very simple linear algebra puzzle

Since I intend to study operations research in my free time, I need to revisit a lot of the linear algebra material I had a better command of a while ago. That is essential to understanding the simplex and related algorithms. In Jim Hefferon's Linear Algebra, there is a really cute puzzle close to the beginning. It's so easy it's almost not worth mentioning, but I still had fun with it for a few minutes. Here it is:

Three truck drivers went into a roadside cafe. One truck driver purchased four sandwiches, a cup of coffee, and ten doughnuts for $8.45. Another driver purchased three sandwiches, a cup of coffee, and seven doughnuts for $6.30. What did the third truck driver pay for a sandwich, a cup of coffee, and a doughnut?

My solution was a bit different from the book's. The first two purchases can be represented as the linear equations and , respectively. They can be reduced to the linear combination . (I never thought of donuts as parameters. I guess I'm in for more surprises when I study topology.)

Anyway, with that knowledge, you can substitute into the equation with the unknown right-hand value (i.e., the cost of the third purchase) and so arrive at . The terms cancel out and so you are left with a constant: two dollars for the third trucker's purchase. Not a bad deal at all, to be honest. Notice that, even with the added constant, there is still a free variable in this system. As such, the menu items can assume infinitely many combinations of prices s.t. they are non-negative and very economical.