I truly hate these problems. There is no ONE correct answer; there are instead an infinite number of correct answers. In this problem, we can get a single answer if the relationship between x and y is stipulated to be strictly linear. I agree with Dr. Peterson that a strictly linear relationship was likely intended by this problem's author, but the student's natural inference is that relationships are likely to be strictly linear, which simply is false.
My objection would not be valid if the problem was stated as: Using the information in the following table, determine whether a linear relationship between x and y is possible and what the missing values will be if that relationship is correct. (Perhaps the "box method" clearly implies this sensible formulation. A very quick Google search did not enlighten me.) Notice that the problem as given does provide the information needed to find whether a linear relation is even possible. Confirmation of that possibility in scientific practice would typically require considerably more than one data point.
Now jumping to hoosie's point. There is an algorithm for finding the equation of a line joining two points. It is not that hard. So teach it. Whatever the box method is, I doubt that anyone intends for it to be the algorithm that is ultimately to be learned by students.
[MATH]\dfrac{y - 11}{x - 6} = \dfrac{11 - 5}{6 - 3} \implies \dfrac{y - 11}{x - 6} = \dfrac{6}{3} \implies [/MATH]
[MATH]3(y - 11) = 6(x - 6) \implies 3y - 33 = 6x - 36 \implies 3y = 6x - 36 + 33 \implies 3y = 6x - 3 \implies y = 2x - 1.[/MATH]
We check the work. 2 * 6 - 1 = 12 - 1 = 11. Checks. 2 * 3 - 1 = 5. Also checks.
Now we try confirming the hypothesis that the relationship may be linear.
[MATH]2 * 11 - 1 = 22 - 1 = 21.[/MATH]
Yes, the relationship may be linear. If that is true, what are the missing numbers in the table?
So much could be taught using tables and hypotheses, but isn't.