I have always loved math, and I don’t think there’s any field within math I love more than algebra. I certainly enjoyed calculus, and there are parts of number theory that still fascinate me (Goldbach’s conjecture, an unproven hypothesis that dates back to 1742, more than anything else), but algebra just speaks to me like nothing else in the domains of math or science. So when I saw this week’s FiveThirtyEight Riddler problem, which boils down to solving two equations for two unknowns, I might have dropped everything I was doing and spent about a half an hour solving it – messily, but I think ultimately getting the right answer.

My own personal love of algebra dates back to when I was ten years old, and my junior high school, lacking an honors math class for sixth grade, decided instead to bump me up to the regular eighth grade math course, where I met a wonderful teacher and some awful kids. (I was three years younger than they were, and of course small for my age anyway.) I was told very little about the move but somehow understood that I’d be learning algebra, so I went to the school library and found a book, long out of print now, called *Realm of Algebra*, by an author I’d never heard of at the time but would later grow to know very well through his science fiction writing, Isaac Asimov. I devoured the book, which I credit to Asimov’s ability to make even abstruse concepts clear to readers, in a weekend, and ended up ahead of where I needed to be for the class. Algebra felt to me like another language, as easy to comprehend as English, maybe even more so – like this was my native tongue and everyone had been hiding it from me. I could always “think” in numbers, but algebra gave me an entire framework for it, and everything I learned that year, especially from Asimov’s book, still directs much of how I think about problems today.

It has also made me an easy mark for puzzles and games that revolve around algebraic questions. I often check FiveThirtyEight’s Riddler questions, but I rarely try to solve them – some look too hard or involved, some just don’t grab me. This week’s question, about finding the area of a missing rectangle, hooked me from the start. (I suppose I should disclose that FiveThirtyEight is part of the ESPN network of sites, and thus I am connected to it as well.) Here’s the question in brief: Find the missing area in the picture below, bearing in mind that it is not to scale.

I tweeted the link to the article last night and got a slew of responses from readers, some right, some I don’t think were right, and a few that gave me more insight into the problem – one of which made me realize my first answer was impossible – so I decided to take a few minutes and explain my method, in case it’s useful to anyone or still contains a mistake.

To figure out the area of the lower-left rectangle, you need to know its height and width, so this is a problem of two unknowns, which means you need (at least) two equations containing those unknowns to be able to solve it. To make the math a little less messy, I defined the height (vertical axis) of the target area as 11-x and the width (horizontal axis) as 14-y, meaning that the height of the upper-left area is x and the width of the lower-right area is y. We know the area of the lower right rectangle is 45, and the area of the upper left rectangle is 32, so using the formula for the area of a rectangle we get the following two equations:

(14 – y)x = 14x – xy = 32

y(11 – x) = 11y – xy = 45

You can then solve one equation for one variable in terms of the other, substitute that back into the second equation, and solve for one of the two variables. I chose to solve the first equation for x, yielding:

x(14 – y) = 32

x = 32/(14 – y)

11y – y(32/(14 – y)) = 45

11y – 32y/(14 – y) = 45

11y(14 – y) – 32y = 45(14 – y)

154y – 11y^{2} – 32y = 630 – 45y

–11y^{2} + 167y – 630 = 0

That, my friends, is a quadratic equation, and if you remember your quadratic formula – where you take the two coefficients and the constant and plug them into the formula to get the two possible solutions – you can solve it from here, or you can just plug them into this site and get your two answers for y, which in this case are 7 or 90/11.

It turns out that both answers produce whole-number, positive results for the area of the lower left rectangle. If y is 7, x is 32/7, and the area is 45. If y is 90/11, x is 11/2 (5.5) and the area is 32. A reader pointed out that the first answer is impossible, however, because of the one number I haven’t mentioned yet: the truncated, upper-right rectangle’s area, which is 34. Because the area of the whole shebang has to be less than 154 (11 * 14), since there’s a piece missing at the extreme upper right, then the lower left area has to be less than (154 – 32 – 34 – 45), which is 43. That leaves 32 as the only possible answer. (EDIT: Fixed the second sentence, where I transposed the two answers.)

I think.