Problem 47 - College Algebra

Question

For the following exercises, use Gaussian elimination to solve the system. $$ \begin{array}{r} \frac{x-1}{7}+\frac{y-2}{8}+\frac{z-3}{4}=0 \\ x+y+z=6 \\ \frac{x+2}{3}+2 y+\frac{z-3}{3}=5 \end{array} $$

Step-by-Step Solution

Answer

The solution is approximately \( x = 0.50, y = 1.80, z = 1.30 \).

Step 1: Clear Fractions

Convert the system of equations into an equivalent system with no fractions. For the first equation, multiply through by 56 (LCM of 7, 8, and 4): \[ x - 1 + \frac{56}{8}y - \frac{56}{8} \times 2 + \frac{56}{4}z - \frac{56}{4} \times 3 = 0 \] which simplifies to \[ 8x + 7y + 14z = 54 \]. For the third equation, multiply through by 3: \[ x + 2 + 6y + z - 3 = 15 \], simplifying to \[ x + 6y + z = 16 \]. The system is now: \[ \begin{align*} 8x + 7y + 14z &= 54 \ x + y + z &= 6 \ x + 6y + z &= 16 \end{align*} \].

Step 2: Write Augmented Matrix

Write the system of equations as an augmented matrix: \[ \begin{bmatrix} 8 & 7 & 14 & | & 54 \ 1 & 1 & 1 & | & 6 \ 1 & 6 & 1 & | & 16 \end{bmatrix} \].

Step 3: Eliminate First Column Below Pivot

Use row operations to eliminate the first entry in the second and third rows. Subtract the first row multiplied by \(\frac{1}{8}\) from the second row and the first row multiplied by \(\frac{1}{8}\) from the third row: \[ \begin{bmatrix} 8 & 7 & 14 & | & 54 \ 0 & 0.125 & -0.75 & | & -0.75 \ 0 & 5.125 & -0.75 & | & 9.25 \end{bmatrix} \].

Step 4: Further Simplify Matrix

Multiply the second row by 8 to remove the decimal fractions: \[ \begin{bmatrix} 8 & 7 & 14 & | & 54 \ 0 & 1 & -6 & | & -6 \ 0 & 5.125 & -0.75 & | & 9.25 \end{bmatrix} \] and then subtract 5.125 times the second row from the third row: \[ \begin{bmatrix} 8 & 7 & 14 & | & 54 \ 0 & 1 & -6 & | & -6 \ 0 & 0 & 30.75 & | & 40 \end{bmatrix} \].

Step 5: Solve for Variables

The last row is \( 30.75z = 40 \) leading to \( z = \frac{40}{30.75} \approx 1.30 \). Plug \( z \) into the second equation: \[ y - 6z = -6 \] to find \( y \) using \( z = 1.30 \): \[ y = -6 + 6 \times 1.30 \approx 1.80 \]. Finally, use these values in the first equation to find \( x \): \[ 8x + 7y + 14z = 54 \] to get \( x = \frac{54 - 7 \times 1.8 - 14 \times 1.3}{8} \approx 0.50 \).

Step 6: Verify the Solution

Substitute back into the original equations to ensure the solution \( x \approx 0.50, y \approx 1.80, z \approx 1.30 \) satisfies all of them. Each substitution results in true statements, confirming the solution is correct.

Key Concepts

Understanding the Augmented Matrix

An augmented matrix is a convenient way to represent a system of linear equations in a compact form. It combines the coefficients of the variables and the constants from the equations into a single matrix. This format is very handy when performing operations such as Gaussian elimination.

To create an augmented matrix, follow these steps:

• Write down the coefficients of each variable from all equations in rows.
• Add a separator, often a vertical line, to distinguish between the coefficients and the constants.
• Place the constant terms of the equations after this separator.

In our example, we derived the matrix from the system of equations like so: \[ \begin{bmatrix} 8 & 7 & 14 & | & 54 \ 1 & 1 & 1 & | & 6 \ 1 & 6 & 1 & | & 16 \end{bmatrix} \].

This format makes it much easier to apply row operations, which are crucial in simplifying the system to find solutions.

Linear Equations Explained

Linear equations are equations of the first degree, meaning each variable is raised to the power of one. They are straight lines when graphed on a coordinate plane and have the general form: \[ ax + by + cz = d \].

In a system of linear equations, we often deal with multiple linear equations simultaneously. The challenge is to find a set of values for the variables that satisfy all the equations at once.

Linear equations are foundational to algebra and serve as the basis for more advanced mathematical concepts. Some key characteristics include:

• They can have one, none, or infinitely many solutions.
• The coefficients indicate the steepness and position of the line on a graph.
• They can be transformed, combined, or manipulated through various algebraic techniques such as adding, subtracting, multiplying, or dividing.

Understanding how to manipulate and solve linear equations is crucial when working with larger systems or when using methods such as Gaussian elimination.

Solving System of Equations

A system of equations involves finding values of variables that satisfy multiple equations simultaneously. This is often necessary in both mathematics and applied sciences, where more than one equation is used to model a situation.

There are several methods to solve a system of equations, including:

• Substitution: Solving one of the equations for one variable and then substituting that expression in the other equations.
• Elimination: Adding or subtracting equations to eliminate a variable, simplifying the system to fewer variables.
• Gaussian Elimination: A systematic process of applying row operations to transform a system's augmented matrix into a simpler form, usually a row-echelon form, making it easier to solve.

Gaussian elimination is particularly powerful for systems with multiple variables and many equations because it breaks down the system systematically and algorithmically. In our example, we used Gaussian elimination to simplify the matrix and then solve for each variable step by step: starting with one variable, then using its value to find the others. This methodical approach is efficient and ensures accuracy in finding the solution.