Extends elimination to three equations in three unknowns, reducing to a triangular system and back-substituting. Interprets each equation as a plane, classifies the possible outcomes, and distinguishes the two geometrically different ways a system can have infinitely many solutions.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 9 — Systems of Equations and Inequalities
§9.2 Systems of Linear Equations: Three Variables, pp. 1077-1089
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1077-1089 — the pages these objectives are drawn from
Warm-up
In two variables it was a line. One more variable changes the picture.
Discussion prompt
What is the graph of a linear equation in three variables?
Hint: How much freedom do the solutions have?
Answer:
Two of the three values can be chosen freely, and the third is then determined. Two degrees of freedom gives a plane, not a line.
So a system of three such equations asks where three planes all meet.
Generically three planes meet at a single point, which is why three equations are the right number for three unknowns. But they can also be arranged to share a line, or to have no common point at all.
Concept
Three equations reduce to two by eliminating one variable twice, and those two reduce to one by the methods of the previous section. The values are then recovered by back-substitution.
\[ 3\text{ equations} \;\to\; 2 \;\to\; 1 \;\to\; \text{back-substitute} \]
The one requirement is that the same variable be eliminated from both pairs. Eliminating a different variable each time leaves two equations with no common pair of unknowns, which is no progress at all.
Figure (svg): A diagram showing the systematic order of elimination in three variables: eliminate one variable from two pairs, then solve the resulting two-variable system
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1077-1081
Section
Section 1
Concept
Pick one variable and eliminate it from two different pairs of equations. That leaves two equations in the remaining two unknowns, which the previous section handles.
Using the same equation in both pairs — say the first with the second and the first with the third — is the standard arrangement. Using each equation at least once matters: pairing the first with the second and then the first with the second again produces nothing new.
Figure (svg): A diagram showing the systematic order of elimination in three variables: eliminate one variable from two pairs, then solve the resulting two-variable system
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1077-1082
Picture it
Two eliminations, then a two-variable system.
Figure (svg): A diagram showing the systematic order of elimination in three variables: eliminate one variable from two pairs, then solve the resulting two-variable system
The caption is the requirement that makes it work. Eliminating different variables from the two pairs leaves two equations that do not share a common pair of unknowns.
Worked example
Eliminate the same variable twice.
\[ \text{Reduce } x+y+z=6,\; 2x-y+z=3,\; x+2y-z=2. \]
Choose a variable
Why: The third, with coefficients 1, 1, -1.
Use the first and third
Why: Add them.
\[ 2 x + 3 y = 8 \]
Use the second and third
Why: Add them.
\[ 3 x + y = 5 \]
Note the result
Why: Two equations, two unknowns.
Figure (svg): A diagram showing the systematic order of elimination in three variables: eliminate one variable from two pairs, then solve the resulting two-variable system
\[ 2x+3y=8, \; 3x+y=5 \]
Verify: check the same variable was eliminated
Why: Both new equations involve only the first two variables, which is what makes them a solvable pair. Had one elimination removed the third variable and the other the second, the two results would share only one unknown and could not be solved together.
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1078-1079
Prediction
You eliminate one variable from two different pairs.
Predict first
What must be true of the two eliminations?
Correct: They must remove the same variable.
Why: Only then do both results involve the same two remaining unknowns, forming a solvable two-variable system. Removing different variables leaves two equations sharing just one unknown, which is no reduction at all.
Worked example
Solve the pair, then work back up.
\[ \text{Complete the solution of that system.} \]
Solve the two-variable system
Why: By elimination.
\[ x = 1, y = 2 \]
Substitute into an original
Why: Use the first equation.
\[ 1 + 2 + z = 6 \]
Solve for the third
Why: Subtract.
\[ z = 3 \]
State the triple
Why: All three values.
\[ (1, 2, 3) \]
Figure (svg): The solution to Worked example finish and back-substitute shown as a ladder of expressions, one row per legal move
\[ (x,y,z)=(1,2,3) \]
Verify: check in all three equations
Why: First: 1 plus 2 plus 3 is 6 — correct. Second: 2 minus 2 plus 3 is 3 — correct. Third: 1 plus 4 minus 3 is 2 — correct. Checking all three is essential, since a triple can satisfy two equations and fail the third.
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1079-1082
Trap
\[ \text{eliminate }z\text{ from equations 1 and 2, then }y\text{ from 1 and 3} \]
Choose whichever variable looks easiest in each pair
Why: The two eliminations remove different variables.
The two results share only one unknown, and no two-variable system has been produced.
Eliminate the same variable both times. Only then do the two results involve the same pair of unknowns.
Pick the variable that is easiest overall — often one with a coefficient of one somewhere, or with coefficients that already cancel.
Decide which variable to eliminate before starting, rather than choosing pair by pair. That single decision is what keeps the reduction coherent.
Faded example
Coefficients on the third variable are 1, 1 and -1.
Fill in the blanks
\text31\text1___\text___z, \text___1+(-___)=0
Why: Coefficients that are already opposites cancel on addition with no scaling. Scanning for such a pair before starting is what makes one variable the obvious choice to eliminate.
Sorting
Each equation should be used and the same variable removed.
Sort into buckets
Sort each plan.
Step zero
You are given three equations in three unknowns.
Discussion prompt
What do you decide before writing anything?
Hint: One choice governs the whole reduction.
Answer:
Which variable to eliminate. That single decision governs both elimination steps, and it must be the same variable for both.
Choose the one with the friendliest coefficients — a variable with coefficients of one, or with a pair already opposite so no scaling is needed.
Then plan the two pairings, using all three equations. Deciding before starting is what prevents the standard failure of eliminating whatever looks easiest in each pair and ending up with nothing.
Section
Section 2
Concept
The reduction can be organised so the system becomes triangular: the last equation has one unknown, the next has two, and the first has three. The values then fall out in sequence.
Triangular form is worth aiming for explicitly rather than reaching by accident. It makes the solving mechanical, it generalises to any number of variables, and it is exactly what row reduction produces in the later matrix sections.
Figure (svg): A worked ladder solving a triangular three-variable system by back-substitution from the bottom equation upward
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1082-1084
Picture it
One new unknown per line, from the bottom upward.
Figure (svg): A worked ladder solving a triangular three-variable system by back-substitution from the bottom equation upward
The mechanical character of this stage is the point. All the thinking went into reaching the triangular form; recovering the values is bookkeeping.
Worked example
Start from the last equation.
\[ \text{Solve } x+2y-z=5,\; y-3z=-4,\; z=2. \]
Read the last equation
Why: One variable already.
\[ z = 2 \]
Substitute into the second
Why: One unknown remains.
\[ y - 6 = -4 \]
Solve
Why: Add six.
\[ y = 2 \]
Substitute both into the first
Why: One unknown remains.
\[ x + 4 - 2 = 5, x = 3 \]
Figure (svg): A worked ladder solving a triangular three-variable system by back-substitution from the bottom equation upward
\[ (3,2,2) \]
Verify: check in all three
Why: First: 3 plus 4 minus 2 is 5 — correct. Second: 2 minus 6 is negative 4 — correct. Third: 2 equals 2 — correct. Each substitution left exactly one unknown, which is what triangular form guarantees.
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1082-1084
Prediction
A system is in triangular form.
Predict first
Which equation do you solve first?
Correct: The last, with one unknown.
Why: Only that equation can be solved as it stands. Each equation above it introduces one more unknown, so substituting upward keeps exactly one unknown in view at every step.
Worked example
Eliminate downward, keeping the shape.
\[ \text{Triangularise } x+y+z=4,\; 2x+3y-z=3,\; 3x-y+2z=7. \]
Eliminate the first variable from equation two
Why: Subtract twice the first.
\[ y - 3 z = -5 \]
Eliminate it from equation three
Why: Subtract three times the first.
\[ -4 y - z = -5 \]
Eliminate the second variable from the last
Why: Add four times the new second.
\[ -13 z = -25 \]
Read off
Why: Triangular.
\[ z = \frac{25}{13} \]
Figure (svg): The solution to Worked example reach triangular form shown as a ladder of expressions, one row per legal move
\[ x+y+z=4,\; y-3z=-5,\; -13z=-25 \]
Verify: check the shape
Why: The first equation has three variables, the second two and the third one — exactly triangular. Each elimination used the equation above it, which is the systematic pattern that row reduction automates in §9.5.
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1084-1084
Error analysis
A student back-substitutes through a triangular system.
Annotate
On: \( x+2y-z=5 \text{ first, before } y \text{ and } z \text{ are known} \)
The whole point of triangular form is that it can be solved in one direction only. Recognising which end has the single unknown takes one glance and settles the direction.
Faded example
With the third variable known to be 2.
Fill in the blanks
y-3(2)=-4 \;\Longrightarrow\; y=2
Why: Substituting the known value leaves a single-variable equation, which solves in one step. Each back-substitution has this shape, which is why the stage is mechanical.
Sorting
Each equation should introduce one new unknown.
Sort into buckets
Sort each system shape.
Explain it to yourself
The reduction aims at a particular shape.
Discussion prompt
Explain what triangular form buys.
Hint: What happens once you have it?
Answer:
It makes the remaining work mechanical: one equation solves immediately, and each substitution upward leaves exactly one new unknown.
No further decisions are needed — no choosing which variable to eliminate, no scaling. The thinking is all in the reduction.
And it generalises: the same shape works for any number of variables, and §9.5's row reduction is precisely a systematic way of producing it. Aiming at the form rather than at the answer is what makes the method scale.
Section
Section 3
Concept
Each equation is a plane. Three planes generically meet at one point, but they can also be arranged so that no point is common to all three, or so that a whole line is.
The 'infinitely many' case now has two distinct pictures — a shared line and a shared plane — where in two variables it had only one. The algebra distinguishes them by how many free variables remain.
Figure (svg): Three tilted planes drawn in perspective meeting at a single point, illustrating the one-solution case in three variables
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1084-1088
Picture it
The generic arrangement, and the one-solution case.
Figure (svg): Three tilted planes drawn in perspective meeting at a single point, illustrating the one-solution case in three variables
Tilting any of the three planes so that it becomes parallel to another, or so that all three intersections coincide, produces the other cases.
Worked example
A false statement appears during elimination.
\[ \text{Solve } x+y+z=2,\; 2x+2y+2z=5,\; x-y=1. \]
Compare the first two
Why: The second is twice the first on the left.
Check the constants
Why: Two doubled is four, not five.
Attempt elimination
Why: Subtract twice the first.
\[ 0 = 1 \]
Conclude
Why: A false statement.
Figure (svg): Three tilted planes drawn in perspective meeting at a single point, illustrating the one-solution case in three variables
\[ \text{inconsistent} \]
Verify: identify the geometry
Why: The first two equations describe parallel planes — same normal direction, different constants — so no point lies on both, regardless of the third equation. One parallel pair is enough to make the whole system inconsistent.
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1085-1087
Sorting
The arrangement of the planes decides.
Sort into buckets
Sort each arrangement.
Worked example
A true statement appears instead.
\[ \text{Solve } x+y+z=3,\; 2x+2y+2z=6,\; x-y=0. \]
Compare the first two
Why: The second is exactly twice the first.
Eliminate
Why: Subtract twice the first.
\[ 0 = 0 \]
Note what remains
Why: Two independent equations, three unknowns.
Describe the solutions
Why: A line in space.
Figure (svg): The solution to Worked example a dependent system shown as a ladder of expressions, one row per legal move
\[ x=y,\; z=3-2y \]
Verify: count the degrees of freedom
Why: Three unknowns with two independent equations leaves one free, which is a line — the intersection of the two distinct planes. Had two equations vanished, two would be free and the solution set would be a whole plane.
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1087-1088
Trap
\[ \text{three equations, three unknowns, so there is one solution} \]
Count equations against unknowns
Why: Matching counts is taken to guarantee a unique answer.
Dependent and inconsistent systems are not anticipated.
The equations must be independent, not merely three in number. A repeated equation contributes nothing.
If one equation is a combination of the others, the effective count drops and either infinitely many solutions or none result.
Counting is necessary but not sufficient. The elimination itself reveals whether the equations were genuinely independent.
Prediction
Elimination leaves two independent equations in three unknowns.
Predict first
What is the solution set?
Correct: A line.
Why: Three unknowns with two independent constraints leaves one degree of freedom, which is a line in space. One independent equation would leave two degrees of freedom and give a plane.
Faded example
Three unknowns and two independent equations.
Fill in the blanks
3-2=1\text___
Why: Each independent equation removes one degree of freedom, so subtracting the number of independent equations from the number of unknowns gives the dimension of the solution set. One free variable is a line and two is a plane.
Explain it
In three variables, infinitely many solutions has two pictures.
Discussion prompt
Explain the difference to a classmate.
Hint: How many equations were genuinely independent?
Answer:
If two of the three equations are independent, they cut the space down to a line — the intersection of two distinct planes.
If only one is independent, the constraint is a single plane, and every point on it is a solution.
So the two cases differ in the number of free variables: one gives a line and two give a plane. The algebra distinguishes them by how many equations survive elimination, which is a more reliable test than trying to visualise the planes.
Section
Section 4
Concept
When solutions form a line, one variable is free and the other two are determined by it. The solution set is written by naming the free variable and expressing the rest.
Any of the variables may be chosen as the free one, giving different-looking but equivalent descriptions of the same line. Choosing the one that makes the expressions simplest is worth a moment's thought.
Figure (svg): Three cards giving the possible outcomes for a system in three variables and what each looks like geometrically
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1088-1089
Picture it
The same algebraic signals as before, with richer geometry.
Figure (svg): Three cards giving the possible outcomes for a system in three variables and what each looks like geometrically
The third card covers both a shared line and a shared plane. Which one is decided by counting how many independent equations survived.
Worked example
One free variable, two expressed in terms of it.
\[ \text{After elimination: } x+y+z=6 \text{ and } y-z=1. \text{ Write the solutions.} \]
Count the free variables
Why: Three unknowns, two equations.
Choose the free variable
Why: The third is convenient.
\[ \text{let } z = t \]
Express the second variable
Why: From the second equation.
\[ y = t + 1 \]
Express the first
Why: From the first equation.
\[ x = 5 - 2 t \]
Figure (svg): Three cards giving the possible outcomes for a system in three variables and what each looks like geometrically
\[ (5-2t,\;t+1,\;t) \]
Verify: substitute the general form back
Why: In the first equation, 5 minus 2t plus t plus 1 plus t is 6 for every value of the parameter — the t terms cancel. In the second, t plus 1 minus t is 1, also always. The whole line satisfies both equations, which is what an infinite solution set means.
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1088-1088
Faded example
From the equation that the second variable minus the third is 1.
Fill in the blanks
\text1z=t: \quad y=t+1, \quad\text___(t+___,\;t)
Why: Setting the third variable free and solving for the second expresses it in terms of the parameter. Every solution is obtained by choosing a value for the parameter.
Worked example
The same line, described another way.
\[ \text{Rewrite that solution set with } y \text{ free.} \]
Set the second variable free
Why: Call it a parameter.
\[ y = s \]
Express the third
Why: From the second equation.
\[ z = s - 1 \]
Express the first
Why: From the first equation.
\[ x = 7 - 2 s \]
Write the triple
Why: In terms of the new parameter.
\[ (7 - 2 s, s, s - 1) \]
Figure (svg): The solution to Worked example a different free variable shown as a ladder of expressions, one row per legal move
\[ (7-2s,\;s,\;s-1) \]
Verify: check the two descriptions agree
Why: Setting the parameter s equal to t plus 1 turns the second description into the first exactly. Both describe the same line, which is expected — the choice of free variable is a labelling decision, not a mathematical one.
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1088-1089
Error analysis
A student reports a solution to a dependent system.
Annotate
On: \( \text{taking }t=0\text{ gives }(5,1,0), \text{ so that is the answer} \)
A dependent system's answer is a set, not a point. Naming a parameter and expressing the others in terms of it is the only way to describe all of them at once.
Prediction
Three unknowns and one independent equation.
Predict first
How many variables are free?
Correct: Two, so the solutions form a plane.
Why: Each independent equation removes one degree of freedom, so one equation leaves two free out of three. Two free variables describe a plane, which happens when all three original equations describe the same plane.
Sorting
A dependent system needs its whole solution set.
Sort into buckets
Sort each answer.
Explain it
Infinite solution sets are written with a parameter.
Discussion prompt
Explain to a classmate what the parameter is doing.
Hint: What does choosing a value for it produce?
Answer:
The parameter stands for the free variable — the one the equations do not pin down. Any value at all is allowed for it.
Once it is chosen, the other two are determined by the surviving equations. So each value of the parameter gives one solution.
Running the parameter through every real number produces every solution, which is why this form describes the whole line at once. A single triple is one point on it, and a good explanation makes clear that reporting one is like naming one point on a line when asked for the line.
Section
Section 5
Concept
Problems with three unknowns need three genuinely different relationships, and finding them is the modelling work that the algebra then completes.
The comparison equation — one quantity being twice another, or two together equalling a third — is the one most easily overlooked. Problems that seem to give only two relationships usually contain a comparison stated in words rather than as an obvious count.
Figure (svg): Three cards giving the possible outcomes for a system in three variables and what each looks like geometrically
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1081-1089
Picture it
The second and third cards are the failure modes in modelling.
Figure (svg): Three cards giving the possible outcomes for a system in three variables and what each looks like geometrically
A dependent system in an application usually means one stated relationship was a consequence of the others, which is a modelling error rather than a computational one.
Worked example
Three counts of three different things.
\[ \text{Three items cost } \$26 \text{ total. The first costs } \$2 \text{ more than the second, and the third is twice the first. Find the prices.} \]
Name the unknowns
Why: The three prices.
Write the total
Why: They sum to 26.
\[ a + b + c = 26 \]
Write the first comparison
Why: Two more than the second.
\[ a = b + 2 \]
Write the second comparison
Why: Twice the first.
\[ c = 2 a \]
Figure (svg): The solution to Worked example set up a three-unknown problem shown as a ladder of expressions, one row per legal move
\[ a=6,\; b=4,\; c=16 \]
Verify: check all three conditions
Why: The three sum to 26 — correct. Six is two more than four — correct. Sixteen is twice six — correct. All three relationships hold, and the three were genuinely independent since none follows from the other two.
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1082-1084
Prediction
A problem has three unknowns and two independent relationships.
Predict first
What is the outcome?
Correct: Infinitely many solutions; more information is needed.
Why: Two independent equations leave one degree of freedom among three unknowns, so a whole line of triples satisfies the conditions. That is a modelling gap rather than a computational failure.
Worked example
One relationship adds nothing.
\[ \text{Three numbers sum to } 12, \text{ the first two sum to } 8, \text{ and the third is } 4. \text{ Can they be found?} \]
Write the three conditions
Why: As equations.
\[ a + b + c = 12, a + b = 8, c = 4 \]
Check independence
Why: The first is the second plus the third.
Count effective equations
Why: Only two independent ones.
Conclude
Why: The first two are not determined.
Figure (svg): Three cards giving the possible outcomes for a system in three variables and what each looks like geometrically
\[ c=4,\; a+b=8,\; a\text{ free} \]
Verify: confirm the dependence
Why: Adding the second and third conditions gives exactly the first, so it carries no new information. The third number is determined but the first two are only constrained to sum to eight — infinitely many pairs do that.
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1084-1089
Trap
\[ a+b+c=12,\; a+b=8,\; c=4 \]
Write down every stated fact as an equation
Why: The three conditions are recorded without checking independence.
The first is the sum of the other two, so only two constraints exist.
Check that each equation is independent of the others before solving.
An equation that is a sum, difference or multiple of the others contributes nothing and leaves the system under-determined.
A dependent system in an application is a modelling problem, not a computational one. More information is needed, and no amount of algebra can supply it.
Sorting
Given that three numbers sum to 12 and the third is 4.
Sort into buckets
Sort each additional statement.
Faded example
The third item costs twice the first.
Fill in the blanks
c=2a \;\text2\; c-___a=0
Why: A comparison between two of the unknowns is a legitimate third equation and is often the one overlooked. Written in standard form it takes its place alongside the other two.
Explain it to yourself
Three unknowns need three independent equations.
Discussion prompt
Explain what goes wrong without independence.
Hint: What does a dependent equation contribute?
Answer:
A dependent equation is a consequence of the others, so it constrains nothing they do not already constrain. Writing it down adds a line but no information.
The effective count of equations is then less than three, leaving at least one variable free and infinitely many solutions.
So the problem is under-specified, and no algebra can fix it — the missing constraint has to come from the situation. Recognising this early saves the effort of solving a system that cannot have a unique answer.
Comparison
Fill the blanks from memory. The method carries over; the geometry gets richer.
Comparison matrix
| two variables | three variables | |
|---|---|---|
| each equation is | a line | a plane |
| the method | one elimination | two eliminations, then back-substitution |
| outcomes | one, none, infinitely many | the same three |
| infinite case looks like | one line | a line or a plane |
The last row is the genuine addition. In three variables the infinite case has two shapes, distinguished by how many independent equations survive.
Pattern
Five steps, and the first governs the next two.
Step 5's check must use all three original equations. A triple can satisfy two and fail the third, and only the full check catches that.
OpenStax Algebra and Trigonometry 2e, §11.2 Systems of Linear Equations: Three Variables §11.2
Check
The systematic order.
Check your understanding
When reducing three equations to two, what must the two eliminations have in common?
Answer: A
Why: Only if the same variable is removed do the two resulting equations involve the same pair of unknowns, forming a solvable two-variable system. Removing different variables leaves two equations sharing just one unknown.
Check
Geometry.
Check your understanding
What does a single linear equation in three variables describe?
Answer: A
Why: Two of the three values can be chosen freely and the third follows, which is two degrees of freedom — a plane. A system of three such equations asks where three planes all meet.
Check
Infinite solution sets.
Check your understanding
Elimination leaves two independent equations in three unknowns. How should the answer be written?
Answer: A
Why: One free variable remains, so the solutions form a line. Naming that variable as a parameter and expressing the other two in terms of it describes every solution at once.
Real world
Balancing a chemical equation is a linear system in as many unknowns as there are compounds.
Discussion prompt
Why does balancing a reaction reduce to solving a system, and why is the answer never unique?
Hint: What is conserved, and what can be scaled?
Answer:
Each element must have the same total count on both sides, which is one linear equation per element. The unknowns are the coefficients on each compound.
But the system is always dependent: doubling every coefficient balances the equation just as well, so there is always at least one free variable.
Chemists resolve it by taking the smallest whole-number solution, which picks one point on the line of solutions by an extra convention. The dependence is intrinsic rather than a defect — it reflects the physical fact that a reaction's proportions matter and its absolute scale does not.
Commit first
State your confidence along with your answer.
Predict first
Why must the same variable be eliminated from both pairs of equations?
Correct: So the two results involve the same two remaining unknowns.
Why: A solvable two-variable system requires both equations to be in the same two variables. Eliminating different variables leaves two equations sharing only one unknown, which is no reduction at all.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why three variables is not really harder than two.
Hint: What is the method?
Answer:
The method is the same elimination, done twice. Reducing three equations to two is one elimination step repeated on two different pairs.
After that it is exactly a §9.1 problem, and once that is solved, two back-substitutions recover the third value.
So the only new thing is bookkeeping, and the one rule that keeps it straight is eliminating the same variable both times. A good explanation adds that this generalises: four variables is three eliminations, and the matrix methods of §9.5 are the same idea written more compactly.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The fourth is where marks are most often lost, since reporting one triple instead of the whole set is a natural mistake. The second is what §9.5 automates, so it repays attention now.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the three-step reduction plan with the rule about eliminating the same variable. Beside it, work one system all the way to a triangular form and back-substitute. Underneath, sketch three planes meeting at a point and three that share a line, and write the algebraic signal for each of the three outcomes.
If your two sketches distinguish the one-solution case from the line case, and your triangular system solves from the bottom upward, the section's structure is on the page.
Recap
Five things, and the first is the only genuinely new rule.
| if you remember one thing | it should be this |
|---|---|
| about the method | eliminate the same variable twice, then it is §9.1 |
| about the form to aim for | triangular: one new unknown per line |
| about the geometry | planes, not lines, so the infinite case has two shapes |
| about infinite answers | a parameter describes them all; one triple describes one |
Section 9.3 turns to nonlinear systems, where a line can meet a curve in more than one place and the solution count is no longer limited to three possibilities.
OpenStax, Precalculus, §9.2 Systems of Linear Equations: Three Variables §9.2, pp. 1077-1089 — everything on these slides traces back here
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