SAT Linear equations in two variables Practice Question 20 (Hard)

Question bank / Linear equations in two variables

Math · AlgebraHard

In the $xy$ -plane, the graph of $kx - 3y = 6$ , where $k$ is a constant, does not pass through Quadrant II. Which of the following could be the value of $k$ ?

$-6$

$-1$

$-\frac{1}{3}$

$2$

Show answer and explanation

Answer: D

Solve for $y$ : $y = \frac{k}{3}x - 2$ . The $y$ -intercept is $-2$ . If the slope is negative, the line rises to the left and enters Quadrant II (where $x < 0$ and $y > 0$ ). So the slope must be nonnegative, meaning $k \ge 0$ . Only $k = 2$ works.

Choices A, B, and C all give negative slopes; with a negative $y$ -intercept, such a line climbs into Quadrant II as $x$ decreases.

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