This distinction matters when two equations have proportional coefficients. Dividing by a zero pivot or returning one convenient pair can hide whether the original system is inconsistent or has infinitely many solutions.
Keep the right-hand side attached
Consider this original real-number system:
x + 2y = 3
2x + 4y = 7
Augmented matrix:
[1 2 | 3]
[2 4 | 7]
Apply R2 ← R2 - 2R1, including the constant column:
[1 2 | 3]
[0 0 | 1]
The second row says 0x + 0y = 1, which is impossible. There is no ordered pair satisfying both original equations. The coefficient matrix has rank 1; the augmented matrix has rank 2. The inconsistency is visible in that extra pivot in the constant column.
An answer that drops the final column can mistakenly call the system dependent. Keep the vertical separator in every displayed step.
Change the constant and describe the whole solution set
Replace 7 with 6 in the second original equation. The same operation now gives a zero row:
[1 2 | 3]
[0 0 | 0]
The remaining condition is x + 2y = 3. Choose y = t for any real t, then x = 3 - 2t. The complete family is (x, y) = (3 - 2t, t), t real.
| t | x | y | First equation | Second equation |
|---|---|---|---|---|
| 0 | 3 | 0 | 3 = 3 | 6 = 6 |
| 1 | 1 | 1 | 3 = 3 | 6 = 6 |
| -1 | 5 | -1 | 3 = 3 | 6 = 6 |
Those rows illustrate members of the family. The parameterization proves more: substituting any real t gives (3 - 2t) + 2t = 3 and twice that expression equals 6. Both ranks are 1, below the two unknowns, leaving one free parameter.
The pair (0, 0) does not satisfy the first equation. A row of zeros is not an instruction to set all variables to zero.
Request the reason for the classification
OpenStax's systems-of-equations chapter, checked October 24, 2026, distinguishes one solution, no solution and infinitely many solutions. A system solution must satisfy every equation, so checking only the reduced nonzero row is insufficient if the reduction was copied incorrectly.
OpenAI's prompting guidance supports giving your attempted work and the response you need. Preserve exact integers or fractions during this exercise so rounding does not manufacture or conceal a zero row.
Review these two real-number linear systems using augmented matrices.
A: x + 2y = 3; 2x + 4y = 7.
B: x + 2y = 3; 2x + 4y = 6.
Show each row operation, including the right-hand side. Explain
R2 <- R2 - 2R1. Read the resulting second row as an equation.
State coefficient rank, augmented rank and solution classification.
For any free variable, describe the full parameterized solution set.
Do not divide by a zero pivot or report only one convenient solution.
Check the family symbolically in all original equations, then show
three example members. Keep exact arithmetic throughout.
Afterward give one different system for me to classify before solving.
The single-variable equation inspector explicitly handles a narrower equation form. This two-variable task needs a joint solution set. Finish by keeping the original equations, the decisive augmented row and the classification together; a final pair of numbers alone cannot explain an inconsistent or dependent system.