Q.15
The solution to the following set of equations is
2x + 3y = 4
4x + 6y = 0
- (A) x = 0, y = 0
- (B) x = 2, y = 0
- (C) 4x = -6y
- (D) No solution
Solving Linear Equations: 2x + 3y = 4 and 4x + 6y = 0 – No Solution Explained
The system of equations 2x + 3y = 4 and 4x + 6y = 0 has no solution because the second equation is exactly twice the first but with a contradictory constant term.
🧮 System Analysis
Observe that the second equation (4x + 6y = 0) equals twice the first (2*(2x + 3y) = 2*4 = 8, but right side is 0 ≠ 8).
For ax + by + c₁ = 0 and a’x + b’y + c₂ = 0,
No solution exists if a/a’ = b/b’ ≠ c/c’
No solution exists if a/a’ = b/b’ ≠ c/c’
Here: a₁=2, b₁=3, c₁=-4 | a₂=4, b₂=6, c₂=0
Ratios: 2/4 = 3/6 = 1/2, but -4/0 undefined
(Inconsistent since 1/2 * 0 ≠ 4)
✅ Correct Answer
(D) No solution is correct
The lines are parallel (same slope -2/3) and never intersect.
📋 Option Breakdown
(A) x = 0, y = 0: Fails first equation (0 + 0 = 0 ≠ 4)
(B) x = 2, y = 0: Satisfies first (4 + 0 = 4) but not second (8 + 0 = 8 ≠ 0)
(C) 4x = -6y: Rearranges second equation exactly, but ignores first; represents one line only, not system solution
(D) No solution: Matches inconsistent parallel lines condition
📖 Key Takeaway
When coefficients of both equations are proportional but constants are not, the system represents parallel lines with no intersection point – hence no solution exists.
