Problem 36 - College Algebra

Question

Solve each system by Gaussian elimination. $$ \begin{array}{l} \frac{1}{40} x+\frac{1}{60} y+\frac{1}{80} z=\frac{1}{100} \\ -\frac{1}{2} x-\frac{1}{3} y-\frac{1}{4} z=-\frac{1}{5} \\ \frac{3}{8} x+\frac{3}{12} y+\frac{3}{16} z=\frac{3}{20} \end{array} $$

Step-by-Step Solution

Answer

Infinitely many solutions parameterized by \(x = 0.4 - 0.6t - 0.5s, y = t, z = s\).

Step 1: Clear Fractions

The first step is to clear the fractions from each equation by multiplying by the least common multiple (LCM) of the denominators. Equation 1 has denominators 40, 60, 80, and 100. The LCM is 1200. Multiplying throughout by 1200 gives:\[ 30x + 20y + 15z = 12 \]Equation 2 has denominators 2, 3, 4, and 5. The LCM is 60. Multiplying throughout by 60 gives:\[ -30x - 20y - 15z = -12 \]Equation 3 has denominators 8, 12, 16, and 20. The LCM is 240. Multiplying throughout by 240 gives:\[ 90x + 60y + 45z = 36 \]

Step 2: Simplify the System

Observe that the system of equations obtained in Step 1 is:\[\begin{align*}1) & \quad 30x + 20y + 15z = 12 \2) & \quad -30x - 20y - 15z = -12 \3) & \quad 90x + 60y + 45z = 36\end{align*}\]Notice that equation 1 and equation 2 are negatives of each other, which means their addition will lead to 0 = 0. Therefore, the first two equations are effectively redundant, and we only need to consider equation 3.

Step 3: Solve the Simplified System

From equation 3:\[ 90x + 60y + 45z = 36 \]We can divide everything in the equation by 9 to simplify:\[ 10x + 6y + 5z = 4 \]This is a single equation with three variables, indicating infinitely many solutions can satisfy this equation.

Step 4: Express Variables Using Parameters

We will express two of the variables in terms of one of them, treating one variable as a parameter.Let's solve for \(x\) in terms of \(y\) and \(z\):\[10x = 4 - 6y - 5z \x = \frac{4 - 6y - 5z}{10} \x = 0.4 - 0.6y - 0.5z\]Let \(y = t\) and \(z = s\), we'll express \(x\):\[x = 0.4 - 0.6t - 0.5s\]

Step 5: General Solution

The solution to the system is expressed in terms of parameters \(t\) and \(s\):\[\begin{align*}x &= 0.4 - 0.6t - 0.5s \y &= t \z &= s\end{align*}\]Where \(t\) and \(s\) can be any real numbers, providing infinitely many solutions.

Key Concepts

Linear Systems of Equations

A linear system of equations is a collection of two or more linear equations involving the same set of variables. In our exercise, we have three equations with three variables \(x\), \(y\), and \(z\). The goal is to find values for these variables that satisfy all equations in the system simultaneously.
Understanding how to solve these systems is critical in algebra, as they appear frequently in mathematical modeling and real-life scenarios. In solving these, we typically use methods like substitution, elimination, or more broadly applicable techniques such as Gaussian elimination.

• **Gaussian Elimination**: This method involves a series of operations to transform a system into upper triangular form, which makes it easier to solve through back-substitution.
• **Applications**: Linear systems can model real-world problems such as resource allocation, network flows, and economic forecasts.

This exercise demonstrated using Gaussian elimination by transforming equations to eliminate fractions and analyze solutions.

Fractions in Algebra

When dealing with equations that include fractions, simplifying these fractions early on can substantially ease the solving process. Fractions add complexity because they require precise operations to maintain their values. In our exercise, we began by clearing the fractions from each equation.
This is accomplished by multiplying each equation by its least common multiple (LCM) of the denominators involved:

• **Removing Fractions**: This step converts the equations into a simpler format by removing fractions, making the calculations more straightforward.
• **Example**: In equation 1, multiplying \( \frac{1}{40}x + \frac{1}{60}y + \frac{1}{80}z = \frac{1}{100} \) by 1200 simplifies it greatly to \(30x + 20y + 15z = 12 \).

By clearing fractions, we achieve whole number coefficients, making the following steps in Gaussian elimination easier.

Infinitely Many Solutions

In mathematics, a system of equations is said to have infinitely many solutions when there is not just one, but countless sets of values that satisfy all equations of the system. This often occurs in a system where at least one equation can be derived from others through addition, subtraction, or scalar multiplication.
This manifests in various forms:

• **Dependent Equations**: Such as when one equation is a multiple of another, leading to redundancy in the system.
• **Example from Exercise**: In the given problems, after simplifying, it became evident that the first two equations were simply negatives of each other, meaning they didn't contribute new information, leading to a dependent system.
• **Parameterization**: Reducing a system like these into a simpler form using a parameterization approach allows representation of solutions in terms of free variables, for example, expressing \(x\), \(y\), and \(z\) with parameters \(t\) and \(s\).

Understanding when and why a system has infinitely many solutions is crucial as it influences how we interpret these solutions within different contexts.