Linear equations in one variable SAT question

The correct answer is $\frac{16}{17}$. It's given that the equation $3(kx+13)=\frac{48}{17}x+36$ has no solution. A linear equation in the form $ax+b=cx+d$, where $a$, $b$, $c$, and $d$ are constants, has no solution only when the coefficients of $x$ on each side of the equation are equal and the constant terms aren't equal. Dividing both sides of the given equation by $3$ yields $kx+13=\frac{48}{51}x+\frac{36}{3}$, or $kx+13=\frac{16}{17}x+12$. Since the coefficients of $x$ on each side of the equation must be equal, it follows that the value of $k$ is $\frac{16}{17}$. Note that 16/17, .9411, .9412, and 0.941 are examples of ways to enter a correct answer.