JEE Main 2025 · 7 April · evening · Single correct · +4 / −1
Let the system of equations
x + 5y - z = 1
4x + 3y - 3z = 7
24x + y + λz = μ
λ, μ ∈ ℝ, have infinitely many solutions. Then the number of the solutions of this system,
if x, y, z are integers and satisfy 7 ≤ x + y + z ≤ 77, is :
- (a) 4
- (b) 5
- (c) 3
- (d) 6
Answer and solution
Answer (c) 3
Solution
Maths sir · 4 steps
Infinitely many solutions means one free variable: write x, y, z in terms of t and count.
- Third equation. For infinitely many solutions, the third equation must equal a(eq 1) + b(eq 2). Matching x and y gives a + 4b = 24 and 5a + 3b = 1, so b = 7 and a = -4. Then λ = 4 - 21 = -17 and μ = -4 + 49 = 45.
- Solve the first two. Subtract three times eq 1 from eq 2 to remove z: x - 12y = 4. Put y = t. Then x = 4 + 12t, and from eq 1, z = x + 5y - 1 = 3 + 17t.
x + y + z = (4 + 12t) + t + (3 + 17t) = 7 + 30t
- Apply the conditions. x, y, z are all integers exactly when t is an integer. The condition 7 ≤ 7 + 30t ≤ 77 gives 0 ≤ t ≤ 7/3, so t = 0, 1, 2.
- So. The solutions are (4, 0, 3), (16, 1, 20) and (28, 2, 37). That makes 3 solutions, option (c).
The idea to keep. When a 3 by 3 system has infinitely many solutions, write every variable in terms of one parameter t. Each condition on x, y, z then becomes a simple condition on t.
Answer from the NTA final answer key. Solution by PrepWiser.