BetterGrades Precalculus · Unit 8 · Lesson

Systems in three variables

Solve and classify three-variable linear systems and interpret their solutions as intersections of planes.

Opening

Start with the situation

A linear equation in three variables describes a plane, and a system seeks their common point, line, plane, or empty intersection.

Systems combine several conditions into one decision. Graphs, equations, inequalities, and matrices are different views of the same requirement: the final result must satisfy every condition at once.

Before you begin

Prerequisite check

Solve equations and systems.
Interpret graphs as solution sets.
Use organized arithmetic and units.

Core explanation

Explanation

Eliminate one variable from two equation pairs, solve the reduced system, and back-substitute.

A contradiction means no solution; a free variable produces an infinite family.

A symbolic answer is not complete by itself. In this lesson, the same claim must also be readable through three-plane gallery, elimination ladder, or another equivalent representation.

Conceptual reading

What the idea is really doing

A system asks for simultaneous truth. Graphs show common intersections, elimination preserves the solution set, and matrices record the same operations compactly; the representation changes, but the solution condition does not.

This lesson narrows that lens to one goal: solve and classify three-variable linear systems and interpret their solutions as intersections of planes. The point is not to memorize an isolated trick; it is to know what evidence makes the conclusion valid and how a second representation can check it.

Reusable method

A reliable route through the problem

Eliminate one variable from two equation pairs.
Solve the reduced system.
Back-substitute.

Verification: Substitute the result into every original equation or inequality. For matrix work, translate the final rows back into statements about variables, pivots, free variables, and consistency.

Foundation walkthrough

Plan before calculating

Problem

Solve $x+y+z=6,x-y+z=2,x+y-z=0$

Plan: Start by identifying the mathematical structure in the prompt. Then use the lesson method rather than guessing from appearance: Eliminate one variable from two equation pairs, solve the reduced system, and back-substitute.
Conclusion: $(1,2,3)$
Why the check works: Pairwise subtraction isolates variables.

Worked examples

See the idea in three forms

foundation example

Solve $x+y+z=6,x-y+z=2,x+y-z=0$

Solution $(1,2,3)$

Pairwise subtraction isolates variables.

representation example

Meaning $0=5$ .

SolutionNo solution.

This example expresses systems in three variables in a second form.

transfer example

Meaning free variable.

SolutionInfinitely many solutions.

A contradiction means no solution; a free variable produces an infinite family.

Three-plane gallery. Follow the foundation example from its given information to the conclusion. The labels identify the mathematical feature that makes the result valid: Pairwise subtraction isolates variables. — **Anchor figure · Three-plane gallery**
Follow the foundation example from its given information to the conclusion. The labels identify the mathematical feature that makes the result valid: Pairwise subtraction isolates variables.

Elimination ladder. Read the numbered reasoning path in order. Each stage preserves the quantities, restrictions, or structural conditions needed for systems in three variables. — **Mechanism figure · Elimination ladder**
Read the numbered reasoning path in order. Each stage preserves the quantities, restrictions, or structural conditions needed for systems in three variables.

Parameterized solution line. Compare the valid path with the tempting shortcut. The figure shows why performing elimination on only one pair and stopping with insufficient information leads to a false conclusion. — **Comparison and error figure · Parameterized solution line**
Compare the valid path with the tempting shortcut. The figure shows why performing elimination on only one pair and stopping with insufficient information leads to a false conclusion.

Common mistake

Find the first invalid move

A frequent error is performing elimination on only one pair and stopping with insufficient information.

Check yourself

One equation in three variables.

Write a complete attempt before opening the exact answer.

Attempt once to unlock the answer

Complete a substantive attempt before revealing the server-held answer.

Practice

Ten concrete questions

Practice 101

One equation in three variables.

Write a complete attempt before opening the exact answer.

Attempt once to unlock the answer

Complete a substantive attempt before revealing the server-held answer.

Practice 202

Meaning $0=5$ .

Write a complete attempt before opening the exact answer.

Attempt once to unlock the answer

Complete a substantive attempt before revealing the server-held answer.

Practice 303

Meaning free variable.

Write a complete attempt before opening the exact answer.

Attempt once to unlock the answer

Complete a substantive attempt before revealing the server-held answer.

Practice 404

Solve $x+y+z=3,x-y+z=1,x+y-z=1$

Write a complete attempt before opening the exact answer.

Attempt once to unlock the answer

Complete a substantive attempt before revealing the server-held answer.

Practice 505

Explain why this conclusion is valid: $(1,2,3)$ . Use the foundation problem as evidence: Solve $x+y+z=6,x-y+z=2,x+y-z=0$ .

Write a complete attempt before opening the exact answer.

Attempt once to unlock the answer

Complete a substantive attempt before revealing the server-held answer.

Practice 606

Solve the representation example, then name the feature of systems in three variables that it illustrates: Meaning $0=5$

Write a complete attempt before opening the exact answer.

Attempt once to unlock the answer

Complete a substantive attempt before revealing the server-held answer.

Practice 707

Correct this reasoning and identify the first unsafe assumption: A frequent error is performing elimination on only one pair and stopping with insufficient information.

Write a complete attempt before opening the exact answer.

Attempt once to unlock the answer

Complete a substantive attempt before revealing the server-held answer.

Practice 808

Connect two representations for this example: Solve $x+y+z=6,x-y+z=2,x+y-z=0$ . Describe what a graph, table, mapping, or algebraic form would have to show.

Write a complete attempt before opening the exact answer.

Attempt once to unlock the answer

Complete a substantive attempt before revealing the server-held answer.

Practice 909

Create a nearby example by changing one number or condition in this prompt: Meaning free variable. Predict the effect, solve your new example, and compare it with the original.

Write a complete attempt before opening the exact answer.

Attempt once to unlock the answer

Complete a substantive attempt before revealing the server-held answer.

Practice 1010

Write a short verification checklist for systems in three variables, then apply it to one worked example from this lesson.

Write a complete attempt before opening the exact answer.

Attempt once to unlock the answer

Complete a substantive attempt before revealing the server-held answer.

Lesson close

Connect forward

The next lesson, Matrix notation and operations, uses this result as part of a larger structure.

Source record

Original BetterGrades manuscript, rights-separated references.

Yoshiwara, Modeling, Functions, and Graphs
Utah College Algebra
Stitz and Zeager, Precalculus

No long source passage is reproduced.

Solvex+y+z=6,x−y+z=2,x+y−z=0x+y+z=6,x-y+z=2,x+y-z=0x+y+z=6,x−y+z=2,x+y−z=0

Solvex+y+z=6,x−y+z=2,x+y−z=0x+y+z=6,x-y+z=2,x+y-z=0x+y+z=6,x−y+z=2,x+y−z=0

Meaning 0=50=50=5.

Meaning free variable.

Original BetterGrades manuscript, rights-separated references.

Solve $x+y+z=6,x-y+z=2,x+y-z=0$

Solve $x+y+z=6,x-y+z=2,x+y-z=0$

Meaning $0=5$ .