Skip to main content

Activities for Intermediate Algebra: Instructors’ Guide

Katherine Yoshiwara

Section 2.4 Gaussian Reduction

This Lesson extends the elimination method to 3x3 linear systems. We concentrate on the process of Gaussian elimination, because this useful technique generalizes to higher order linear systems. Students usually have no trouble mastering back-substitution in a triangular system. Although there is a lot of grumbling initially at the number of steps in Gaussian reduction, if you show students how to keep their work organized, after the second example they feel more comfortable.

Activity 2.4.1. Triangular Systems.

Use back-substitution to solve the system.
\begin{align*} 3x + y - 2x \amp = 10\\ 2y + 3x \amp = 4\\ 6x \amp = -24 \end{align*}
Solution:

Activity 2.4.2. Gaussian Elimination.

  1. Follow the steps given in your textbook to solve the system.
    \begin{align*} \text{(1)}~~~~~~x - 2y + 4z \amp = -3\\ \text{(2)}~~~~~~3x + y - 2z \amp = 12\\ \text{(3)}~~~~~~2x + y - 3z \amp = 11 \end{align*}
    1. There are no fractions in the equations, so we go on to Step 2.
    2. Eliminate \(x\) from Equations (1) and (2). Call your new equation (A).
    3. Eliminate \(x\) from Equations (1) and (3). Call your new equation (B).
    4. Form a 2x2 system from Equations (A) and (B). Eliminate \(x\) from that system.
      (Hint: Divide Equation (A) by 7 first.)
    5. Form a triangular system, and solve that system by back-substitution.
      Solution:
  2. Solve the system below. Begin by clearing the fractions from each equation.
    \begin{align*} \text{(1)}~~~~~~x + 2y + \frac{1}{2} z \amp = 0\\ \text{(2)}~~~~~~x + \frac{3}{5} y - \frac{2}{5} z \amp = \frac{1}{5}\\ \text{(3)}~~~~~~4x - 7y - 7z \amp = 6 \end{align*}

Activity 2.4.3. Making the Smart Choice.

  1. Sometimes a wise choice of steps can shorten the process. Consider the system below.
    \begin{align*} \text{(1)}~~~~~~\amp 3y + z = 3\\ \text{(2)}~~~~~~\amp -2x + 3y = 7\\ \text{(3)}~~~~~~\amp 3x + 2z = -6 \end{align*}
    Notice that there is no \(x\)-term in Equation (1), so it can serve as Equation (A), and Step 2 is done. What is the smart choice for Step 3?
  2. Now finish the solution.

Activity 2.4.4. An Application.

A manufacturer of office supplies makes three types of file cabinet: two-drawer, four-drawer, and horizontal. The manufacturing process is divided into three phases: assembly, painting, and finishing.
  • A two-drawer cabinet requires 3 hours to assemble, 1 hour to paint, and 1 hour to finish.
  • The four-drawer model takes 5 hours to assemble, 90 minutes to paint, and 2 hours to finish.
  • The horizontal cabinet takes 4 hours to assemble, 1 hour to paint, and 3 hours to finish.
The manufacturer employs enough workers for 500 hours of assembly time, 150 hours of painting, and 230 hours of finishing per week. How many of each type of file cabinet should he make in order to use all the hours available?
  1. Represent the number of each model of file cabinet by a different variable, for example:
    Number of two-drawer cabinets:\(\hphantom{00} x\)
    Number of four-drawer cabinets:\(\hphantom{00} y\)
    Number of horizontal cabinets: \(\hphantom{00} z\)
  2. Organize the time constraints into a table. (Convert all times to hours.)
    2-Drawer 4-Drawer Horizontal Total Available
    Assembly \(\hphantom{0000000000}\) \(\hphantom{00000}\) \(\hphantom{00000}\) \(\hphantom{00000}\)
    Painting
    Finishing
  3. Write three equations describing the time constraints in each of the three manufacturing phases. For example, the assembly phase requires 3 hours for the two-drawer cabinets, 3 hours for the four-drawer cabinets, and 4 hours for the horizontal cabinets, and the sum of these times should be the time available, 500 hours.
    (Assembly time) (1)
    (Painting time) (2)
    (Finishing time) (3)
  4. Solve the system. Follow the steps suggested below.
    1. Clear the fractions from the second equation.
    2. Subtract Equation (1) from 3 times Equation (3) to obtain a new Equation (4).
    3. Subtract Equation (2) from twice Equation (3) to obtain a new Equation (5).
    4. Equation (4) and (5) form a 2x2 system in \(y\) and \(z\text{.}\) Subtract Equation (5) from Equation (4) to obtain a new Equation (6).
    5. Form a triangular system with equations (3), (4), and (6). Use back-substitution to complete the solution.
    You should find the following solution: The manufacturer should make 60 two-drawer cabinets, 40 four-drawer cabinets, and 30 horizontal cabinets.

Subsection 2.4.1 Check Your Understanding

  1. What is a triangular system?
  2. In Gaussian elimination, after you eliminate one of the variables from a pair of equations, what should you do next?
  3. What should you do if one of the original equations has only two variables?
  4. What is the difference between an inconsistent system and a dependent system?

Subsection 2.4.2 Wrap Up

In this Lesson, we worked on the following skills and goals related to linear models:
  • Use back-substitution to solve a triangular system
  • Use Gaussian elimination to solve a 3x3 linear system
  • Model a problem with three linear equations in three unknowns

Subsection 2.4.3 Questions for Writing or Discussion

  1. What is a solution to a 3x3 linear system?
  2. Why do we try to reduce a 3x3 linear system to triangular form?
  3. Explain how you know immediately that the following system has no solution:
    \begin{align*} \text{(1)}~~~~~~ x + y + z = \amp 4\\ \text{(2)}~~~~~~2x + 2y + 2z = \amp 5\\ \text{(3)}~~~~~~x - y - 3z = \amp 6 \end{align*}
  4. Explain how you know immediately that the following system is dependent:
    \begin{align*} \text{(1)}~~~~~~ x + y + z = \amp 3\\ \text{(2)}~~~~~~2x + 2y + 2z = \amp 6\\ \text{(3)}~~~~~~-x - y - z = \amp -3 \end{align*}

Concept Questions.

  1. What sort of system can be solved by back-substitution?
    1. Any linear system
    2. A Gaussian system
    3. A triangular system
    4. A dependent system
  2. After we eliminate one variable from a pair of equations, what is the next step?
    1. Eliminate a different variable from the same pair of equations.
    2. Eliminate the same variable from a different pair of equations.
    3. Eliminate a different variable from a different pair of equations.
    4. Substitute into the third equation.
  3. What does it mean if you obtain an equation of the form \(~0x+0y+0z=k,~k \not=0\text{?}\)
    1. The system has no solution.
    2. The system is dependent.
    3. The solution is \((0,0,0)\text{.}\)
    4. The system is triangular.
  4. What should you do if one of the original equations in a 3x3 system is "missing" one of the variables?
    1. Throw out that equation.
    2. Add a variable to both sides of the equation.
    3. Eliminate that variable from the other two equations.
    4. Nothing: there is no solution
PrevTopNext