A system of linear equations has three possible answers, not one. Either exactly one set of values satisfies every equation, or none does, or infinitely many do. Most calculators handle the first case and treat the other two as errors, which is why they so often produce a shrug where an answer was a...
A determinant of zero does not mean no solution. It means the answer is not unique — and that leaves two possibilities which the determinant cannot tell apart. x + y = 2 with x + y = 3 has none; x + y = 2 with 2x + 2y = 4 has infinitely many. Both have a determinant of zero. Separating them needs the constants as well, which is why this page reports two ranks rather than one number. Everything is computed in exact fractions: coefficients differing in the tenth decimal place are enough to make a floating-point solver reach a different kind of answer, not just a different digit.
Type them as you would write them — 2x + 3y = 6. Variables x, y and z; fractions and decimals are fine.
THE SOLUTION
x = 3/2, y = 1
One point satisfies every equation, and substituting it back reproduces each right-hand side exactly.
PUT BACK INTO THE ORIGINAL EQUATIONS
2x + 3y = 6 → 6 = 6 ✓
x - y = 1/2 → 1/2 = 1/2 ✓
DETERMINANT
-5
not zero — so exactly one solution
RANK
2 of 2
every variable is pinned down
WITH THE CONSTANTS
2 of 2
the same, so the equations agree
SUBSTITUTED BACK
2 of 2
every equation is satisfied
As decimals. x = 1.5, y = 1. Whole-number coefficients rarely give whole-number answers — among invertible integer systems, 92.5% do not. The fractions above are the exact values; these are for reading.
WHAT IS HAPPENING
2 3 | 6 · 1 -1 | 1/2
Only the coefficients matter to the arithmetic, so the letters can be set aside. The bar separates the left-hand sides from the constants — and that bar is what distinguishes no solution from infinitely many later on.
1 3/2 | 3 · 1 -1 | 1/2
Scaling a whole equation leaves its solutions untouched, and a leading 1 makes the next step a subtraction rather than a juggle.
1 3/2 | 3 · 0 -5/2 | -5/2
Subtracting a multiple of one equation from another is the elimination step. It is legitimate because the two equations are both true at once, so any combination of them is true as well.
1 3/2 | 3 · 0 1 | 1
The same move again, on a row that elimination has already changed — dividing by -5/2 is fine even though that number arrived part-way through, because it is still a whole equation being scaled.
1 0 | 3/2 · 0 1 | 1
Elimination again, now upward as well as downward: clearing y out of the rows above is what turns a triangle into a table where each row names a single variable.
x = 3/2 · y = 1
Every column has a leading 1 and nothing else, so each row now says exactly what one variable equals. Nothing is left to rearrange.
Why two ranks and not one determinant. The rank of the coefficients says how many independent constraints there are. The rank with the constants attached says whether those constraints can all be met at once. Equal ranks mean the equations agree; a higher rank with the constants means one of them contradicts the rest, and that is what no solution is. A determinant collapses all of this into a single number and necessarily loses the distinction.
Exact fractions throughout · the answer is substituted back into the equations as written
Type each equation the way you would write it — 2x + 3y = 6, or x = 4 − y. Variables x, y and z, on either side, with fractions and decimals both allowed. Brackets are the one thing to expand yourself.
Add or remove equations freely. There is no requirement to have as many equations as unknowns; more usually means no solution and fewer always means infinitely many, and the page will tell you which you have.
Read the verdict first. It says one solution, none, or infinitely many — and when there are infinitely many, it gives the whole family in terms of a free parameter rather than picking one member arbitrarily.
Check the two ranks underneath. Equal ranks mean the equations agree; a higher rank once the constants are attached is exactly what no solution looks like, and it is the distinction the determinant cannot make.
Take 2x + 3y = 6 together with x − y = 1/2. Written as a table of numbers, the coefficients are 2, 3 and 1, −1, with 6 and 1/2 on the right. Divide the first row by 2 to get a leading 1, then subtract that row from the second to clear the x column. Scale what is left of the second row so its y coefficient is 1, and subtract the right multiple of it from the first row to clear the y column. What remains reads x = 3/2 and y = 1. Substituting back is the only check worth doing: 2(3/2) + 3(1) = 6, and 3/2 − 1 = 1/2. Both hold exactly, so the arithmetic did not slip. Note that the answer is not a pair of whole numbers despite every coefficient being one — among invertible systems with small integer coefficients, 92.5% behave that way, which is why keeping fractions through the working matters. Now x + y = 2 together with x + y = 3. Eliminating gives a row that reads 0 = 1. No values of x and y can make that true, so there is no solution. Drawn as lines, the two are parallel and never meet. The determinant here is zero. Now x + y = 2 together with 2x + 2y = 4. Eliminating gives a row that is entirely zero, including the constant. The second equation was the first one doubled and told us nothing new, so one equation constrains two unknowns and a whole line of points satisfies both. The answer is the family x = 2 − t, y = t. The determinant here is also zero — the same number as the previous case, with a completely different meaning. That is the point worth carrying away. Of all the coefficient matrices with entries between −6 and 6, 4.6% have a zero determinant, and every one of them is either the second case or the third. The determinant says which pile they are in; only the constants say which of the two. Finally, a case that looks harmless: x + y = 1 with x + 1.0000000001y = 1. The determinant is one ten-billionth — small, but not zero, so there is exactly one solution. A floating-point solver comparing that against a 1e-9 tolerance concludes the system is singular and reports infinitely many. The arithmetic here is exact, so the answer is x = 1, y = 0.
| Rule | What it says | Why |
|---|---|---|
| Three possible answers | one · none · infinitely many | Every linear system lands in exactly one of these. Most calculators only handle the first. |
| One solution | the lines cross once | Full rank. Every variable is pinned down and substituting back reproduces each equation. |
| No solution | parallel and apart | Elimination reaches a row saying 0 = something. The equations contradict each other. |
| Infinitely many | the same line twice | The equations agree but do not pin everything down. A family, not a failure. |
| The determinant | zero means not unique | It cannot tell you which of the other two. That needs the constants as well. |
| The two ranks | compare A with A augmented | Equal means solvable. Higher with the constants means no solution. |
| Free variables | variables minus rank | How many can be chosen freely. Each one gives a parameter in the answer. |
| Elimination | add a multiple of one row | Legitimate because both equations are true, so any combination of them is. |
| Row swaps | change nothing | The equations are a set, not a sequence. Order is a convenience. |
| Scaling a row | changes nothing | Multiplying a whole equation by a non-zero number keeps the same solutions. |
| More equations than unknowns | usually no solution | Not always — repeated or dependent equations cost nothing. |
| Fewer equations than unknowns | never unique | There is always at least one free variable, so none or infinitely many. |
| Substituting back | the only real check | Recompute each left-hand side from the answer. It catches arithmetic, not method. |
| Floating point | changes the kind of answer | A near-singular system can be called singular, and vice versa. Exact fractions avoid both. |
Exactly three possibilities: one, none, or infinitely many. There is never a case of two or seventeen. Geometrically, two lines either cross once, run parallel without meeting, or lie exactly on top of one another. The same three outcomes hold with any number of equations and unknowns, which is why the classification is worth doing before hunting for a number.
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Last updated: August 5, 2026 · Exact fractions throughout, never floating point · One solution, none, or infinitely many — the determinant alone cannot say which.