For what value of k, the system of equations kx - 15y + 7 = 0 and 7x - 21y - 19 = 0 has NO solution?

Correct — C. a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 have no solution when a₁⁄a₂ = b₁⁄b₂ ≠ c₁⁄c₂ — parallel lines that are not the same line.

Ratio of the y-coefficients: −15⁄−21 = 5⁄7
Match it in x: k⁄7 = 5⁄7 → k = 5

Constant check: 7⁄(−19) is not 5⁄7, so the two lines stay distinct and never meet.
k = 5 → option (c).

Why the others are wrong

• (a)
  12 — 12 makes the x-ratio 12⁄7 against a y-ratio of 5⁄7. Unequal ratios mean different slopes, so the lines cross once and the system has a unique solution.
• (b)
  25 — 25 makes the x-ratio 25⁄7, five times the y-ratio 5⁄7. Again unequal, so the lines intersect at a single point instead of never meeting.
• (d)
  6 — 6 gives 6⁄7, which is close to 5⁄7 but not equal to it. −15⁄−21 cancels by 3 to 5⁄7, so 6 leaves the slopes different and a solution exists.

A pair of linear equations in two variables has exactly one of three fates, and the coefficient ratios decide which.

a₁⁄a₂ ≠ b₁⁄b₂ — different slopes, the lines cross once: a unique solution.
a₁⁄a₂ = b₁⁄b₂ ≠ c₁⁄c₂ — same slope, different intercept: parallel, no solution.
a₁⁄a₂ = b₁⁄b₂ = c₁⁄c₂ — one line written twice: infinitely many solutions.

'No solution' is the middle case, so you match the x- and y-ratios and then confirm the constants disagree.

Key facts

• No solution requires a₁⁄a₂ = b₁⁄b₂ ≠ c₁⁄c₂, which describes parallel and distinct lines.
• −15⁄−21 reduces to 5⁄7: the two minus signs cancel and 3 divides both numbers.
• With k = 5 the equations are 5x − 15y + 7 = 0 and 7x − 21y − 19 = 0, both of slope 1⁄3.

Common traps

• Dropping a minus sign and taking the y-ratio as −5⁄7.
• Skipping the constant check, which is what separates 'no solution' from 'infinitely many'.
• Comparing coefficients before moving every constant to the same side of the equation.

No directly related past PYQ was found.