For which of the following values of a and b do the given equations, have NO solution? x - ay = 2 - a (1 - a)x + 6y = a + b

Correct — D. A pair of linear equations has no solution when the x and y coefficients are in the same ratio but the constants are not: a₁⁄a₂ = b₁⁄b₂ ≠ c₁⁄c₂.

x − ay = 2 − a gives coefficients 1 and −a, constant 2 − a
(1 − a)x + 6y = a + b gives coefficients 1 − a and 6, constant a + b

Equate the coefficient ratios:
1⁄(1 − a) = −a⁄6
6 = −a(1 − a) = a² − a
a² − a − 6 = 0 → (a − 3)(a + 2) = 0

The roots are a = 3 and a = −2, and a = 3 is the one the options offer.

Now force the constants to disagree at a = 3:
coefficient ratio = 1⁄(1 − 3) = −1⁄2
constant ratio = (2 − 3)⁄(3 + b) = −1⁄(3 + b)
They would match only if 3 + b = 2, so b ≠ −1 → option (d)

Why the others are wrong

• (a)
  a = 3,b ≠ 1 — The a = 3 half is right, but the barred value is wrong. At a = 3 the two ratios coincide when 3 + b = 2, that is at b = −1 — so barring b = 1 rules out nothing.
• (b)
  a = -3,b ≠ 1 — a = −3 fails the coefficient test: 1⁄(1 + 3) = 1⁄4 while −a⁄6 = 3⁄6 = 1⁄2. Unequal coefficient ratios mean the lines cross, so the system has a solution.
• (c)
  a = -3,b ≠ -1 — a = −3 again, so the same coefficient ratios 1⁄4 and 1⁄2 — not equal, and the pair meets at exactly one point. The condition on b is never reached, whatever it says.

Three cases, one comparison. For a₁x + b₁y = c₁ and a₂x + b₂y = c₂, compare a₁⁄a₂, b₁⁄b₂ and c₁⁄c₂.

Unequal first two ratios means the lines are not parallel and there is exactly one solution. All three equal means the two equations are the same line and there are infinitely many. First two equal but the third different means parallel and distinct — no solution.

Here the parameter sits inside a coefficient, so the first condition becomes a quadratic in a; the second condition then becomes an inequality in b.

Key facts

• One solution when a₁⁄a₂ ≠ b₁⁄b₂.
• No solution when a₁⁄a₂ = b₁⁄b₂ ≠ c₁⁄c₂.
• Infinitely many solutions when a₁⁄a₂ = b₁⁄b₂ = c₁⁄c₂.
• Here the coefficient condition reduces to a² − a − 6 = 0, with roots 3 and −2.

Common traps

• Treating 2 − a and a + b as coefficients rather than as the constants.
• Solving for a and stopping, without asking what b has to avoid.
• Assuming the quadratic has only one root and never testing a = −2 against the options.

No directly related past PYQ was found.