Problem 13 - Intermediate Algebra

Question

Solve each of the following systems. If the solution set is \(\varnothing\) or if it contains infinitely many solutions, then so indicate. $$ \left(\begin{array}{l} 3 x-2 y+4 z=6 \\ 9 x+4 y-z=0 \\ 6 x-8 y-3 z=3 \end{array}\right) $$

Step-by-Step Solution

Answer

The system has a unique solution: \(x = \frac{187}{132}\), \(y = -\frac{43}{22}\), \(z = \frac{6}{11}\).

Step 1: Analyze the System of Equations

We have a system of three linear equations with three variables: \(x\), \(y\), and \(z\). The equations are:\[\begin{cases} 3x - 2y + 4z = 6 \ 9x + 4y - z = 0 \ 6x - 8y - 3z = 3 \end{cases}\] Our task is to find if there is a unique solution, infinitely many solutions, or no solution (\(\varnothing\)).

Step 2: Transform to Row Echelon Form

Use Gaussian elimination to convert the system into row echelon form. First, let's subtract twice the first equation from the second and third equations. This gives us:\[\begin{cases} 3x - 2y + 4z = 6 \ 0 \cdot x + 8y - 9z = -12 \ 0 \cdot x - 4y - 11z = -9 \end{cases}\]

Step 3: Simplify the System

Now subtract half of the second equation from the third equation to eliminate the \(y\) term in the third equation. This results in:\[\begin{cases} 3x - 2y + 4z = 6 \ 0 \cdot x + 8y - 9z = -12 \ 0 \cdot x + 0 \cdot y - 5.5z = -3 \end{cases}\] Simplifying the third equation gives us \(-5.5z = -3\) or \(z = \frac{6}{11}\).

Step 4: Back Substitution

Substitute \(z = \frac{6}{11}\) back into the second equation: \(8y - 9\left(\frac{6}{11}\right) = -12\). Solve for \(y\): \(8y = -12 + \frac{54}{11}\), and \(y = \frac{-186}{88}\). Simplify further to find \(y = -\frac{43}{22}\).

Step 5: Solve for x

Substitute \(y = -\frac{43}{22}\) and \(z = \frac{6}{11}\) into the first equation: \(3x - 2(-\frac{43}{22}) + 4\left(\frac{6}{11}\right) = 6\). Simplify the equation to solve for \(x\): \(3x = 6 - \frac{86}{22} - \frac{24}{11}\), which gives \(x = \frac{187}{132}\).

Step 6: Conclusion

Since we have found unique values for \(x\), \(y\), and \(z\), the system has a unique solution.

Key Concepts

Linear Equations

Linear equations are equations where the highest power of the variable is one. They form straight lines when graphed on a coordinate plane.
In this exercise, we have three linear equations with three variables: \(x\), \(y\), and \(z\):

• \(3x - 2y + 4z = 6\)
• \(9x + 4y - z = 0\)
• \(6x - 8y - 3z = 3\)

Linear equations are often used to model real-life situations where relationships are proportional.
To solve a system of linear equations like this, we need to find the values of \(x, y,\) and \(z\) that make all equations true simultaneously.
One effective way to find these values is by turning these equations into a different, more manageable form known as the row echelon form.

Row Echelon Form

The row echelon form is a way of organizing a system of linear equations to make them easier to solve.
Using Gaussian elimination, we transform the original system into a triangular layout which makes back substitution straightforward.We begin by making the leading coefficient of the first equation into one, referred to as the pivot element.
Then, eliminate the x terms beneath the pivot. Our new system looks like this:

• \(3x - 2y + 4z = 6\)
• \(0 \cdot x + 8y - 9z = -12\)
• \(0 \cdot x - 4y - 11z = -9\)

We continue to use row operations to eliminate the y terms in the third equation:

• \(0 \cdot x + 0 \cdot y - 5.5z = -3\)

This results in a clear path to solve for \(z\) first.
Achieving row echelon form simplifies the system, setting the stage for back substitution.

Back Substitution

Back substitution is the process of solving for variables in a system of equations that is already in row echelon form.
Once we have transformed the system, back substitution begins with solving the equation for the last variable first.In our system, we derived from row echelon form:

• Calculate \(z\) from the third equation: \(-5.5z = -3\), yielding \(z = \frac{6}{11}\).
• Use the value of \(z\) to solve for \(y\) in the second equation: \(8y - 9(\frac{6}{11}) = -12\).
• Once \(y\) is found, \(y = -\frac{43}{22}\), substitute both \(y\) and \(z\) into the first equation to find \(x\).
• The final solution for \(x\) is \(x = \frac{187}{132}\).

By tackling the system from bottom to top, we efficiently find each variable in order.
This systematic approach ensures that we find unique solutions or identify if there are infinitely many or none.
Back substitution completes the process in handling linear systems solved through Gaussian elimination.