SAT Systems of two linear equations in two variables Practice Question 19 (Hard)

Question bank / Systems of two linear equations in two variables

Math · AlgebraHard

$kx + 2y = 7$
$8x + ky = 14$
In the system of equations above, $k$ is a constant. For which of the following values of $k$ does the system have no solution?

$-4$

$-2$

$2$

$4$

Show answer and explanation

Answer: A

The lines are parallel when $\frac{k}{8} = \frac{2}{k}$ , so $k^2 = 16$ and $k = 4$ or $k = -4$ . If $k = 4$ , the second equation is twice the first ( $8x + 4y = 14$ ), so there are infinitely many solutions. If $k = -4$ , the first equation times $-2$ is $8x - 4y = -14$ , which conflicts with $8x - 4y = 14$ , so there is no solution.

Choice D makes the lines parallel, but they coincide, giving infinitely many solutions instead of none.

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