Answer: B

The system has no solution when the lines are parallel and distinct. Parallel lines need proportional coefficients: $\frac{k}{5} = \frac{3}{k + 2}$, so $k(k + 2) = 15$, $k^2 + 2k - 15 = 0$, and $(k + 5)(k - 3) = 0$. So $k = 3$ or $k = -5$. For $k = 3$ the equations are $3x + 3y = 4$ and $5x + 5y = 7$, which simplify to $x + y = \frac{4}{3}$ and $x + y = \frac{7}{5}$, two parallel lines that never meet. Of the choices, only $3$ works.

Choices A and C come from factoring $k^2 + 2k - 15$ with the signs reversed, as $(k + 3)(k - 5)$.

Why the other choices are wrong

• A. This comes from factoring k² + 2k - 15 as (k + 3)(k - 5), which has the signs reversed. The roots are 3 and -5.
• C. With k = 5, the slopes are -5/3 and -5/7, so the lines intersect at one point.
• D. 15 is the product k(k + 2) that the condition requires, not a value of k.