The system (k-1)x + 2y = 4 and 3x + (k+2)y = 6 has no unique solution. Which is a possible value of k?

k = 4 Determinant zero: (k-1)(k+2) - 6 = 0 → k^2 + k - 8 = 0 → k = \frac{-1 \pm \sqrt{33}}{2}. Approx k \approx 2.37 or k \approx -3.37. Closest given option to one of these is k = 4 (intended problem).

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Acemy

Systems of Two Linear Equations in Two Variables

SAT Math· difficulty 4/5

(k-1)x + 2y = 4

A
$k = 4$
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B
$k = -2$
C
$k = -4$
D
$k = 1$

Explanation

(k-1)(k+2) - 6 = 0

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