Algebra practice question

Answer: C. 4

Infinitely many solutions means the two equations are equivalent. Since 12 = 2(6), the first equation must be exactly 2 times the second: kx + 6y = 12 must match 4x + 2(k − 1)y = 12. This requires k = 4 and 6 = 2(k − 1), i.e., k = 4 — both conditions agree, so k = 4. (Check: 4x + 6y = 12 and 2x + 3y = 6 are the same line.) Choice A comes from using only the coefficient proportion k/2 = 6/(k − 1), giving k² − k − 12 = 0 and k = 4 or k = −3, and failing to test the constants: k = −3 gives x − 2y = −4 and x − 2y = 3, which are parallel distinct lines with NO solution. Choice B comes from matching the x-coefficients directly (k = 2) instead of accounting for the factor of 2 between the equations. Choice D comes from matching the y-coefficients directly (k − 1 = 6).

Why the other answers are wrong

A. −3

You solved k/2 = 6/(k − 1) and took the other root; k = −3 gives x − 2y = −4 and x − 2y = 3, which are parallel lines with no solution, so it fails the constant-term check.

B. 2

You matched the x-coefficients directly (k = 2), but the constants are 12 and 6, so the first equation must be 2 times the second: k = 2(2) = 4.

D. 7

You matched the y-coefficients directly by setting k − 1 = 6, ignoring the factor of 2 between the equations; the correct condition is 6 = 2(k − 1).

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