In short
YES. If both equations are homogeneous — meaning the constant on the right side is 0, so both lines pass through the origin — then plugging in x = 0, y = 0 gives 0 = 0 for both. So (0, 0) is always a solution. It is called the trivial solution. The real question now becomes: is (0, 0) the only solution, or are there others? That depends on whether the two lines have different slopes (only origin works) or the same slope (the entire shared line is solutions).
You are looking at a system like this:
Notice the right side. Both are zero. A linear equation ax + by = c becomes "homogeneous" when c = 0, and a homogeneous equation always has the origin sitting on its line — because plugging x = 0, y = 0 gives a(0) + b(0) = 0, which matches the right side perfectly.
So the doubt the question is poking at is real and important: does the origin always solve such a system? Yes. Always. Let us see exactly why, and then ask the more interesting follow-up.
Why (0, 0) always works
Take any homogeneous linear equation:
Substitute x = 0 and y = 0:
The left side equals the right side. So (0, 0) satisfies this equation no matter what a and b are. Why: every term on the left has an x or y multiplied by a coefficient; when x and y are both zero, every term collapses to zero, and the right-hand-side zero matches automatically.
Now suppose you have a system — two such equations:
The argument runs twice. (0, 0) satisfies the first, (0, 0) satisfies the second, so (0, 0) satisfies both. It is a shared solution. That is precisely what "solution of a system" means in the parent article on systems of linear equations.
This shared point (0, 0) is called the trivial solution — "trivial" not because it is unimportant, but because you did not have to compute anything to find it. It comes for free from the structure of the system.
Is it the only solution?
Here is where it gets interesting. Geometrically, two lines that both pass through the origin look like two spokes of a wheel meeting at the hub. There are exactly two cases.
Case 1 — different directions: the two lines cross only at the origin. Anywhere else, they pull apart. So the origin is the unique meeting point, and (0, 0) is the only solution.
Case 2 — same direction: the two lines are actually the same line (or one is a non-zero multiple of the other). Then every point on that line satisfies both equations, and there are infinitely many solutions, including (0, 0) as one of them.
Different slopes — only $(0, 0)$ works
Slope of line 1: rearrange 3y = -2x so y = -\tfrac{2}{3}x. Slope is -\tfrac{2}{3}.
Slope of line 2: rearrange y = 5x. Slope is 5.
Different slopes, so the lines cross at exactly one point. Both pass through the origin (constant terms are zero), so the unique crossing point is (0, 0).
Quick verification by elimination: from the second equation, y = 5x. Substitute into the first: 2x + 3(5x) = 0 \implies 17x = 0 \implies x = 0. Then y = 5(0) = 0. The only solution is (0, 0).
Same slope — infinitely many solutions
Look at the second equation. Divide everything by 2: 2x + 3y = 0. That is literally the first equation. The two equations describe the same line.
So every point on 2x + 3y = 0 satisfies both. Examples: (3, -2), (-3, 2), (6, -4), (0, 0). Infinite solutions. The trivial solution (0, 0) is one of them, but not the only one.
Verifying the difference by substitution
Take a non-trivial point on the line 2x + 3y = 0, say (3, -2). Check: 2(3) + 3(-2) = 6 - 6 = 0\;\checkmark.
Now test it in equation 2 of the first example: 5x - y = 5(3) - (-2) = 15 + 2 = 17 \neq 0. So (3, -2) fails equation 2 — confirming that in the different-slopes case, no non-trivial point works. Only (0, 0) survives both equations.
The classification — homogeneous systems are special
Recall the standard trichotomy from the parent article: a 2 \times 2 linear system has either one solution, no solution, or infinitely many. For a homogeneous system — both right-hand sides zero — that classification simplifies dramatically.
| System type | General system | Homogeneous system |
|---|---|---|
| One solution | possible | possible (only (0, 0) — the trivial solution) |
| No solution | possible (parallel non-coincident) | impossible |
| Infinitely many | possible | possible (entire line of solutions) |
A homogeneous system is always consistent — it always has at least one solution, namely (0, 0). Why no "no-solution" case: parallel-but-distinct lines never share a point, but two homogeneous lines both pass through the origin, so they cannot be parallel-but-distinct — if they are parallel, they must coincide.
So the question for a homogeneous system is never "does a solution exist?" The answer is always yes. The real question is: is the trivial solution the only one, or are there non-trivial solutions too?
That single question — "are there non-trivial solutions?" — is the seed of one of the most important ideas in higher mathematics: linear independence. In Class 12 and JEE Advanced you will meet vectors \vec{v_1}, \vec{v_2}, \ldots and ask whether the equation
has only the trivial solution c_1 = c_2 = \cdots = c_n = 0, or whether some non-trivial choice of coefficients also makes the sum zero. Vectors are called linearly independent exactly when only the trivial solution exists. Linearly dependent when a non-trivial one exists too. The two-line case you just understood is the simplest possible instance of this idea — two homogeneous equations, two unknowns, asking whether the trivial solution is alone.
Cricket analogy: imagine two fielders running along two straight paths from the centre of the pitch. They both start from the same point. If their paths point in different directions, they meet only at the starting point. If their paths happen to point the same way, they share the entire run. That is exactly Case 1 vs Case 2 above — and exactly the geometric content of linear independence.
Quick check — slopes vs the determinant
Two homogeneous lines a_1 x + b_1 y = 0 and a_2 x + b_2 y = 0 have the same slope when \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}, equivalently a_1 b_2 - a_2 b_1 = 0. The expression a_1 b_2 - a_2 b_1 is the determinant of the coefficient matrix.
- Determinant \neq 0 → different slopes → only (0, 0).
- Determinant = 0 → same slope → infinitely many.
For the first example: a_1 b_2 - a_2 b_1 = (2)(-1) - (5)(3) = -2 - 15 = -17 \neq 0, so only (0, 0). For the second: (2)(6) - (4)(3) = 12 - 12 = 0, so infinitely many. The determinant test is the algebraic shortcut for the geometric question. You will meet it formally in the determinants chapter.