Solve for the Missing Numerator in (?)/(3x-1) = 3

Let's examine the problem, first we'll write the expression on the right side as a fraction (using the fact that dividing a number by 1 doesn't change its value):

$\frac{?}{3x-1}=3 \\ \downarrow\\ \frac{?}{3x-1}=\frac{3}{1}$

Now let's think logically, and remember the fraction reduction operation,

For the fraction on the left side to be reducible, we want all terms in its denominator to have a common factor, therefore first we'll check if it can be factored, however we identify that it cannot be factored using finding a common factor- because there is no common factor between its two terms, moreover - it cannot be factored in any other way (trinomial, shortened multiplication formulas),

We also notice that in the right side - the numerator has the number 3, and the denominator has the number 1, therefore we can conclude that the expression in the denominator on the left side needs to reduce completely, since it doesn't appear on the right side at all, and therefore the only choice left for the missing expression in the numerator on the left side is the expression:

$3(3x-1)$since the binomial in the denominator will reduce with the expression in parentheses, and after reduction the number 3 will remain,

Let's verify that with this choice we indeed get the expression on the right side: (reduction sign)

$\frac{?}{3x-1}=\frac{3}{1} \\ \downarrow\\ \frac{\textcolor{red}{3(3x-1)}}{3x-1}\stackrel{?}{= }\frac{3}{1} \\ \downarrow\\ \boxed{\frac{3}{1} \stackrel{!}{= }\frac{3}{1} }$

therefore choosing the expression:

$3(3x-1)$is indeed correct.

We'll use the distributive property to open the parentheses, and identify that the correct answer is answer B.