For what value(s) of k does the system 2x + ky = 4 and 4x + 6y = k have NO solution?

k = 3 and k \neq 8 Coefficient ratio: \frac{2}{4} = \frac{k}{6} → k = 3. Constant ratio: \frac{4}{k}. No solution requires \frac{4}{k} \neq \frac{1}{2}, i.e., k \neq 8. So k = 3 and k \neq 8.

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Acemy

Systems of Two Linear Equations in Two Variables

SAT Math· difficulty 5/5

k

A
$k = 8$
B
$k = 3$
C
$k = 3$
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D
$k = -3$

Explanation

\frac{2}{4} = \frac{k}{6}

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