Find the value of 'k' for which the system of equations 4x+6y=7 and 6x+(k+4)y-21=0 has a unique solution.

Correct — A. Put both equations in the form ax + by = c, then compare coefficient ratios.

4x + 6y = 7 gives a₁ = 4, b₁ = 6
6x + (k + 4)y − 21 = 0 rearranges to 6x + (k + 4)y = 21, so a₂ = 6, b₂ = k + 4

A unique solution needs a₁⁄a₂ ≠ b₁⁄b₂:
4⁄6 ≠ 6⁄(k + 4) → 4(k + 4) ≠ 36 → k + 4 ≠ 9

So k ≠ 5 → option (a)

Why the others are wrong

• (b)
  k ≠ 7 — k ≠ 7 bars the wrong number. At k = 7 the ratio is 6⁄11, not 2⁄3, so k = 7 works — and the condition would still admit k = 5, where no solution exists.
• (c)
  k = 5 — k = 5 names the one value that must be excluded. There 4⁄6 = 6⁄9 = 2⁄3 while 7⁄21 = 1⁄3, so the lines are parallel and the system has no solution.
• (d)
  k = 7 — k = 7 pins k to a single number. Every value except 5 gives a unique solution, so the answer has to be an inequality, not one value.

A pair of linear equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ has three possible fates, and the coefficient ratios name each one.

a₁⁄a₂ ≠ b₁⁄b₂ — the lines cross once: a unique solution.
a₁⁄a₂ = b₁⁄b₂ ≠ c₁⁄c₂ — parallel lines: no solution.
a₁⁄a₂ = b₁⁄b₂ = c₁⁄c₂ — the same line twice: infinitely many solutions.

Only the first is asked here, and it is an inequality — so the answer is a value of k to avoid, not a value to hit.

Key facts

• A two-variable linear system has a unique solution when a₁⁄a₂ ≠ b₁⁄b₂.
• Here 4⁄6 equals 6⁄(k + 4) at k = 5, the single value the condition excludes.
• At k = 5 the ratios read 4⁄6 = 6⁄9 = 2⁄3 against 7⁄21 = 1⁄3, so that case is parallel lines.

Common traps

• Answering k = 5, the value the condition rules out
• Comparing ratios while the second equation still carries − 21 on the left
• Testing the c-ratio as well, which matters only for the parallel and coincident cases

No directly related past PYQ was found.