Linear equations in three variables and their graphs as planes, ordered triples as solutions, solving a three-equation system by eliminating one variable twice, the degenerate cases, and modelling a situation with three unknowns.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 3 — Linear Systems and Matrices
Solve Systems of Linear Equations in Three Variables
Objectives
Five outcomes. The second is the whole method, and it is Lesson 3.2 applied twice.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.4 Solve Systems of Linear Equations in Three Variables §3.4, pp. 178-181 — the lesson these objectives are drawn from
Warm-up
Lesson 3.2 eliminated one variable from two equations. This lesson does the same thing twice.
Discussion prompt
You have three equations in x, y and z. If you could get rid of z from two different pairs of them, what would you be left with, and would you know how to finish?
Hint: Count what remains after the eliminations.
Answer:
Two equations in x and y alone — which is exactly a Lesson 3.2 problem. So the whole of this lesson is: eliminate one variable twice, solve the two-by-two system that remains, then work backwards for the third variable.
Nothing new is being invented. The only genuinely new habit is choosing the same variable to eliminate both times, since eliminating different ones leaves you no better off.
Concept
A linear equation in three variables graphs as a plane in three-dimensional space. Three such equations give three planes, and the system's solutions are the points lying on all three at once.
ordered triple — A solution of a system in three variables, written as three coordinates x, y and z whose values make every equation true.
\[ ax + by + cz = d, \quad a, b, c \text{ not all zero} \]
The three possible outcomes match Lesson 3.1's exactly: one solution, infinitely many, or none. Only the pictures change, from lines in a plane to planes in space.
Figure (svg): Three parallelograms standing for planes, arranged to meet at a single point, along a line, and not at all
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.4 Solve Systems of Linear Equations in Three Variables §3.4, pp. 178-178
Section
Section 1
Concept
A solution is an ordered triple whose three values satisfy every equation in the system. Checking one means three substitutions, not one, exactly as a two-variable system needed two.
linear equation in three variables — An equation of the form ax plus by plus cz equals d, with a, b and c not all zero. Its graph is a plane in three-dimensional space.
The graph of a single equation is a whole plane, which contains infinitely many triples. It takes all three equations together to narrow that down.
Figure (svg): An ordered triple shown as three coordinates, with each equation of the system tested against it
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.4 Solve Systems of Linear Equations in Three Variables §3.4, pp. 178-178
Picture it
The triple (2, 1, -3) tested against each equation of Example 1's system.
Figure (svg): An ordered triple shown as three coordinates, with each equation of the system tested against it
A triple satisfying two of the three equations lies on the line where those two planes meet — close, but not a solution of the system.
Worked example
Example 1's answer, verified before it is derived.
\[ \text{Is } (2, 1, -3) \text{ a solution of } 4x+2y+3z=1, \; 2x-3y+5z=-14, \; 6x-y+4z=-1? \]
Substitute into the first equation
Why: Four times two is eight, two times one is two, three times negative three is negative nine.
\[ 8 + 2 - 9 = 1 \]
Substitute into the second
Why: Four, minus three, minus fifteen.
\[ 4 - 3 - 15 = -14 \]
Substitute into the third
Why: Twelve, minus one, minus twelve.
\[ 12 - 1 - 12 = -1 \]
Compare each with its right side
Why: All three match, so the triple satisfies the whole system.
Figure (svg): The solution to Worked example check an ordered triple shown as a ladder of expressions, one row per algebraic move
\[ (2, 1, -3) \quad \text{is a solution} \]
Verify: ask what two checks would have proved
Why: Passing only the first two equations would place the triple on the line where those two planes meet — a whole line of triples do that. It is the third equation that cuts the line down to a single point, which is why all three checks are needed.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.4 Solve Systems of Linear Equations in Three Variables §3.4, pp. 179-179
Sorting
Test each triple against 4x + 2y + 3z = 1, 2x - 3y + 5z = -14 and 6x - y + 4z = -1.
Sort into buckets
Sort each ordered triple.
Two of the four wrong answers use exactly the right three numbers. Order is not a formality in an ordered triple.
Worked example
A candidate that passes two equations and fails the third.
\[ \text{Test } (1, 2, -1) \text{ in the same three equations.} \]
Substitute into the first equation
Why: Four plus four minus three is five, but the right side is 1.
\[ 5\text{ is not } 1 \]
Substitute into the second
Why: Two minus six minus five is negative nine, and the right side is negative 14.
\[ -9\text{ is not } -14 \]
Substitute into the third
Why: Six minus two minus four is zero, and the right side is negative 1.
\[ 0\text{ is not } -1 \]
Conclude
Why: It fails all three, so it is not on any of the planes, let alone all of them.
Figure (svg): The solution to Worked example a triple that nearly works shown as a ladder of expressions, one row per algebraic move
\[ (1, 2, -1) \quad \text{is not a solution} \]
Verify: compare with the genuine solution
Why: The true solution is (2, 1, -3), so this candidate has x and y swapped and z wrong. Swapping coordinates is a common slip when writing a triple, and checking is what catches it — the arithmetic of the solving may have been perfect.
Trap
\[ \text{Solving gives } z = -3, \; x = 2, \; y = 1. \]
Write the triple in the order the values were found
Why: The answer is recorded in the order of discovery rather than in the standard order.
\[ (-3, 2, 1) \quad \text{(wrong)} \]
Substituting into the first equation gives -12 + 4 + 3, which is -5, not 1.
\[ \text{Solving gives } z = -3, \; x = 2, \; y = 1. \]
Write the triple in the order x, y, z regardless of the order found
Why: An ordered triple is ordered: the position of each number carries its meaning.
\[ (2, 1, -3) \]
Elimination usually finds the variables in an order that has nothing to do with x, y, z, so this reordering step is a real one rather than a formality.
Prediction
Commit before reasoning.
Predict first
What is the graph of the single equation 2x plus y minus z equals 5?
Correct: A plane in three-dimensional space.
The same counting explains Lesson 3.1: two equations in two variables meet at a point, because each removes one of the two degrees of freedom.
Why: One linear equation in two variables gives a line — a one-dimensional set inside a two-dimensional space. One linear equation in three variables gives a plane, a two-dimensional set inside a three-dimensional space. Each equation removes one degree of freedom, which is why three equations are needed to pin down a single point in three-dimensional space.
Matching
Three planes can be arranged in several ways.
Match the pairs
Why: The three outcomes are the same as for two lines in a plane, but the infinitely-many case now comes in two flavours: the solutions can form a line or a whole plane. The no-solution case also has more shapes — planes can be parallel, or can pairwise intersect in three parallel lines that never all meet, which is the situation a triangular prism describes.
Explain it to yourself
Two equations in three variables would still be a system.
\[ \begin{cases} 2x + y - z = 5 \\ 3x - 2y + z = 16 \end{cases} \]
Discussion prompt
Explain what the solution set of this two-equation system in three variables would be, and why a third equation is normally needed to get a single answer.
Hint: Think about what two planes have in common.
Answer:
Two non-parallel planes intersect in a line, so the system has infinitely many solutions — every point on that line.
A third equation cuts that line with a third plane, and a line meets a plane at one point unless it happens to lie inside it or run parallel to it. So three equations normally pin down exactly one triple, which is the counting behind the whole lesson.
Section
Section 2
Concept
Choose one variable and eliminate it from two different pairs of equations. What remains is two equations in the other two variables — a system you already know how to solve.
\[ \begin{cases} 4x + 2y + 3z = 1 \\ 2x - 3y + 5z = -14 \\ 6x - y + 4z = -1 \end{cases} \]
Use the same equation in both pairings when you can. Here the third equation, with its coefficient of negative one on y, is the natural partner for both of the others.
Figure (svg): The three-variable system reduced to a two-variable one by eliminating the same variable twice
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.4 Solve Systems of Linear Equations in Three Variables §3.4, pp. 179-179 — Use the elimination method, Step 1
Picture it
Example 1: eliminating y from equations 1 and 3, and again from 2 and 3.
Figure (svg): The three-variable system reduced to a two-variable one by eliminating the same variable twice
Both new equations contain only x and z, which is what makes the pair a two-by-two system. Eliminating y from one pair and z from the other would leave two equations with three variables between them and no progress at all.
Worked example
Example 1, step one. The variable y is chosen because equation 3 has a coefficient of negative one on it.
\[ \text{Eliminate } y \text{ from equations 1 and 3, then from equations 2 and 3.} \]
Choose the variable and the pivot equation
Why: Equation 3 has y with coefficient negative 1, so multiplying it is easy and creates no fractions.
\[ \text{target } y,\text{ pivot on equation } 3 \]
Add 2 times equation 3 to equation 1
Why: Two times equation 3 is 12x minus 2y plus 8z equals negative 2. Adding cancels the y terms.
\[ 16 x + 11 z = -1 \]
Add -3 times equation 3 to equation 2
Why: Negative three times equation 3 is negative 18x plus 3y minus 12z equals 3. Adding cancels the y terms again.
\[ -16 x - 7 z = -11 \]
Note what is left
Why: Two equations in x and z alone, which is a Lesson 3.2 problem.
Figure (svg): The solution to Worked example the first elimination shown as a ladder of expressions, one row per algebraic move
\[ \begin{cases} 16x + 11z = -1 \\ -16x - 7z = -11 \end{cases} \]
Verify: check that y really has vanished from both
Why: The first new equation has terms in x and z only, and so does the second. Had either retained a y, the multiplier was wrong. It is also worth noticing that x now has coefficients 16 and negative 16, so the next elimination will need no multiplication at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.4 Solve Systems of Linear Equations in Three Variables §3.4, pp. 179-179
Discrimination
Scan the coefficients for a 1 or negative 1, or for the smallest least common multiple.
Sort into buckets
Sort each system by the variable that is cheapest to eliminate.
Worked example
The same system, scanned before any work is done.
\[ \text{Which variable is cheapest to eliminate from } 4x+2y+3z=1, \; 2x-3y+5z=-14, \; 6x-y+4z=-1? \]
Look at the x coefficients
Why: Four, two and six. The least common multiple is 12, so multipliers of 3, 6 and 2 would be needed.
\[ x: \operatorname{lcm} 12 \]
Look at the y coefficients
Why: Two, negative three and negative one. The coefficient of negative one makes equation 3 a free pivot.
\[ y: a\text{ coefficient of } -1 \]
Look at the z coefficients
Why: Three, five and four. The least common multiple is 60, which is much worse.
\[ z: \operatorname{lcm} 60 \]
Choose y
Why: A coefficient of one or negative one anywhere means that variable can be eliminated by simple multiplication of a single equation.
Figure (svg): The solution to Worked example choosing which variable to eliminate shown as a ladder of expressions, one row per algebraic move
\[ \text{eliminate } y \text{ using equation 3} \]
Verify: confirm the choice by the size of the multipliers
Why: Eliminating y needed multipliers of 2 and negative 3, both small whole numbers. Eliminating z would have needed multipliers like 5 and 3 and then 4 and 5, producing coefficients in the hundreds. The scan takes ten seconds and saves considerably more.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.4 Solve Systems of Linear Equations in Three Variables §3.4, pp. 179-179
Trap
\[ \text{Eliminate } y \text{ from equations 1 and 3; eliminate } z \text{ from equations 2 and 3.} \]
Get rid of whichever variable looks easiest in each pair
Why: Each pair is treated as a separate problem rather than as part of one plan.
The result is one equation in x and z, and another in x and y — three variables between two equations, so nothing has been reduced.
\[ \text{Eliminate } y \text{ from BOTH pairs.} \]
Choose one variable and remove it everywhere
Why: The point is to leave two equations in the SAME two variables, which only happens if the same variable goes both times.
\[ 16x + 11z = -1 \qquad -16x - 7z = -11 \]
Both remaining equations involve x and z, which is precisely what makes them a solvable two-by-two system.
Fill the middle
Example 1's first elimination.
Fill in the blanks
4x + 2y + 3z = 1 \;\text12x - 2y + 8z = -2\; ___ \;\Longrightarrow\; 16x + 11z = -1
Why: Two times equation 3 gives 12x minus 2y plus 8z equals negative 2. Adding it to equation 1 cancels the y terms — 2y plus negative 2y is zero — and combines the rest into 16x plus 11z equals negative 1. Note that the constant was multiplied too: negative 1 times 2 is negative 2, and forgetting it is the same slip as in Lesson 3.2.
Ranking
Solving a three-variable system by elimination.
Put in order
Why: Choosing the variable first is what keeps both eliminations aimed at the same target, which is the one place this method goes wrong. The two-variable system cannot be solved before it exists, and the back-substitution needs both of its values. The final check uses all three originals because errors made during the eliminations would otherwise survive undetected.
Prediction
Commit before reasoning.
Predict first
You have three equations in three variables and you eliminate one variable from two different pairs. How many equations do you have, in how many variables?
Correct: Two equations in two variables.
\[ 3 \text{ eq}, 3 \text{ var} \;\xrightarrow{\text{eliminate } y \text{ twice}}\; 2 \text{ eq}, 2 \text{ var} \]
Why: Each elimination combines two equations into one, so two eliminations produce two new equations. Both had the same variable removed, so both contain only the other two. That is a Lesson 3.2 problem, which is the point of the whole step — three by three has become two by two, and the method for two by two is already known.
Section
Section 3
Concept
The two-variable system is solved exactly as in Lesson 3.2. Its two values are then substituted into any original equation to recover the variable that was eliminated.
\[ \begin{cases} 16x + 11z = -1 \\ -16x - 7z = -11 \end{cases} \]
Substitute back into an ORIGINAL equation rather than one of the new ones. It is simpler, and it partly checks the eliminations at the same time.
Figure (svg): The back-substitution chain: z found first, then x, then y
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.4 Solve Systems of Linear Equations in Three Variables §3.4, pp. 179-179 — Steps 2 and 3
Picture it
Example 1, steps two and three.
Figure (svg): The back-substitution chain: z found first, then x, then y
Each line uses something already found, so the arithmetic gets easier rather than harder as you go.
Worked example
The two-by-two system from the previous section, solved and back-substituted.
\[ \begin{cases} 16x + 11z = -1 \\ -16x - 7z = -11 \end{cases} \quad \text{then find } y. \]
Add the two equations
Why: The x coefficients are already opposites, so no multiplication is needed and the x terms cancel.
\[ 4 z = -12 \]
Solve for z
Why: Negative twelve divided by four.
\[ z = -3 \]
Substitute z into either new equation
Why: Sixteen x plus 11 times negative 3 is negative 1, so 16x equals 32.
\[ x = 2 \]
Substitute x and z into an original equation
Why: Using the third: 6 times 2 minus y plus 4 times negative 3 equals negative 1, so 12 minus y minus 12 is negative 1.
\[ y = 1 \]
Figure (svg): The solution to Worked example finish Example 1 shown as a ladder of expressions, one row per algebraic move
\[ (2, 1, -3) \]
Verify: check in all three originals
Why: First: 8 plus 2 minus 9 is 1 — correct. Second: 4 minus 3 minus 15 is negative 14 — correct. Third: 12 minus 1 minus 12 is negative 1 — correct. All three hold, which is the only thing that makes this a solution of the system rather than of part of it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.4 Solve Systems of Linear Equations in Three Variables §3.4, pp. 179-179
Fill the middle
Example 1, with x equal to 2 and z equal to negative 3.
Fill in the blanks
6(2) - y + 4(-3) = -1 \;\Longrightarrow\; -y = -1 \;\Longrightarrow\; y = 1
Why: Twelve and negative twelve cancel, leaving negative y equals negative 1, so y is 1. Choosing equation 3 for this step was deliberate: its y coefficient is negative one, so no division is needed. Substituting into equation 1 would have worked too, giving 8 plus 2y minus 9 equals 1 and then 2y equals 2 — the same answer with one more step.
Worked example
A system where the eliminations need no multiplication at all.
\[ \begin{cases} x + y + z = 6 \\ 2x - y + z = 3 \\ x + 2y - z = 2 \end{cases} \]
Choose z, since two equations have coefficient 1 and one has negative 1
Why: Adding the first and third cancels z immediately.
Add equations 1 and 3
Why: Two x plus 3y equals 8.
\[ 2 x + 3 y = 8 \]
Add equations 2 and 3
Why: Three x plus y equals 5.
\[ 3 x + y = 5 \]
Solve the two-by-two system
Why: Multiplying the second by negative 3 and adding gives negative 7x equals negative 7, so x is 1 and then y is 2.
\[ x = 1, y = 2 \]
Back-substitute into equation 1
Why: One plus 2 plus z equals 6, so z is 3.
\[ z = 3 \]
Figure (svg): The solution to Worked example a system with a coefficient of one everywhere shown as a ladder of expressions, one row per algebraic move
\[ (1, 2, 3) \]
Verify: check in all three originals
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. Note how the coefficients of one meant no multiplication was needed for either elimination, which is exactly what the scanning step is looking for.
Error analysis
A student finds x and z, then looks for y in one of the new equations.
Annotate
On: \( 16x + 11z = -1 \quad \text{with } x = 2, \; z = -3 \;\Longrightarrow\; \text{now find } y \)
The eliminated variable lives only in the original equations. Choose whichever of them has the simplest coefficient on it.
Sorting
You have x and z and need y from the system of Example 1.
Sort into buckets
Sort each equation by how convenient it is for finding y.
Prediction
Commit before reasoning.
Predict first
In Example 1, after eliminating y the two remaining equations are 16x + 11z = -1 and -16x - 7z = -11. Which variable is found first, and why?
Correct: z, because the x coefficients are already opposites and cancel on addition.
\[ 16x + 11z = -1 \;\text{ plus }\; -16x - 7z = -11 \;\Longrightarrow\; 4z = -12 \]
Why: Sixteen and negative sixteen add to zero, so simply adding the two equations removes x with no multiplication at all and leaves 4z equal to negative 12. Solving for x first would require multiplying both equations to match the z coefficients of 11 and negative 7, whose least common multiple is 77. The order is chosen by the arithmetic, not by the alphabet.
Explain it
A classmate is intimidated by three equations.
Discussion prompt
Tell them what the method is, why it works, and reassure them that nothing new is required beyond Lesson 3.2. Include the one choice they have to get right.
Hint: The choice is which variable to eliminate.
Answer:
Pick one variable and get rid of it from two different pairs of the equations. That leaves two equations in the other two variables, which is a system they already know how to solve.
Once they have those two values, put them into any original equation to recover the third. The one choice that matters is eliminating the SAME variable both times — a different variable each time leaves three variables spread across two equations and no progress at all.
Section
Section 4
Concept
If every variable cancels during the eliminations, read what is left. A true statement means infinitely many solutions; a false one means none. The rule is exactly Lesson 3.2's.
\[ 0 = 0 \;\Rightarrow\; \text{infinitely many} \qquad 0 = 7 \;\Rightarrow\; \text{none} \]
Geometrically, infinitely many solutions means the planes share a line or coincide; no solution means no point lies on all three, which can happen even when every pair of planes does intersect.
Figure (svg): Two columns contrasting the true and false leftover statements in a three-variable system
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.4 Solve Systems of Linear Equations in Three Variables §3.4, pp. 178-179
Picture it
The same two outcomes as for two lines, with richer geometry behind them.
Figure (svg): Two columns contrasting the true and false leftover statements in a three-variable system
The no-solution case is more varied in three dimensions: three parallel planes give it, but so do three planes meeting pairwise in three parallel lines, like the faces of a triangular prism.
Worked example
Three planes with no common point, detected algebraically.
\[ \begin{cases} x + y + z = 3 \\ 2x + 2y + 2z = 7 \\ x - y + z = 1 \end{cases} \]
Compare the first two equations
Why: Every coefficient of the second is twice the first, but 7 is not twice 3.
Eliminate to confirm
Why: Multiplying the first by negative 2 and adding to the second gives 0 equals 1.
\[ 0 = 1 \]
Read the leftover statement
Why: Zero equals one is false for every triple.
State the conclusion
Why: No triple can satisfy the first two equations, so none can satisfy all three.
Figure (svg): The solution to Worked example a system with no solution shown as a ladder of expressions, one row per algebraic move
\[ 0 = 1 \text{ is false} \;\Longrightarrow\; \text{no solution} \]
Verify: say what the geometry is
Why: The first two planes are parallel and distinct, since their normal directions match but their constants do not. Two parallel planes never meet, so a third plane cannot rescue the system — the third equation was never even used, which is itself a signal that the first two settled it.
Sorting
In each case the variables have all cancelled during elimination.
Sort into buckets
Sort each leftover statement.
Only the truth of the statement matters, never the particular numbers. Four equals four means the same as zero equals zero.
Worked example
Three planes sharing a whole line.
\[ \begin{cases} x + y + z = 6 \\ 2x + 2y + 2z = 12 \\ x - y + z = 2 \end{cases} \]
Compare the first two equations
Why: Every coefficient AND the constant of the second is twice the first, so the second says nothing new.
Eliminate to confirm
Why: Multiplying the first by negative 2 and adding gives 0 equals 0.
\[ 0 = 0 \]
Reduce the system to what is left
Why: Only two genuinely different equations remain: x plus y plus z equals 6 and x minus y plus z equals 2.
Describe the solutions
Why: Subtracting gives 2y equals 4, so y is 2, and then x plus z equals 4 with x free.
\[ y = 2, x + z = 4 \]
Figure (svg): The solution to Worked example a system with infinitely many shown as a ladder of expressions, one row per algebraic move
\[ y = 2, \quad x + z = 4 \quad \text{(a line of solutions)} \]
Verify: produce two different solutions explicitly
Why: Take x equal to 0: then z is 4, giving (0, 2, 4). Check: 0 plus 2 plus 4 is 6, and 0 minus 2 plus 4 is 2 — both correct. Take x equal to 3: then z is 1, giving (3, 2, 1), and 3 plus 2 plus 1 is 6 while 3 minus 2 plus 1 is 2. Two distinct solutions confirms there is not exactly one.
Error analysis
A student reaches 0 equals 0 and reports one of the solutions as the answer.
Annotate
On: \( 0 = 0 \;\Longrightarrow\; \text{try } x = 0: \; \text{the solution is } (0, 2, 4) \)
When a system has infinitely many solutions, the answer is a description, not an example. Naming one solution is like answering which numbers are less than five by saying three.
Prediction
Commit before doing any algebra.
Predict first
What will happen when you solve x + y + z = 3, 2x + 2y + 2z = 7, x - y + z = 1?
Correct: No solution — the first two equations contradict each other.
\[ 2(x+y+z) = 6 \neq 7 \quad \text{for every triple} \]
Why: The coefficients of the second equation are exactly twice those of the first, so the two describe parallel planes; but the constant 7 is not twice 3, so the planes are distinct. Two parallel distinct planes share nothing, and no third equation can create a common point. Spotting a proportional coefficient row before starting can settle a system in five seconds.
Comparison
Fill the blanks. The rule is unchanged; the geometry is richer.
Comparison matrix
| Feature | Two equations, two variables | Three equations, three variables |
|---|---|---|
| Graph of one equation | a line | a plane |
| Solution is | an ordered pair | an ordered triple |
| 0 = 0 means | infinitely many | infinitely many |
| Infinitely many looks like | one line | a line or a whole plane |
| Number of possible outcomes | three | three |
The third and fifth rows are identical, which is the point: the algebra transfers unchanged. Only the fourth row gains a second possibility.
Counterexample
A classmate offers a rule about three-variable systems.
\[ \text{if every pair of planes intersects, all three must meet at a point} \]
Discussion prompt
Describe an arrangement of three planes in which every pair meets in a line and yet no point lies on all three. What everyday object has that shape?
Hint: Think about three walls that do not all meet.
Answer:
Take three planes whose pairwise intersections are three parallel lines. Every pair meets, but the three lines never come together, so no point is on all three planes.
The everyday shape is a triangular prism: extend its three rectangular faces and they meet pairwise along the three parallel edges, and nowhere all together.
Algebraically this produces a false statement like 0 equals 7, exactly as parallel planes do — which is why the algebra is more reliable than trying to picture the arrangement.
Section
Section 5
Concept
A situation with three unknown quantities needs three independent pieces of information. Each becomes an equation, and the system is solved by the method of this lesson.
\[ \begin{cases} x + y + z = 100 \\ \text{a value condition} \\ \text{a relationship} \end{cases} \]
Counting the unknowns and counting the given facts before writing anything is the same discipline as Lesson 1.5, extended by one.
Figure (svg): The three-variable system reduced to a two-variable one by eliminating the same variable twice
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.4 Solve Systems of Linear Equations in Three Variables §3.4, pp. 178-181
Picture it
The same reduction, whatever the situation.
Figure (svg): The three-variable system reduced to a two-variable one by eliminating the same variable twice
The modelling is the difficult part. Once the three equations exist, the solving is mechanical.
Worked example
Three prices, three facts, one system.
\[ \text{100 tickets sold for } \$1000. \text{ Adult } \$15, \text{ student } \$8, \text{ child } \$5. \text{ There were twice as many students as children.} \]
Name the three unknowns
Why: Let a, s and c be the numbers of adult, student and child tickets.
Write the count condition
Why: The three numbers total 100 tickets.
\[ a + s + c = 100 \]
Write the money condition
Why: Fifteen a plus 8s plus 5c is 1000 dollars.
\[ 15 a + 8 s + 5 c = 1000 \]
Write the relationship
Why: Twice as many students as children means s equals 2c.
\[ s = 2 c \]
Substitute the relationship into the other two
Why: The third equation is already solved for s, so substitution is free: a plus 3c equals 100, and 15a plus 21c equals 1000.
\[ a + 3 c = 100, 15 a + 21 c = 1000 \]
Figure (svg): The solution to Worked example a three-unknown ticket problem shown as a ladder of expressions, one row per algebraic move
\[ \begin{cases} a + 3c = 100 \\ 15a + 21c = 1000 \end{cases} \]
Verify: check the substitution preserved the meaning
Why: The first reduced equation says adults plus three times the children equals 100, which is right because students are twice the children, so students plus children is three times the children. The second says the money adds up with 8 times 2c plus 5c, which is 21c. Both reductions are faithful to the situation.
Real world
A shop sells three sizes of candle. Small cost 4 dollars, medium 7, large 12. One day 30 candles sold for 224 dollars, and twice as many small sold as large.
Discussion prompt
Name the variables, write all three equations, reduce to two, and solve. Then check your answer against every fact in the problem.
Hint: The relationship is the equation that is easiest to substitute.
Answer:
\[ \begin{cases} s + m + l = 30 \\ 4s + 7m + 12l = 224 \\ s = 2l \end{cases} \]
\[ 3l + m = 30, \quad 20l + 7m = 224 \;\Longrightarrow\; l = 14, \; m = -12 \]
The medium count comes out negative, which is impossible. So the stated totals are inconsistent with a real sale — the algebra is fine and the data is not, which is exactly the kind of thing checking against the situation catches and checking against the equations does not.
Worked example
The two-by-two system, solved and interpreted.
\[ \begin{cases} a + 3c = 100 \\ 15a + 21c = 1000 \end{cases} \]
Eliminate a
Why: Multiplying the first by negative 15 gives negative 15a minus 45c equals negative 1500. Adding cancels a.
\[ -24 c = -500 \]
Solve for c
Why: Five hundred divided by 24 is about 20.83.
\[ c = 20.83 \]
Notice the answer is not a whole number
Why: Ticket counts must be whole, so no combination of whole tickets matches these totals exactly.
Report honestly
Why: The system is consistent but has no solution in whole numbers, which is a fact about the stated totals rather than an arithmetic error.
Figure (svg): The solution to Worked example finish the ticket problem shown as a ladder of expressions, one row per algebraic move
\[ c = \tfrac{125}{6} \;\Longrightarrow\; \text{no whole-number solution} \]
Verify: test the nearest whole values
Why: With c equal to 21, s is 42 and a is 37: the money is 555 plus 336 plus 105, which is 996 dollars, not 1000. With c equal to 20, s is 40 and a is 40: 600 plus 320 plus 100 is 1020. Neither hits 1000, confirming that the fractional answer reflects the data rather than a slip.
Error analysis
A student models a three-unknown problem with two equations.
Annotate
On: \( \begin{cases} a + s + c = 100 \\ 15a + 8s + 5c = 1000 \end{cases} \quad \text{with three unknowns} \)
Count the unknowns, then count the facts. If the counts do not match, either something in the problem has been overlooked or the answer will be a description rather than a triple.
Matching
Each sentence from a word problem becomes one equation.
Match the pairs
Why: A count condition adds the quantities; a value condition weights each by its price. The two relationships are the ones worth reading twice: twice as many students as children means students is the doubled one, and five more adults than children means the 5 is added on the children's side of the balance.
Estimation
One hundred tickets brought in 1000 dollars, at prices of 15, 8 and 5 dollars.
Predict first
Roughly what does the average ticket price tell you about the mix?
Correct: The average is 10 dollars, so the mix leans toward the middle and top prices.
\[ \frac{1000}{100} = 10 \text{ dollars per ticket, against prices of } 5, 8, 15 \]
Why: One thousand divided by 100 is 10 dollars a ticket. That sits between 8 and 15 and above both the child and student prices, so there must be a substantial number of adult tickets to pull the average up. Computing the average first gives a sanity check on any answer: a solution with only a handful of adult tickets would be immediately suspect.
Socratic
One question, and nothing else on this slide.
\[ \begin{cases} a + s + c = 100 \\ 15a + 8s + 5c = 1000 \end{cases} \]
Discussion prompt
Suppose the problem really did give only these two facts about three unknowns. Is the problem unanswerable, or does it have a different kind of answer? What could you still say, and what could you not?
Hint: Think about what two planes have in common.
Answer:
It is not unanswerable — it simply has infinitely many solutions, lying along the line where the two planes meet. The answer is a description rather than a triple.
You could still say a great deal: solving for two variables in terms of the third gives a formula, and the requirement that all three counts be non-negative whole numbers cuts the line down to a finite list of possibilities you could enumerate.
What you could not do is name THE answer, because there isn't one. Recognising when a problem has under-determined its unknowns is as much a part of modelling as solving it.
Comparison
Fill the blanks. The method scales; the vocabulary follows.
Comparison matrix
| Feature | Lesson 3.2 (two variables) | This lesson (three variables) |
|---|---|---|
| Equations needed | two | three |
| Answer is | an ordered pair | an ordered triple |
| Eliminations required | one | two, of the same variable |
| Graph of one equation | a line | a plane |
| 0 = 0 means | infinitely many | infinitely many |
The one row that genuinely differs is the third, and it names the single choice this lesson asks you to get right.
Pattern
One routine solves any three-by-three linear system.
Step one is worth the ten seconds it costs, and step two contains the only genuinely new decision: the same variable, both times.
OpenStax Algebra and Trigonometry 2e, §11.2 Systems of Linear Equations: Three Variables §11.2
Check
A triple must satisfy all three equations.
Check your understanding
Is (2, 1, -3) a solution of 4x + 2y + 3z = 1, 2x - 3y + 5z = -14 and 6x - y + 4z = -1?
Answer: A
Why: Substituting gives 8 + 2 - 9 = 1, then 4 - 3 - 15 = -14, then 12 - 1 - 12 = -1. Every equation matches its right side.
Check
The key decision of the method.
Check your understanding
When solving a three-variable system by elimination, what must be true of your two eliminations?
Answer: A
Why: Eliminating the same variable from two different pairs leaves two equations in the other two variables — a solvable two-by-two system. Different variables would leave three variables spread across two equations.
Check
A degenerate outcome. Read the statement.
Check your understanding
Eliminating variables from a three-variable system leaves the statement 0 = 7. What does the system have?
Answer: A
Why: Zero equals seven is false for every triple, so no triple can satisfy the equations that produced it. The three planes have no common point and the system is inconsistent.
Real world
A parabola y equals ax squared plus bx plus c passes through the points (1, 4), (2, 3) and (3, 6).
Discussion prompt
Explain why finding a, b and c is a three-variable linear system even though the curve is not a line. Write the three equations and solve them.
Hint: The unknowns are a, b and c — not x and y.
Answer:
\[ \begin{cases} a + b + c = 4 \\ 4a + 2b + c = 3 \\ 9a + 3b + c = 6 \end{cases} \]
Each point gives one equation, and although x appears squared, a, b and c each appear to the first power — so the system is linear in the unknowns, which is all the method requires.
\[ a = 2, \; b = -7, \; c = 9 \;\Longrightarrow\; y = 2x^2 - 7x + 9 \]
Checking at x equal to 2: 8 minus 14 plus 9 is 3 — correct. This is exactly how Lesson 4.10 will fit a quadratic model to three data points, and it is why three-variable systems are worth having before Chapter 4.
Commit first
Answer, then rate your confidence honestly.
Predict first
Can a system of three linear equations in three variables have exactly two solutions?
Correct: No — the only possibilities are one, none, or infinitely many.
The argument is worth keeping: linearity means the solution set is a point, a line, a plane, all of space, or empty — never a scattering of isolated points.
Why: If two distinct triples both solve the system, then so does every point on the line joining them, because the equations are linear. So two solutions immediately forces infinitely many. This is the same structural fact as for two lines in Lesson 3.1, and it means any method you use must land on one of exactly three outcomes.
Explain it
They can solve two-by-two systems and are alarmed by three equations.
Discussion prompt
In four sentences or fewer, explain the method, why nothing new is needed, and the one decision they must not get wrong.
Hint: The decision is about which variable.
Answer:
Pick one variable and eliminate it from two different pairs of the three equations. That leaves two equations in the remaining two variables, which is a system they already know how to solve, and then one substitution into any original equation recovers the variable they removed.
The decision that matters is eliminating the SAME variable both times. Choosing a different one each time leaves three variables spread across two equations, which is no simpler than where they started.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For choosing, scan all nine coefficients for a 1 or negative 1. For keeping them aimed together, write the target variable at the top of your page before starting. For back-substitution, remember the eliminated variable appears only in the originals. For ordering, write x, y, z as headings and fill values in underneath rather than in the order found. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top of a page write a three-variable system of your own in which at least one variable has a coefficient of 1 somewhere. Circle the variable you will eliminate and write beside it why you chose it. Below, show both eliminations in full, with the multiplier written next to each equation you scale, and box the two-variable system that results. Solve that system, then back-substitute into an original equation, marking which original you used and why. Write the final answer as an ordered triple and check it in all three equations, ticking each. Finally, in a margin, sketch three parallelograms meeting at a point, three meeting along a line, and three that do not all meet, labelling each with its number of solutions.
If the variable you circled did not have a coefficient of 1 or negative 1 somewhere, try again with one that does and compare how much smaller the numbers stay.
Recap
Five things, and the second is the only genuinely new one.
| If you see | Then |
|---|---|
| A coefficient of 1 or -1 | Eliminate that variable |
| Two equations with proportional coefficients | Check the constants before solving |
| Two eliminations of different variables | Start over; nothing was reduced |
| 0 = 0 | Infinitely many; describe the set |
| 0 = a nonzero number | No solution |
Lesson 3.5 introduces matrices, which are the bookkeeping that lets a system like this be written and solved without carrying the letters around.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.4 Solve Systems of Linear Equations in Three Variables §3.4, pp. 178-181 — everything on these slides traces back here
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