In short

You are eliminating x and y from a pair of equations and you end up with something like 0 = 5. Both variables have vanished from the equation, but they have NOT been "found" to equal 0. What you have produced is a contradiction — a statement that is false for every value of x and y. The correct conclusion is "the system has no solution" (the two lines are parallel and never meet). The wrong conclusion — written almost weekly in CBSE Class 10 board copies — is "x = 0, y = 0". The variables disappeared because the system is inconsistent; the leftover constants 0 = 5 are the proof of that, not the answer. The clean way to remember it: when both variables vanish, look only at the constants. 0 = 0 means infinite solutions; 0 = c with c \neq 0 means no solution; nothing here ever means x = y = 0.

You have already worked through systems of linear equations and met all three cases — one solution, no solution, infinitely many solutions. The first case is the comfortable one: elimination kills one variable, the other variable falls out with a clean number, and you back-substitute to finish.

But every now and then, elimination kills both variables in a single move. The equation collapses to something like 0 = 3 or 0 = -7 — pure constants on each side, no x or y in sight. The exam clock is ticking. Panic. Pen moves. And out comes the answer: "x = 0, y = 0". Two-mark question lost.

This article is the gate that stops that pen. The mistake is so common in CBSE Class 10 board exams that NCERT specifically calls it out — the trap question has appeared in board papers and sample papers many times, and graders see the wrong answer more often than the right one.

What is actually happening when both variables vanish

Take a pair of equations. Multiply one (or both) by some number so that the coefficients of x match. Subtract. The x terms cancel by design. If you are lucky, the y term remains and you solve for y. If you are unlucky — meaning the equations were secretly multiples of each other — then the y term cancels too. Both variables disappear simultaneously.

What's left is a statement of the form

0 = c

where c is some number — possibly zero, possibly not.

Why "vanish" is not the same as "equal zero": A variable vanishes from the equation when its coefficient becomes 0. A variable equals zero when, after solving, its value turns out to be 0. These are completely different events. 0 \cdot x is 0 for every value of x — the term carries no information about x at all. It does not pin x to 0; it pins x to nothing. The same goes for 0 \cdot y.

So when both coefficients become zero, the equation is no longer an equation about x and y — it is just an arithmetic statement about the leftover constants. And that statement is the only thing you should look at.

The two-case rule

Once both variables vanish, you have 0 = c. There are exactly two cases.

Notice what is not on this list: "x = 0, y = 0." That option does not exist. Nowhere in the elimination did you derive x = 0 or y = 0 — you derived an equation where x and y no longer appear.

After elimination, both $x$ and $y$ vanish. You are left with $0 = c$. Look at the constant $c$. $c = 0$ $c \neq 0$ $0 = 0$ is always true. Infinite solutions — lines coincide. $0 = c$ is never true. No solution — lines are parallel. NEVER write "$x = 0$, $y = 0$". The variables vanished — they were not solved.
Once both variables disappear, the only thing that matters is the leftover constant. There is no third path that leads to $x = y = 0$.

Three worked examples

The classic trap (no solution)

Solve

2x + 3y = 5 \qquad \text{and} \qquad 4x + 6y = 7.

Multiply the first equation by 2 to match the x coefficient of the second:

4x + 6y = 10.

Subtract the second equation 4x + 6y = 7 from this:

(4x + 6y) - (4x + 6y) = 10 - 7
0x + 0y = 3
0 = 3.

Wrong conclusion (the trap): "x = 0 and y = 0." Then check: 2(0) + 3(0) = 0 \neq 5. The "answer" doesn't even satisfy the first equation. Two marks gone.

Right conclusion: 0 = 3 is a contradiction — false for every (x, y). The system has no solution. Why: the ratios \tfrac{a_1}{a_2} = \tfrac{2}{4} = \tfrac{1}{2}, \tfrac{b_1}{b_2} = \tfrac{3}{6} = \tfrac{1}{2}, \tfrac{c_1}{c_2} = \tfrac{5}{7}. Two ratios match, the third does not — the lines are parallel.

A second contradiction

Solve

x - y = 1 \qquad \text{and} \qquad 2x - 2y = 3.

Multiply the first by 2:

2x - 2y = 2.

Subtract the second:

(2x - 2y) - (2x - 2y) = 2 - 3
0 = -1.

Again 0 = c with c \neq 0. No solution. Geometrically, x - y = 1 and 2x - 2y = 3 have the same slope (both rearrange to y = x - 1 and y = x - \tfrac{3}{2}) but different intercepts — parallel lines that never cross.

The mistake to avoid: writing "x = 0, y = -1" or "x = 0, y = 0" because the leftover looked like a value. It is not a value. It is a verdict on the system itself.

The contrast (infinite solutions)

Solve

x + y = 5 \qquad \text{and} \qquad 2x + 2y = 10.

Multiply the first by 2:

2x + 2y = 10.

Subtract the second:

(2x + 2y) - (2x + 2y) = 10 - 10
0 = 0.

Now the leftover is 0 = 0, which is always true. Every pair (x, y) that satisfies x + y = 5 also satisfies 2x + 2y = 10 (the second equation is just twice the first). Infinite solutions — the two equations describe the same line.

The mistake to avoid here is the opposite error: writing "no solution" because both variables vanished. They vanished, yes, but the leftover is true, not false. Why this case feels strange: you didn't get a single point because there isn't a single point — there is a whole line of them. Any (x, 5-x) works: (0,5), (1,4), (2,3), \ldots all satisfy both equations.

Why students fabricate "x = 0, y = 0"

The mistake is psychological, not arithmetical. Here is the script that runs in the student's head:

  1. The equation now reads 0x + 0y = 5.
  2. "The coefficients of x and y are zero. So x and y must be zero."
  3. Write "x = 0, y = 0." Move on.

Step 2 is the slip. The coefficient being zero means x's contribution to this particular equation is zero — not that x's value is zero. Why "vanish" feels like "equal zero": both end with the symbol x disappearing from your page. But disappearing from the page can happen for two completely different reasons. (a) You solved for x and the answer happened to be 0 — then x is known, with value 0. (b) The coefficient in front of x became 0 during elimination — then x is unconstrained by this equation, free to be anything. Case (b) is what happened here. The x on your page is gone but the x in the original problem is still unknown.

There is also a second, sneakier reason students write x = y = 0: the leftover 0 = 5 feels wrong, and the brain wants to "fix" it. Writing x = 0, y = 0 feels like producing an answer, even a wrong one, instead of the seemingly defeatist "no solution". But "no solution" is the answer — the question was "what pair (x,y) satisfies both?" and the honest reply is "no such pair exists." That is a complete, correct, full-marks response.

How to never get caught again

A two-line check, before you write any final answer.

  1. After elimination, did both variables disappear? If only one disappeared, you are in the normal case — solve for the other variable as usual. If both disappeared, go to step 2.
  2. Look at what remains. Is it 0 = 0? Then infinite solutions. Is it 0 = c for some nonzero c? Then no solution. There is no third option, and "x = 0, y = 0" is never the answer.

For an even quicker pre-check, before doing elimination at all: compute the three ratios \tfrac{a_1}{a_2}, \tfrac{b_1}{b_2}, \tfrac{c_1}{c_2} from a_1 x + b_1 y = c_1 and a_2 x + b_2 y = c_2. If all three are equal — infinite solutions. If the first two are equal but the third differs — no solution. If the first two differ — unique solution. The full theorem and its geometric meaning live in the parent article, systems of linear equations, and in the related article on linear equations in two variables.

The board exam version of this trap is almost always question 1 of the systems section. Spend the extra ten seconds. Two marks are not worth a wrong word.

References

  1. NCERT Mathematics Class 10, Chapter 3 — Pair of Linear Equations in Two Variables — the official chapter that introduces the consistency cases and explicitly warns about misreading 0 = c.
  2. Khan Academy — Number of solutions to a system of equations — a video walkthrough of the same trap with similar examples.
  3. Wikipedia — Consistent and inconsistent equations — formal definitions of consistent, inconsistent, and dependent systems.
  4. Paul's Online Math Notes — Linear Systems with Two Variables — clear treatment of the elimination method including the contradiction and identity cases.