In short
You added (or subtracted) two linear equations and both x and y disappeared at the same time. That is not bad luck — it is a signal. The two equations were not independent: one was a constant multiple of the other in its x- and y-coefficients. Now look at the number left on the right-hand side. If you see 0 = 0, the two equations describe the same line — infinite solutions. If you see 0 = c with c \neq 0, the two equations describe parallel lines — no solution. In both cases the system is under-determined: two equations dressed up as two, but really only carrying the information of one (or one and a contradiction).
You are sitting with a pair of linear equations, planning to use elimination. You scale one equation, you subtract, and instead of the x term cancelling and a clean y falling out — both variables walk off the page together. The line on your paper now reads something like 0 = 0 or 0 = -1.
A sibling article handles the panic move that happens next — students writing "x = 0, y = 0" and losing two marks. This article handles the deeper question: why did both variables vanish in the first place, and what is the system trying to tell you when that happens?
What "both vanish" really tells you
A pair of linear equations is supposed to give you two independent pieces of information about (x, y). Independent means: the second equation tells you something the first one didn't. When the two equations are independent, you can pin (x, y) down to a single point — the lines cross.
But what if your second equation is just the first one in disguise? Multiply 2x + 3y = 5 on both sides by 2 and you get 4x + 6y = 10. These look like two equations, they take up two lines on your page — but they carry exactly one piece of information. The second one is redundant.
When you eliminate using a redundant pair, both variables vanish at once. That is the algebraic fingerprint of redundancy.
Why "linear dependence" means exactly this: two equations a_1 x + b_1 y = c_1 and a_2 x + b_2 y = c_2 are linearly dependent in their coefficients if there is some non-zero number k such that a_2 = k\, a_1 and b_2 = k\, b_1. The whole left-hand side of equation 2 is just k times the left-hand side of equation 1. So when you scale equation 1 by k and subtract equation 2, both sides of the left cancel completely. The variables don't disappear by accident — they disappear because the left-hand sides were always copies of each other.
There are two kinds of redundancy, and they look identical on the left-hand side. The right-hand side is what separates them:
- Honest redundancy (c_2 = k\, c_1 also). The full equation 2 is k times equation 1, right-hand side and all. Same line, infinite solutions.
- Dishonest redundancy (c_2 \neq k\, c_1). The left sides are copies, but the right sides disagree. Parallel lines, no solution — the equations are now contradictory.
So "both variables vanish" by itself never tells you the full story. It only tells you the left sides were dependent. You have to glance at the constant to learn which case you are in.
The diagnostic procedure
Here is the recipe, kept short enough to use in a board exam.
Write your two equations in standard form:
Step 1. Pick a multiplier k that makes the x-coefficient of equation 2 match equation 1 — that is, k = a_1 / a_2 (or scale equation 1 by a_2 and equation 2 by a_1 if you prefer integer arithmetic). Compute k \cdot (\text{equation 2}) and subtract from equation 1.
Step 2. Look at the result.
- If only the x term cancels and a y term survives — you are in the normal case, solve for y. Stop, you don't need this article.
- If both the x and y terms cancel — you have just discovered that \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}. The left-hand sides are linearly dependent. Continue to step 3.
Step 3. Look at the constant left over. Compare c_1 to k \cdot c_2.
- If c_1 = k\, c_2 (equivalently \dfrac{c_1}{c_2} = \dfrac{a_1}{a_2}), you are left with 0 = 0. Same line. Infinite solutions.
- If c_1 \neq k\, c_2, you are left with 0 = c for some non-zero c. Parallel lines. No solution.
That's it. The CBSE Class 10 "ratio test" — \tfrac{a_1}{a_2} = \tfrac{b_1}{b_2} = \tfrac{c_1}{c_2} versus \tfrac{a_1}{a_2} = \tfrac{b_1}{b_2} \neq \tfrac{c_1}{c_2} — is just a fast way to detect this without doing the elimination. The ratios are checking the same dependence we just walked through.
Honest redundancy — same line, infinite solutions
Solve
Multiply equation 1 by 2 to match the x-coefficient of equation 2:
Subtract equation 2 from this:
Both variables gone. Constant on the right is 0. Same line — infinitely many solutions. Why: equation 2 is literally 2 \times equation 1, right-hand side included. The two equations are not telling you two different things; they are telling you one thing twice. One equation in two unknowns gives a whole line of solutions, not a point.
The ratio check confirms it: \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}{10} = \tfrac{1}{2}. All three ratios equal — the CBSE Class 10 fingerprint of coincident lines.
Contradictory pair — parallel lines, no solution
Now change just the right-hand side of equation 2:
Same multiplier — multiply equation 1 by 2:
Subtract equation 2:
Both variables gone again, but the constant is now non-zero. Parallel lines — no solution. Why: the left-hand side of equation 2 is still 2 \times the left-hand side of equation 1, so the lines have the same slope. But the right-hand side has been bumped — they now sit at different heights. Same direction, different intercept = parallel and never meet.
Ratio check: \tfrac{a_1}{a_2} = \tfrac{1}{2}, \tfrac{b_1}{b_2} = \tfrac{1}{2}, \tfrac{c_1}{c_2} = \tfrac{5}{11}. The first two match, the third does not — parallel lines, the inconsistent case.
Verifying the geometry of the "same line" case
Take the first example again — \{2x + 3y = 5,\; 4x + 6y = 10\}. The claim was that every solution of equation 1 is also a solution of equation 2. Let's check.
A quick rearrangement of equation 1 gives y = \tfrac{5 - 2x}{3}, so any x produces a valid solution. Try x = 1: y = \tfrac{5 - 2}{3} = 1. Point (1, 1).
Plug (1, 1) into equation 2: 4(1) + 6(1) = 4 + 6 = 10. Tick.
Try a different one — x = 4: y = \tfrac{5 - 8}{3} = -1. Point (4, -1).
Plug into equation 2: 4(4) + 6(-1) = 16 - 6 = 10. Tick.
Try a fractional one, x = -2: y = \tfrac{5 + 4}{3} = 3. Point (-2, 3).
Equation 2: 4(-2) + 6(3) = -8 + 18 = 10. Tick.
Every point on equation 1 sits on equation 2 because they are the same line. Picking a "solution" gives you nothing more than picking a point on that one line. The pair (x, y) is not pinned — it slides freely along the line.
Why it matters
Here is the conceptual punchline. Two equations in two unknowns should be enough to pin down a unique (x, y) — that is the whole reason you set up two equations in the first place. When both variables vanish in a single elimination step, you have just discovered that you don't actually have two equations. You have one equation, written twice (the 0 = 0 case), or one equation followed by a contradiction (the 0 = c case).
In the language used later in linear algebra, the system is under-determined — the rank of the coefficient matrix is 1 instead of 2. Two rows, one row's worth of information. You cannot solve for two unknowns with one equation; geometry agrees by handing you a whole line of solutions or none at all.
Practically, this matters in three ways:
- It is a feature, not a bug. The vanishing tells you something true and important — that the data you were handed is redundant or contradictory. Treat the leftover 0 = c as a verdict on the system, not a value for the variables.
- In word problems — a Diwali-sweets shopkeeper who tells you "I sold 3 boxes of laddoos and 2 boxes of barfi for ₹500" and then "I sold 6 boxes of laddoos and 4 boxes of barfi for ₹1000" has given you the same information twice. You need a third, genuinely different statement to find unique prices. If instead the second receipt said ₹1100, the receipts contradict each other — somebody made an error.
- The CBSE ratio test is a shortcut to this diagnosis. Computing \tfrac{a_1}{a_2}, \tfrac{b_1}{b_2}, \tfrac{c_1}{c_2} before doing any elimination tells you which case you'll land in. All three equal → infinite solutions. First two equal but third different → no solution. First two different → unique solution and the elimination will work cleanly.
So the next time both variables vanish in one move, don't panic. Glance at the constant. The system has just told you something about itself — and now you know how to read what it said.
References
- NCERT Mathematics Class 10, Chapter 3 — Pair of Linear Equations in Two Variables — the chapter where the ratio test for consistent, inconsistent, and dependent systems is introduced.
- Wikipedia — System of linear equations: Independence — formal treatment of independent versus dependent equations and what redundancy means.
- Wikipedia — Linear independence — the general definition that underlies "the equations are not independent".
- Khan Academy — Solutions to systems of equations: consistent vs. inconsistent — worked examples of all three cases including the both-variables-vanish situation.
- Paul's Online Math Notes — Linear Systems with Two Variables — clear walkthrough of elimination including the contradiction and identity outcomes.