JEE Main 2025 · 22 January · evening · Single correct · +4 / −1

If the system of linear equations :
x+y+2 z=6 ; 2 x+3 y+az=a+1 ; -x-3 y+b z=2 b
where a, b ∈ R, has infinitely many solutions, then 7 a+3 b is equal to :

  1. (a) 12
  2. (b) 9
  3. (c) 22
  4. (d) 16
Answer and solution

Answer (d) 16

Solution

Maths sir · 4 steps

Eliminate until one equation is left in z alone, then make both sides of it zero.

  1. Clear x. Do R₂ - 2R₁ and R₃ + R₁. This gives y + (a-4)z = a - 11 and -2y + (b+2)z = 2b + 6.
  2. Clear y. Add twice the first of these to the second. The y terms cancel and one equation in z alone is left.

    (2a + b - 6) z = 2a + 2b - 16

  3. Infinitely many. Every value of z must satisfy this equation, so both sides must be zero: 2a + b = 6 and a + b = 8. Subtracting gives a = -2, and then b = 10.
  4. So. 7a + 3b = -14 + 30 = 16, which is option (d).

The idea to keep. For infinitely many solutions, setting Δ = 0 is not enough. Reduce the system to 0 · z = 0: the coefficient and the right side must both vanish.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 190 times in JEE Main since 2020. This question is from 2025.