In short

You ran elimination on a 2x2 system, both variables vanished, and the line on your page reads 0 = 0. Do not write "no solution" — that is the opposite answer. 0 = 0 means the two equations describe the same line and the system has infinitely many solutions. The thinking habit you need is parameterisation: pick one variable (say y) as a free parameter t, then express the other variable in terms of t using either original equation. Your final answer is a set of points — the entire line — written as \{(\,\text{something with } t,\; t) : t \in \mathbb{R}\}. That set, not a single (x, y), is what the system is asking you for.

You're attempting a board-exam question on a Tuesday afternoon. The system reads

x + y = 5, \qquad 2x + 2y = 10.

You scale equation 1 by 2, you subtract from equation 2, and the page tells you 0 = 0. A sibling article on this wiki — Both Variables Vanish When You Add the Equations? Here's What That Means — explains why this happened (linear dependence) and how to tell the 0 = 0 case apart from the 0 = c case. A second sibling — What If I Get 0 = 0 at the End — Do I Have Infinite Solutions in One-Variable? — handles the same outcome but in a one-variable equation, where "infinite solutions" means every real x works.

This article picks up where those two leave off. You've correctly diagnosed that you have a dependent 2x2 system. Now what do you actually write? You cannot write a single point (x, y), because there isn't one — there is a whole line of them. You also cannot write "infinite solutions" and stop, because that doesn't say what the solutions are. The thinking-process habit you need is parameterisation: name a free variable, express the other in terms of it, and hand back the whole solution set in one line.

The habit, in one sentence

When elimination collapses to 0 = 0, let one variable be a free parameter t, and use either original equation to write the other variable in terms of t.

That's it. The whole technique fits in a single instruction. The reason it works is geometric — the solution set is a line in the (x, y)-plane, and a line is one-dimensional, so it can be traced out by a single number t moving along it. Each value of t pins down a different point on the line. As t ranges over all real numbers, you visit every solution exactly once.

Why one parameter is enough: a single linear equation in two variables, like x + y = 5, defines a line. Lines are one-dimensional — you only need one number to specify your position along them. The 0 = 0 verdict tells you the two equations of your system are really one equation in disguise, so the joint solution is also a line, and one parameter t traces it perfectly.

Why parameterisation captures all the solutions

Pick the system x + y = 5. Suppose you set y = t and write x = 5 - t. Now consider any solution (x_0, y_0) of x + y = 5. Set t = y_0. Then the formula gives x = 5 - y_0, which (because x_0 + y_0 = 5) equals x_0. So the solution (x_0, y_0) is exactly the point (5 - t, t) for the choice t = y_0. Why this matters: the parameterisation isn't just "a way to make new solutions" — it is a complete inventory. Every actual solution corresponds to exactly one value of t, and every value of t produces an actual solution. No solution is missed, no fake solution is invented. That two-way correspondence is what makes \{(5 - t, t) : t \in \mathbb{R}\} a valid description of the entire solution set.

In one direction you have to check "does every t give a solution?" — yes, plug (5 - t, t) into x + y: it gives (5 - t) + t = 5. In the other direction, "does every solution come from some t?" — yes, just set t equal to the y-coordinate of that solution. Both directions checked, so the parameterisation is the full story.

The parameter line, drawn

Here is what "y = t, x = 5 - t" looks like geometrically. The solution set is the line x + y = 5 on the plane, and t is a coordinate along that line.

The solution line x + y = 5 with several values of the parameter t marked as pointsA coordinate plane with x and y axes. A diagonal line representing x + y = 5 runs from the upper left to the lower right. Five points are marked along the line at parameter values t equals minus 1, 0, 1, 2, and 5, with their corresponding (x, y) coordinates labelled. x y 0 1 2 3 4 5 6 1 2 3 5 t = −1, (6, −1) t = 0, (5, 0) t = 1, (4, 1) t = 2, (3, 2) x + y = 5 (solution line) parameter t increases this way →
The solution set of $\{x + y = 5,\; 2x + 2y = 10\}$ is the entire line $x + y = 5$. Pick any value of $t$, and the corresponding solution is $(5 - t, t)$. Five sample values of $t$ are marked. As $t$ slides continuously along the real line, the point $(5 - t, t)$ slides continuously along the solution line, hitting every solution exactly once.

Three worked examples

Example 1: $\{x + y = 5,\; 2x + 2y = 10\}$ — the textbook collapse

Step 1. Eliminate. Multiply equation 1 by 2:

2x + 2y = 10.

Subtract equation 2 from this:

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

Why: equation 2 is 2 \times equation 1 right down to the constant. The lines are coincident — the same line written twice.

Step 2. Don't write "no solution". The 0 = 0 verdict means infinitely many solutions. Switch into parameterisation mode.

Step 3. Pick a free parameter. Let y = t. Why y and not x? Either works — the choice is yours. y tends to be slightly cleaner here because the coefficient of y is 1 in equation 1, so isolating x is one subtraction.

Step 4. Use equation 1 (the simpler form) to express x:

x + t = 5 \implies x = 5 - t.

Step 5. Write the solution set:

\boxed{\{(x, y) = (5 - t,\; t) : t \in \mathbb{R}\}.}

Sanity check. Try three values of t:

  • t = 0 \implies (x, y) = (5, 0). Equation 1: 5 + 0 = 5 ✓. Equation 2: 10 + 0 = 10 ✓.
  • t = 2 \implies (x, y) = (3, 2). Equation 1: 3 + 2 = 5 ✓. Equation 2: 6 + 4 = 10 ✓.
  • t = -1 \implies (x, y) = (6, -1). Equation 1: 6 - 1 = 5 ✓. Equation 2: 12 - 2 = 10 ✓.

All three points lie on the line x + y = 5, and the parameterisation hits each one for the appropriate t.

Example 2: $\{2x + 3y = 6,\; 4x + 6y = 12\}$ — fractions appear, technique unchanged

Step 1. Eliminate. Multiply equation 1 by 2:

4x + 6y = 12.

Subtract equation 2: 0 = 0. Same situation — infinitely many solutions.

Step 2. Let y = t. Use equation 1:

2x + 3t = 6 \implies 2x = 6 - 3t \implies x = \frac{6 - 3t}{2}.

Step 3. Solution set:

\boxed{\left\{(x, y) = \left(\frac{6 - 3t}{2},\; t\right) : t \in \mathbb{R}\right\}.}

Why fractions are fine: the parameter t ranges over all reals — including ones that make x a fraction or a decimal. You don't need to pick "nice" values of t. (3, 0), (1.5, 1), (0, 2), (-1.5, 3) are all legitimate solutions, sitting on the same line.

Sanity check. t = 0 gives (3, 0) — check: 2(3) + 3(0) = 6 ✓. t = 2 gives (0, 2) — check: 0 + 6 = 6 ✓. t = -2 gives ((6 + 6)/2, -2) = (6, -2) — check: 12 - 6 = 6 ✓.

A tiny variant on the same idea: if you'd rather avoid the fraction, parameterise the other way. Set x = s and solve equation 1 for y: 3y = 6 - 2s, so y = (6 - 2s)/3. Solution set: \{(s, (6 - 2s)/3) : s \in \mathbb{R}\}. The same line, traced by a different parameter — identical answer, just relabelled.

Example 3: Word-problem context — what the parameter actually *means*

A college student in Bengaluru has ₹100 to spend tomorrow on commuting. Each ride-share trip costs ₹10 and each auto trip costs ₹10. Let x be the number of ride-shares and y the number of autos. Two equally-true sentences:

x + y = 10 \quad (\text{total trips}), \qquad 10x + 10y = 100 \quad (\text{total spend}).

Eliminate — multiply equation 1 by 10 and subtract: 0 = 0. Infinite solutions.

Why? The two sentences are saying the same thing — every trip costs the same, so "I made 10 trips" and "I spent ₹100" are the same constraint dressed differently.

Parameterise: let y = t, then x = 10 - t. Solution set: \{(10 - t,\; t) : t \in \mathbb{R}\}.

But here the real-world meaning of the parameter matters. t is "number of auto trips taken", which has to be a non-negative integer at most 10. So the physically sensible solutions are

t \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\},

giving exactly 11 allowed combinations: (10, 0), (9, 1), \ldots, (0, 10).

Why this matters in CBSE word problems: the math says "any combination", but the context says "must be a whole number, must be non-negative". The free parameter t is the dial you twist; the context is what tells you which positions of the dial are physically meaningful. Always check the context after you parameterise. The mathematical answer is the line; the contextual answer is the subset of the line that respects the real-world constraints.

The shopkeeper-Diwali-receipts example in the sibling article is the same idea — when two receipts encode the same constraint, you need a third independent statement to pin down a unique price pair. Until then, the parameterised set is the most honest answer you can give.

How to write it on your CBSE answer sheet

Markers want to see three things, in order:

  1. The elimination step that produced 0 = 0 — show your work, don't just announce the verdict.
  2. A sentence in words: "Since 0 = 0, the system is dependent and has infinitely many solutions."
  3. The parameterised solution set: "Let y = t where t \in \mathbb{R}; then x = \ldots, so the solution set is \{(\ldots, t) : t \in \mathbb{R}\}."

CBSE Class 10 introduces the diagnostic ("dependent system, infinite solutions") but typically does not insist on a parameterised set — a sentence and one or two sample solutions usually suffice. CBSE Class 11, JEE preparation, and any linear-algebra context tighten the standard: you are expected to hand back the full parameterised solution set, because it is the only compact way to describe an infinite collection of points.

A common trap to avoid

Do not write the answer as "x = \text{anything},\; y = \text{anything}". That is too loose — it suggests x and y are independently free, which is false. Once you fix y, the equation x + y = 5 pins down x = 5 - y. There is one degree of freedom (one parameter t), not two.

The whole point of parameterisation is to capture this: one number t is free to roam, and the other coordinate is locked to t by a formula. Your answer must show that lock — that's what makes it a line, not the whole plane.

References

  1. NCERT Mathematics Class 10, Chapter 3 — Pair of Linear Equations in Two Variables — the chapter where dependent systems and the ratio test are introduced.
  2. NCERT Mathematics Class 11, Chapter 6 — Linear Inequalities and Chapter on systems extends the idea of parameterised solution sets.
  3. Wikipedia — System of linear equations: Solution set — formal definition of the general/parameterised solution.
  4. Wikipedia — Parametric equation — the broader idea of using a parameter to trace a curve or line.
  5. Khan Academy — Solutions to systems of equations: dependent — worked examples and the language of "infinitely many solutions" written as a parameterised set.
  6. Paul's Online Math Notes — Linear Systems with Two Variables — clear walkthrough including the dependent case and how to write the answer.