a single expression, making it simple to identify specific solutions by varying the parameter. In this exercise, we treat \(z\) as a parameter for this purpose.

• Expressing \(x\) in terms of \(z\): By solving the equation \(23x + 9z = 5\), we arrive at the expression \(x = \frac{5 - 9z}{23}\). This tells us how \(x\) changes with \(z\).
• Expressing \(y\) similarly, we take the expression \(y = \frac{50z - 15}{115}\), showing its dependence on \(z\).
• The parameter \(z\) remains as is, meaning that it can be any real number.

Consequently, our set of parametric solutions, \([x = \frac{5 - 9z}{23}, y = \frac{50z - 15}{115}, z = z]\), encapsulates the idea that each solution corresponds to a particular \(z\). This method effectively highlights the infinite nature of solutions typical for consistent and dependent systems.