How this instrument works
Reducing a system of equations toward row echelon form means systematically eliminating variables from lower rows using the rows above them, until each row has fewer nonzero leading terms than the one before it. For a 2-equation system, that's a single step: subtract a multiple of row 1 from row 2, chosen so row 2's own x term cancels out to exactly 0, leaving row 2 with only a y term and a constant.
The multiple needed — called the elimination factor — is simply row 2's x coefficient divided by row 1's own x coefficient. Subtracting that many copies of row 1 from row 2, term by term, is what actually zeroes out row 2's x term while correctly adjusting its y coefficient and constant to match.
This site's companion Gauss-Jordan Elimination calculator continues further than row echelon form does — instead of stopping once row 2's x term is cleared, it keeps reducing until BOTH variables are fully isolated, reading the solution directly off the reduced constants rather than needing a separate back-substitution step afterward.
- Enter row 1's x coefficient, y coefficient, and constant.
- Enter row 2's x coefficient, y coefficient, and constant the same way.
- Read Elimination factor: row 2's x coefficient divided by row 1's.
- Read the new row 2's y coefficient and constant — the x term should now read exactly 0.
Worked example — reducing [x+2y=3; 4x+5y=6]
For the system 1x+2y=3 and 4x+5y=6: the elimination factor is row 2's x coefficient divided by row 1's, 4÷1=4. Subtracting 4 copies of row 1 from row 2 gives (4−4×1, 5−4×2, 6−4×3) = (0, −3, −6) — row 2's x term reads exactly 0, leaving −3y=−6, ready for back-substitution (y=2, then x=3−2×2=−1).
For [2x+y=5; 6x+4y=8]: the factor is 6÷2=3, giving a new row 2 of (6−3×2, 4−3×1, 8−3×5)=(0,1,−7). For [x+0y=4; 3x+y=10]: the factor is 3÷1=3, giving a new row 2 of (3−3×1, 1−3×0, 10−3×4)=(0,1,−2).
Questions
What is row echelon form?
A form of a system of equations where each row has strictly fewer leading nonzero terms than the row above it — reached by systematically eliminating variables from lower rows using the ones above, exactly the step this page demonstrates for a 2-equation system.
How is the elimination factor found?
By dividing row 2's own coefficient for the variable being eliminated by row 1's coefficient for that same variable — that ratio tells you exactly how many copies of row 1 to subtract from row 2.
Why must row 1's leading coefficient not be zero?
Dividing by it to find the elimination factor would be undefined — if row 1's own leading coefficient happens to be zero, the two rows need to be reordered first so a genuine pivot leads.
How is this different from the Gauss-Jordan Elimination calculator on this site?
This page performs just the single elimination step that reaches row echelon form, leaving one variable still needing back-substitution; that page continues reducing until both variables are fully isolated, reading the solution directly off the result.
What confirms the elimination step was done correctly?
The eliminated variable's coefficient in the new row should read exactly 0 — if it doesn't, the elimination factor or the subtraction itself has an error somewhere.