Can a 3-variable linear system have exactly two solutions?

In short

No. A linear system — in any number of variables — has exactly 0, 1, or infinitely many solutions. Never two. Never five. Never any other finite count. This rule is called the linear trichotomy, and the reason is short and beautiful: if two different solutions exist, then every point on the straight line joining them is also a solution. So the moment you have two, you have infinitely many. There is no middle ground between "one" and "infinity" for linear systems. Finite-but-greater-than-one solution counts only show up when at least one of the equations is non-linear (a quadratic, a circle, a cubic, etc.).

You are working through a 3-variable system in your CBSE Class 12 matrices chapter, and a question pops into your head: "I've seen one solution. I've seen no solution. I've seen infinite solutions. But what if a system had exactly two solutions? Like the corner of the room and one other random point?" It is a great question — and the answer is a clean no, for a reason that, once you see it, you can never un-see.

This satellite expands a single point from the parent article on systems of linear equations.

The trichotomy: 0, 1, or infinity

Here is the rule, stated as plainly as possible. Take any system of linear equations — two unknowns, three unknowns, or a thousand unknowns. The number of solutions is always one of exactly three options:

The three doors a linear system is allowed to walk through. Every other count is forbidden.

Whether the unknowns are (x, y) or (x, y, z) or (x_1, x_2, \ldots, x_{100}), the answer always lives in one of those three boxes. The only thing that changes with more variables is what "infinitely many" looks like geometrically — a line, a plane, a higher-dimensional flat — but the count is still infinite.

Why "two" collapses into "infinity"

This is the heart of the satellite. Suppose, just for argument, that a 3-variable linear system did have exactly two solutions. Call them (p_1, q_1, r_1) and (p_2, q_2, r_2). We will show this immediately forces a third solution to exist — and then a fourth, and a fifth, and a continuum.

Pick any number t between 0 and 1 and form the linear combination

\big(t \cdot p_1 + (1-t) \cdot p_2,\ \ t \cdot q_1 + (1-t) \cdot q_2,\ \ t \cdot r_1 + (1-t) \cdot r_2\big).

Geometrically, as t slides from 0 to 1, this point slides along the straight line segment joining the two original solutions. We claim every such point also satisfies the system.

Take any one equation in the system, say ax + by + cz = d. Plug in the new point:

a\big(t p_1 + (1-t) p_2\big) + b\big(t q_1 + (1-t) q_2\big) + c\big(t r_1 + (1-t) r_2\big)

Group the t pieces and the (1-t) pieces together:

= t\big(a p_1 + b q_1 + c r_1\big) + (1-t)\big(a p_2 + b q_2 + c r_2\big)

Why this grouping works: multiplication distributes over addition, and the coefficients a, b, c are constants — they pull through the t and (1-t) factors freely. This is exactly what the word "linear" buys you.

But (p_1, q_1, r_1) is a solution, so the first bracket equals d. And (p_2, q_2, r_2) is a solution, so the second bracket also equals d. Hence the whole expression equals

t \cdot d + (1-t) \cdot d = d.

Why we are done: the new point satisfies one equation. But the argument used nothing about which equation — every equation in the system passes the same test. So the new point satisfies them all.

Since t can be any real number (not just between 0 and 1 — extend the line in both directions), you have produced an entire infinite line of solutions. The original "two" was a lie. There were never just two; there were always infinitely many, and you only spotted two of them.

This argument needs nothing about the dimension. Two variables, three variables, fifty — the proof is the same. That is why the trichotomy is a theorem, not just an observation.

The geometric picture in 3D

In three variables, every linear equation is a plane in 3D (see the sibling article on three planes meeting at a point). Three planes can intersect in exactly four ways:

At a single point — one solution. (Floor + back wall + side wall meet at one corner.)
Along a line — infinitely many solutions, parameterised by one variable.
On a whole plane — infinitely many solutions, parameterised by two variables. (All three equations are really the same plane.)
Not at all — no solution. (Two of the planes are parallel and offset, or all three planes form a triangular prism with no common point.)

Notice what is missing from this list: "at exactly two isolated points". You cannot draw three flat sheets in 3D so that they touch in exactly two specks of dust. Flatness forbids it. The moment two solution points coexist, the entire line through them is forced into the intersection.

Worked examples

No solution — not two

The system

\begin{cases} x + y + z = 6 \\ x + y + z = 7 \end{cases}

claims a single quantity equals both 6 and 7. Subtract the first from the second:

0 = 1.

This is a contradiction. The two planes are parallel and offset — like the floor of one storey and the floor of the storey above. They never meet. Solutions: 0. Not two.

Infinite solutions — not two

The system

\begin{cases} x + y = 5 \\ 2x + 2y = 10 \end{cases}

looks like two equations, but the second is just the first multiplied by 2. In 3D (where z is free) both equations describe the same vertical plane. Every point on that plane works: (5, 0, z) for any z, (3, 2, z) for any z, and so on. Solutions: infinite. Not two.

Where "exactly two" actually shows up

Now break linearity. Consider

\begin{cases} x^2 + y^2 = 25 \\ x + y = 5 \end{cases}

The first equation is a circle of radius 5; the second is a straight line. A line and a circle can cross at 0, 1, or 2 points. Substituting y = 5 - x into the circle gives x^2 + (5-x)^2 = 25, which simplifies to 2x^2 - 10x = 0, so x = 0 or x = 5. The two solutions are (0, 5) and (5, 0).

Where did the "two" come from? From the degree-2 term x^2 + y^2. A quadratic equation has up to two roots. Cubic systems can have up to three. Linear systems — degree 1 — can have only zero, one, or infinitely many. The finite-but-greater-than-one count is a fingerprint of non-linearity.

The takeaway

If a problem in your CBSE Class 12 matrices exercise — or in the linear-algebra portion of JEE Mains — ever appears to give you "exactly two solutions" to a linear system, you have made an arithmetic mistake. Recheck. The real answer is one of three: a unique point, no point, or a whole flat infinity of points. Linearity allows nothing in between.

This generalises cleanly. For an n \times n linear system written as the matrix equation A\mathbf{x} = \mathbf{b}, the trichotomy is encoded in the rank of A and the augmented matrix [A \mid \mathbf{b}] — a result you will meet under the name Rouché–Capelli theorem. The theorem is just the formal statement of the geometric story you have just worked through.

References

NCERT, Mathematics Class 12 — Matrices (chapter 3).
Wikipedia, System of linear equations — overview of the consistency trichotomy.
Wikipedia, Rouché–Capelli theorem — the rank criterion behind 0/1/∞.
Gilbert Strang, MIT 18.06 Linear Algebra — Lecture 1 — geometry of linear systems.
Wikipedia, Affine combination — the t \cdot p_1 + (1-t) \cdot p_2 construction.