SAT Systems of two linear equations in two variables Practice Question 16 (Hard)

Question bank / Systems of two linear equations in two variables

Math · AlgebraHard

$3x - 5y = 7$
$kx + 10y = 4$
In the system of equations above, $k$ is a constant. For what value of $k$ does the system have no solution?

$-6$

$-\frac{3}{2}$

$\frac{3}{2}$

$6$

Show answer and explanation

Answer: A

No solution means the lines are parallel but distinct. Multiplying the first equation by $-2$ gives $-6x + 10y = -14$ . Matching $y$ -coefficients requires $k = -6$ . Since $-14 \ne 4$ , the lines are distinct, so there is no solution.

Choice D ignores the negative sign: multiplying by $-2$ changes the sign of the $x$ -coefficient as well.

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