Section 2.4 Gaussian Reduction
Subsection 2.4.1 \(3\times 3\) Linear Systems
A solution to an equation in three variables, such as
is an ordered triple of numbers that satisfies the equation. For example, \((0, 2, 0)\) and \((1, 0, 1)\) are solutions to the equation above, but \((1, 1, 1)\) is not. You can verify this by substituting the coordinates into the equation to see if a true statement results.
A linear equation in three variables has infinitely many solutions.
An ordered triple \((x, y, z)\) can be represented geometrically as a point in space using a threedimensional Cartesian coordinate system, as shown in the figure. The graph of a linear equation in three variables is a plane. A solution to a system of three linear equations in three variables is an ordered triple that satisfies each equation in the system. That triple represents a point where all three planes intersect.
If the planes intersect in a single point, the system has a unique solution.
If there is no point that lies on all three planes (for instance, if at least two of the planes are parallel), the system is inconsistent.
If all three planes are the same, or if they intersect in a line, the system is dependent.
It is impractical to solve \(3\times 3\) systems by graphing. Even when technology for producing threedimensional graphs is available, we cannot read coordinates on such graphs with any confidence. Thus, we will restrict our attention to algebraic methods of solving such systems.
Checkpoint 2.4.1. QuickCheck 1.

A solution to an equation in three variables is an
A) ordered pair that satisfies the equation.
B) ordered triple that satisfies the equation
C) arbitrary number that satisfies the equation
The graph of a linear equation in three variables is a
point
line
plane
parabola
A linear \(3 \times 3\) system has a unique solution if the three planes intersect
a single point
a line
a plane
Three planes may also intersect
in exacty 2 points
in exactly 3 points
in a line
be the same plane
in a parabola
in a circle
Subsection 2.4.2 BackSubstitution
The strategy for solving a \(3 \times 3\) system is the same as the strategy for \(2 \times 2\) systems: we would like to reduce the system to an equation in a single variable. Once we have found the value for that variable, we substitute its value into the other equations to find the remaining unknowns.
A special case of this technique is called backsubstitution. It works when one of the equations involves exactly one variable, and a second equation involves that same variable and just one other variable. A \(3 \times 3\) system with these properties is said to be in .
Example 2.4.2.
Solve the system
We begin by solving the third equation to find \(z = 1\text{.}\) Then we substitute \(\alert{1}\) for \(z\) in the second equation and solve for \(y\text{.}\)
Finally, we substitute \(\alert{1}\) for \(z\) and \(\alert{3}\) for \(y\) into the first equation to find \(x\text{.}\)
The solution is the ordered triple \((1, 3, 1)\text{.}\) You should verify that this triple satisfies all three equations of the system.
Checkpoint 2.4.3. Practice 1.
Use backsubstitution to solve the system
Subsection 2.4.3 Gaussian Reduction
The method for solving a general \(3 \times 3\) linear system is called Gaussian reduction, after the German mathematician Carl Gauss. We use linear combinations to reduce the system to triangular form, and then use backsubstitution to find the solutions.
Checkpoint 2.4.4. QuickCheck 2.
The method for solving a \(3 \times 3\) linear system is called
completing the square.
Gaussian reduction.
induction method.
A special case of this method is called
backsubstitution.
negative reciprocals.
linear regression.
pointslope.

The special case works on systems
A) when one equation involves exactly one variable, and a second equation involves that same variable and just one other variable.
B) with more equations than unknowns.
C) having only two unknowns.
To reduce a system to the special form, we use
guessandcheck.
linear combinations.
the discriminant.
\(\text{Gaussian reduction.}\)
\(\text{backsubstitution.}\)
\(\text{A) when ... other variable.}\)
\(\text{ linear combinations.}\)
Gaussian reduction
backsubstitution
when one equation involves exactly one variable, and a second equation involves that same variable and just one other variable
linear combinations
To obtain the triangular form, we eliminate one of the variables from each of thethree equations by considering them in pairs. This results in a \(2\times 2\) system that we can solve using elimination.
Example 2.4.5.
Solve the system:
We can choose any one of the three variables to eliminate first. For this example, we will eliminate \(x\text{.}\) Next, we choose two of the equations, say (1) and (2), and use a linear combination: We multiply Equation (1) by \(2\) and add the result to Equation (2) to produce Equation (4).
Now we have an equation involving only two variables. But we need two equations in two unknowns to find the solution. So we choose a different pair of equations, say (1) and (3), and eliminate \(x\) again. We multiply Equation (1) by \(3\) and add the result to Equation (3) to obtain Equation (5).
We now form a \(2\times 2\) system with our new Equations (4) and (5).
Finally, we eliminate one of the remaining variables to obtain an equation in a single variable. We choose to eliminate \(y\text{,}\) so we add 4 times Equation (4) to \(5\) times Equation (5) to obtain Equation (6).
Now we are ready to form a triangular system. We choose one of the original equations (in three variables), one of the equations from our \(2\times 2\) system, and our final equation in one variable. We choose Equations (1), (4), and (6).
This new system has the same solutions as the original system, and we can solveit by backsubstitution. We first solve Equation (6) to find \(z = 3\text{.}\) Substituting 3 for \(z\) in Equation (4), we find
So \(y = 2\text{.}\) Finally, we substitute \(\alert{3}\) for \(z\) and \(\alert{2}\) for \(y\) into Equation (1) to find
The solution to the system is the ordered triple \((1, 2, 3)\text{.}\) You should verify that this triple satisfies all three of the original equations.
We summarize the method for solving a \(3\times 3\) linear system as follows.
Steps for Solving a \(3\times 3\) Linear System.
Clear each equation of fractions and put it in standard form.
Choose two of the equations and eliminate one of the variables by forming a linear combination.
Choose a different pair of equations and eliminate the same variable.
Form a \(2\times 2\) system with the equations found in steps (2) and (3). Eliminate one of the variables from this \(2\times 2\) system by using a linear combination.
Form a triangular system by choosing among the previous equations. Use backsubstitution to solve the triangular system.
Checkpoint 2.4.6. QuickCheck 3.
Suppose you eliminate \(y\) from two equations as Step 2 of Gaussian reduction. What should you do for Step 3?
Form a triangular system.
Eliminate either \(x\) or \(z\text{.}\)
Eliminate \(y\) again from a different pair of equations.
Substitute the value for \(y\) into the equations.
Checkpoint 2.4.7. Practice 2.
Use Gaussian reduction to solve the system
Follow the steps:
Step 1: Clear the fractions from Equation (2).
Step 2: Eliminate \(z\) from Equations (1) and (2).
Step 3: Eliminate \(z\) from Equations (1) and (3).
Step 4: Eliminate \(x\) from your new \(2\times 2\) system.
Step 5: Form a triangular system and solve by backsubstitution.
Subsection 2.4.4 Inconsistent and Dependent Systems
If, at any step in forming linear combinations, we obtain an equation of the form
then the system is inconsistent and has no solution. If we donβt obtain such an equation, but we do obtain one of the form
then the system is dependent and has infinitely many solutions.
Example 2.4.8.
Solve the systems.
\(\displaystyle \begin{alignedat}[t]{5} 3x\amp{}+{}\amp y\amp{}{}\amp 2z\amp{}={}1 \amp\qquad\amp (1)\\ 6x\amp{}+{}\amp 2y\amp{}{}\amp 4z\amp{}={}5 \amp\qquad\amp (2)\\ 2x\amp{}{}\amp y\amp {}+{}\amp 3z\amp{}={}1 \amp\qquad\amp (3) \end{alignedat} \)
\(\displaystyle \begin{alignedat}[t]{5} x\amp{}+{}\amp 3y\amp{}{}\amp z\amp{}={}2 \amp\qquad\amp (1)\\ 2x\amp{}+{}\amp y\amp{}{}\amp 4z\amp{}={}1 \amp\qquad\amp (2)\\ 2x\amp{}{}\amp 6y\amp {}+{}\amp 2z\amp{}={}4 \amp\qquad\amp (3) \end{alignedat} \)

To eliminate \(y\) from Equations (1) and (2), we multiply Equation (1) by \(2\) and add the result to Equation (2).
\begin{alignat*}{4} 6x \amp {}{} \amp 2y \amp {}+{} \amp 4z \amp {}={} \amp 2\\ \underline{\hphantom{}6x\vphantom{y}} \amp \underline{{}+{}\vphantom{y}} \amp \underline{2y} \amp \underline{{}{}\vphantom{y} }\amp \underline{4z\vphantom{y}} \amp {}={} \amp \underline{\hphantom{}5}\\ 0x\amp {}+{}\amp{0y} \amp {}+{} \amp 0z \amp {}={} \amp 3~ \end{alignat*}Because the resulting equation has no solution, the system is inconsistent.

To eliminate \(x\) from Equations (1) and (3), we multiply Equation (1) by \(2\) and add Equation (3).
\begin{alignat*}{4} 2x \amp {}+{} \amp 6y \amp {}{} \amp 2z \amp {}={} \amp 4\\ \underline{\hphantom{}2x\vphantom{y}} \amp \underline{{}{}\vphantom{y}} \amp \underline{6y} \amp \underline{{}+{}\vphantom{y} }\amp \underline{2z\vphantom{y}} \amp {}={} \amp \underline{\hphantom{}4}\\ 0x\amp {}+{}\amp{0y} \amp {}+{} \amp 0z \amp {}={} \amp 0~ \end{alignat*}Because the resulting equation vanishes, the system is dependent and has infinitely many solutions.
Checkpoint 2.4.9. Practice 3.
Checkpoint 2.4.10. QuickCheck 4.
True or false.
If a system is dependent, it has no solutions.
True
False
The equation \(0x + 0y + 0z = 0\) has no solution
True
False
If a \(3 \times 3\) system is dependent, all three equations are the same..
True
False
Gaussian reduction will reveal whether a system is dependent or inconsistent.
True
False
Subsection 2.4.5 Applications
Here are some problems that can be modeled by a system of three linear equations.
Example 2.4.11.
One angle of a triangle measures \(4\degree\) less than twice the second angle, and the third angle is \(20\degree\) greater than the sum of the first two. Find the measure of each angle.

Step 1:.
We represent the measure of each angle by a separate variable.
\begin{align*} \amp\text{First angle: }\amp\amp x\\ \amp\text{Second angle: }\amp\amp y\\ \amp\text{Third angle:}\amp\amp z \end{align*} 
Step 2:.
We write the conditions stated in the problem as three equations.
\begin{align*} \amp x\text{ is } 4\degree \text{ less than }2 \text{ times } y: \amp\amp x = 2y  4\\ \amp z\text{ is } 20\degree \text{ more than } x+ y: \amp\amp z = x + y + 20\\ \amp \text{the sum of the angles of a triangle is } 180\degree : \amp\amp x + y + z =180 \end{align*} 
Step 3:.
We follow the steps for solving a \(3\times 3\) linear system.
1. We write the three equations in standard form.
\begin{alignat*}{5} x\amp {}{}\amp 2y\amp \amp \amp =\amp 4\amp\hphantom{blank} \amp(1)\\ x\amp {}+{}\amp y\amp {}{}\amp z\amp =\amp 20\amp\hphantom{blank} \amp(2)\\ x\amp {}+{}\amp y\amp {}+{}\amp z\amp =\amp 180\amp\hphantom{blank} \amp(3) \end{alignat*}2β3. Because Equation (1) has no \(z\)term, it will be most efficient to eliminate \(z\) from Equations (2) and (3). We add these two equations.
\begin{alignat*}{5} x \amp {}+{} \amp y \amp {}{} \amp z \amp {}={} \amp 20\qquad\amp\amp (2)\\ \underline{x\vphantom{y}} \amp \underline{{}+{}\vphantom{y}} \amp \underline{\hphantom{2}y} \amp \underline{{}+{}\vphantom{y} }\amp \underline{\hphantom{~}\vphantom{y}z} \amp {}={} \amp \underline{180\vphantom{y+y}}\qquad\amp\amp (3)\\ 2x\amp {}+{}\amp{2y} \amp \amp \amp {}={} \amp 160\qquad\amp\amp (4) \end{alignat*}4. We form a \(2\times 2\) system from Equations (1) and (4). We add the two equations to eliminate the variable \(y\text{,}\) yielding
\begin{alignat*}{4} x \amp {}{} \amp 2y \amp {}={} \amp 4\qquad\amp\amp (1)\\ \underline{2x\vphantom{y}} \amp \underline{{}+{}\vphantom{y}} \amp \underline{2y} \amp {}={} \amp \underline{160\vphantom{y+y}}\qquad\amp\amp (4)\\ 3x\amp\amp \amp {}={} \amp 156\qquad\amp\amp (5) \end{alignat*}5. We form a triangular system using Equations (2), (1), and (5). We use backsubstitution to complete the solution.
\begin{alignat*}{4} x\amp {}+{}\amp y\amp {}+{}\amp z\amp =\amp 180\hphantom{blankblank} \amp(2)\\ x\amp {}{}\amp 2y\amp \amp\amp =\amp 4\hphantom{blankblank} \amp(1)\\ 3x\amp\amp\amp\amp\amp = \amp156\hphantom{blankblank} \amp(5) \end{alignat*}We divide both sides of Equation (5) by \(3\) to find \(x = 52\text{.}\) We substitute \(52\) for \(x\) in Equation (1) and solve for \(y\) to find
\begin{align*} \alert{52}  2y \amp = 4\\ y \amp= 28 \end{align*}We substitute \(\alert{52}\) for \(x\) and \(\alert{28}\) for \(y\) in Equation (3) to find
\begin{align*} \alert{52} + \alert{28} + z \amp= 180\\ z\amp=100 \end{align*} 
Step 4:.
The angles measure 52Β°, 28Β°, and 100Β°.
Exercises 2.4.6 Problem Set 2.4
Warm Up
Exercise Group.
For Problems 1 and 2, solve the system by elimination.
1.
\((3,5) \)
2.
3.
Karen has $2000, part of it invested in bonds paying 10%, and the rest in a certificate account at 8%. Her annual income from the two investments is $184. How much did Karen invest at each rate?

Choose variables for the unknown quantities, and fill in the table.
Principal Interest rate Interest Bonds Certificate Total ββ Write one equation about the amount Karen invested.
Write a second equation about Kaaren's annual interest.
Principal Interest rate Interest Bonds \(x\) \(0.10\) \(0.10x\) Certificate \(y\) \(0.08\) \(0.08y\) Total \(2000\) ββ \(184\) \(\displaystyle x+y=2000\)
\(\displaystyle 0.10x+0.08y=184\)
4.
The pharmacist at Glenoaks Hospital was asked to supply 10 liters of a 20% solution of iodine. She has on hand iodine solutions in 15% strength and 40% strength. How much of each should she mix to make the required solution?

Choose variables for the unknown quantities, and fill in the table.
Liters Strength Amount of iodine 15% Solution 40% solution 20% solution Write one equation about the number of liters of solution.
Write a second equation about the amount of iodine.
Skills Practice
Exercise Group.
For Problems 5 and 6, use backsubstitution to solve the system.
5.
\(\begin{alignedat}[t]{5} 2x\amp{}+{}\amp 3y\amp{}{}\amp z\amp{}={}7\\ \amp \amp y\amp{}{}\amp 2z\amp{}={}6\\ \amp \amp \amp \amp 5z\amp{}={}15 \end{alignedat}\)
\((2,0,3) \)
6.
\(\begin{aligned}[t] 2x + z\amp = 5\\ 3y + 2z \amp =6\\ 5x \amp = 20 \end{aligned}\)
Exercise Group.
For Problems 7β12, use Gaussian reduction to solve the system.
7.
\(\begin{alignedat}[t]{5} x\amp{}+{}\amp y\amp{}+{}\amp z\amp{}={}0\\ 2x \amp{}{} \amp 2y\amp{}+{}\amp z\amp{}={}8\\ 3x \amp{}+{} \amp 2y\amp {}+{}\amp z\amp{}={}2 \end{alignedat}\)
\((2,2,0) \)
8.
\(\begin{alignedat}[t]{5} x\amp{}{}\amp 2y\amp{}+{}\amp 4z\amp{}={}3\\ 3x \amp{}+{} \amp y\amp{}{}\amp 2z\amp{}={}12\\ 2x \amp{}+{} \amp y\amp {}{}\amp z\amp{}={}11 \end{alignedat}\)
9.
\(\begin{alignedat}[t]{5} 3x\amp{}{}\amp 4y\amp{}+{}\amp 2z\amp{}={}20\\ 4x \amp{}+{} \amp 3y\amp{}{}\amp 3z\amp{}={}4\\ 2x \amp{}{} \amp 5y\amp {}+{}\amp 5z\amp{}={}24 \end{alignedat}\)
\((2,3,1) \)
10.
\(\begin{aligned}[t] 4x + z\amp = 3\\ 2x  y \amp =2\\ 3y+2z \amp = 0 \end{aligned}\)
11.
\(\begin{alignedat}[t]{5} 4x\amp{}+{}\amp 6y\amp{}+{}\amp 3z\amp{}={3} \\ 2x \amp{}{} \amp 3y\amp{}{}\amp 2z\amp{}=5 \\ 6x \amp{}+{} \amp 6y\amp {}+{}\amp 2z\amp{}={5} \end{alignedat} \)
\(\left(\dfrac{1}{2}, \dfrac{2}{3},3 \right) \)
12.
\(\begin{alignedat}[t]{5} x\amp{}{}\amp \dfrac{1}{2} y\amp{}{}\amp \dfrac{1}{2}z\amp{}={}4 \\ x \amp{}{} \amp \dfrac{3}{2}y\amp{}{}\amp 2z\amp{}={}3 \\ \dfrac{1}{4}x \amp{}+{} \amp \dfrac{1}{4}y\amp {}{}\amp \dfrac{1}{4}z\amp{}={}0 \end{alignedat} \)
Exercise Group.
For Problems 13 and 14, when you decide which variable to eliminate first, take advantage of the fact that one of the variables is already missing from each equation.
13.
\(\begin{alignedat}[t]{5} x\amp {}={} {y} \\ x \amp{}+{} z=\dfrac{5}{6} \\ y \amp{}{} 2z = \dfrac{7}{6} \end{alignedat} \)
\(\left(\dfrac{1}{2}, \dfrac{1}{2},\dfrac{1}{3} \right) \)
14.
\(\begin{aligned}[t] x\amp = y+\dfrac{1}{2} \\ y \amp =z+\dfrac{5}{4} \\ 2z \amp = x\dfrac{7}{4} \end{aligned} \)
Exercise Group.
For Problems 15 and 16, decide whether the system is inconsistent or dependent.
15.
\(\begin{alignedat}[t]{5} 3x\amp{}{}\amp 2y\amp{}+{}\amp z\amp{}={6} \\ 2x \amp{}+{} \amp y\amp{}{}\amp z\amp{}=2 \\ 4x \amp{}+{} \amp 2y\amp {}{}\amp 2z\amp{}={3} \end{alignedat} \)
Dependent
16.
\(\begin{aligned}[t] x\amp = 2y 7 \\ y \amp =4z+3 \\ x \amp  8z = 1 \end{aligned} \)
Applications
Exercise Group.
Use a system of equations to solve Problems 1722.
17.
The perimeter of a triangle is 155 inches. Side \(x\) is 20 inches shorter than side \(y\text{,}\) and side \(y\) is 5 inches longer than side \(z\text{.}\) Find the lengths of the sides of the triangle.
\(x=40\) in, \(y=60\) in, \(z=55\) in
18.
One angle of a triangle measures 10Β° more than a second angle, and the third angle is 10Β° more than six times the measure of the smallest angle. Find the measure of each angle.
19.
The Java Shoppe sells a house brand of coffee that is only 2.25% caffeine for $6.60 per pound. The house brand is a mixture of Colombian coffee that sells for $6 per pound and is 2% caffeine, French roast that sells for $7.60 per pound and is 4% caffeine, and Sumatran that sells for $6.80 per pound and is 1% caffeine. How much of each variety is in a pound of house brand?
\(\dfrac{1}{2} \) lb Colombian, \(\dfrac{1}{4} \) lb French, \(\dfrac{1}{4} \) Sumatran
20.
Vegetable Medley is made of carrots, green beans, and cauliflower. The package says that 1 cup of Vegetable Medley provides 29.4 milligrams of vitamin C and 47.4 milligrams of calcium. One cup of carrots contains 9 milligrams of vitamin C and 48 milligrams of calcium. One cup of green beans contains 15 milligrams of vitamin C and 63 milligrams of calcium. One cup of cauliflower contains 69 milligrams of vitamin C and 26 milligrams of calcium. How much of each vegetable is in 1 cup of Vegetable Medley?
21.
Reliable Auto Company wants to ship 1700 Status Sedans to three major dealers in Los Angeles, Chicago, and Miami. From past experience Reliable figures that it will sell twice as many sedans in Los Angeles as in Chicago. It costs $230 to ship a sedan to Los Angeles, $70 to Chicago, and $160 to Miami. If Reliable Auto has $292,000 to pay for shipping costs, how many sedans should it ship to each city?
22.
A farmer has 1300 acres on which to plant wheat, corn, and soybeans. The seed costs $6 for an acre of wheat, $4 for an acre of corn, and $5 for an acre of soybeans. An acre of wheat requires 5 acrefeet of water during the growing season, an acre of corn requires 2 acrefeet, and an acre of soybeans requires 3 acrefeet. If the farmer has $6150 to spend on seed and can count on 3800 acrefeet of water, how many acres of each crop should she plant in order to use all her resources?