SAT Linear inequalities in one or two variables Practice Question 16 (Hard)

Question bank / Linear inequalities in one or two variables

Math · AlgebraHard

$y \ge ax + 3$
$y \le 4x - 1$
In the system of inequalities above, $a$ is a constant. For which of the following values of $a$ does the system have no solution?

$-4$

$\frac{1}{4}$

$3$

$4$

Show answer and explanation

Answer: D

If $a \ne 4$ , the boundary lines cross, and points near the intersection on the correct sides satisfy both inequalities, so solutions exist. If $a = 4$ , the system requires $4x + 3 \le y \le 4x - 1$ , which is impossible because $4x + 3 > 4x - 1$ for every $x$ . So $a = 4$ .

Choice A makes the slopes opposites, but those lines still intersect, so the system has solutions.

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