 Jordanus Numerarius lived when Roman numerals were still in common use. So for example, Book 1, problem 1, separate 10 into two parts with difference 2. So the idea of using a letter to represent an unknown quantity is counterintuitive. Nevertheless, Jordanus did it. For example, in problem 3 of Denumerus Datus, Jordanus solves a familiar problem if a given number is separated into two parts with a given product, the parts can be found. Jordanus' solution, using his verbal algebra, let the given number ABC be separated into AB and C, whose product is D. Let the square root of the given number be E. Let F be 4 times D. Subtract from E, leaving G. The square root of G will be B, the difference between AB and C. Again, Jordanus' purely verbal algebra can be a little hard for modern readers to follow. So in modern notation, we have the following. We have a given number ABC, and we're separating that into two parts, AB and C. So what Jordanus is calling AB is one of our unknowns, we'll call that X, and C is the other unknown, we'll call that Y. And we also have the product, which Jordanus is calling D. So D is XY, the square of the given number, that's E, that's X plus Y squared, and we'll expand that 4 times D, that's 4 XY, and we'll subtract that from E. So E minus F gives us G. And note that G is actually X minus Y squared. So the square root of G will be B, that's the difference between AB and C. So as an example, Jordanus separated 10 into two numbers, whose product was 21. For ease of readability, we'll use our modern number system, instead of trying to write these in Roman numerals, which is what Jordanus did. So the square of our sum of our given number is 100. 4 times the product of the two numbers is 84. The difference is 16. The root is 4. And now we have the problem of separating 10 into two numbers with a difference of 4. But that was actually the first problem in de Numerous Datus, a given number is separated into two numbers with a given difference. So using the solution for problem one of de Numerous Datus, we subtract the difference from the sum to get 6, divide in half to get 3, the smaller number, add the difference to get 7, the larger number, and so the solution to our original problem is 3 and 7.