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Linear systems in two variables and degenerate constraints

On this page
  1. Mathematical statement
  2. Representative formula (exact TeX)
  3. Assumptions
  4. Symbol scope
  5. Intuition, definition, and misconceptions
  6. Adopted derivation guide
  7. Prerequisites
  8. Verification boundary
  9. Next obligation
  10. Same-source structured data

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.

A solution must satisfy two real linear constraints simultaneously. A nonzero determinant gives a unique solution; singular systems and zero rows require separate consistency checks.

mko-system-linear-equations · definition · v0.10.0

Mathematical statement

For real unknowns x and y, require a x+b y=c and d x+e y=f simultaneously. Define L(x,y)=(a x+b y−c,d x+e y−f) and z=(0,0), so the solution set satisfies L(x,y)=z. If D=ae−bd≠0, the unique solution is x=(ce−bf)/D and y=(af−cd)/D. If D=0, check consistency: there may be no solution or infinitely many solutions.

Representative formula (exact TeX)

L(x,y)=z

Assumptions

  • All coefficients and right-side constants are real. Exercises use integer coefficients and exact rational candidates.
  • Zero-coefficient rows are allowed as linear-system constraints: 0=0 imposes no restriction, while 0=c≠0 is inconsistent. Do not apply the unique-solution division when D=0.

Symbol scope

  • L — vector of the two linear-constraint residuals
  • x — first real unknown
  • y — second real unknown
  • z — zero vector (0,0)

Intuition, definition, and misconceptions

To verify a rational candidate pair, first check nonzero denominators and substitute into both rows. Passing only one row does not make it a solution of the system.

Use the unique-solution formula only for D≠0. When D=0, coefficient rank and consistency of the augmented data determine no solution or infinitely many solutions; first check inconsistent zero rows. If all coefficients and right sides are zero, all of R² is the solution set; if all coefficients are zero and any right side is nonzero, there is no solution.

The exercises use exact two-by-two minors and zero-row cases to check consistency. D and both Cramer numerators being zero are insufficient by themselves to establish infinitely many solutions. Exact solutions are not guessed from intersections in a sketch.

Common errors include substituting into only one row, dividing when D=0, or overlooking an inconsistent row 0=nonzero.

Adopted derivation guide

Linear systems in two variables: the unique-solution formula, checking both rows, and consistency when D=0

The original MKO is retained. The guide has its own append-only revision and verification boundaries, not a new Lean proof.

Prerequisites

Verification boundary

Explicit two-row systems in two variables can be checked exactly; general rank theorems and formal correctness of a solver await Evidence.

not_formalized

This does not mean the content is false; it means no replayable computational or formal evidence is registered yet.

Next obligation

Bind solution sets of linear constraints, zero rows, the two-by-two determinant, rank and augmented consistency, and the solution formula for nonzero D.

Same-source structured data

This page and the interactive view use the same object and existing language edition; this adds no proof or human review.

Canonical MKO JSON · Plain Markdown