How would you go about solving this word problem:
John is twice as old as Jim; the sum of their ages is that of Jerry's, while the ages of the three of them add up to 36. How old is each?
You are likely to start with algebra: let's the ages of John, Jerry and Jim be . Then we have three equations:
Solving the equations is not difficult. But here is a method that depends more on arithmetic than algebra. This is the method that would have been used in the ancient Babylon or Egypt.
Let's make a guess. Since John's age could be expressed in terms of Jerry's, we shall start with guessing Jerry's age. Let's assume that Jerry is . Then John is and Jim is . The sum of their ages comes to . An undershot; the first "wrong" implied in the caption is . Let's then guess again and won't be afraid to get a second guess wrong. Assume that Jerry is years old. This make John () and Jim (). Their ages add up to . Hm, an overshot. The second "wrong" is .
So far we got two guesses: and that led to two wrong calculations: and . Now an ancient text could advise to multiply the first guess by the second wrong; the second guess by the first wrong; and divide the difference by the difference of the wrongs. Formally:
Let's see whether this indeed works: the formula gives
So we get a suggestion: guess Jerry's age as . Doing that gives John's age as and Jim's as , with the sum of , as required.
The method - as described - often goes by its Latin name: Regula Falsi, the Rule of the false position. In modern numerical analysis (which is the branch of mathematics concerned with finding approximate solutions to equations) it goes under the moniker of "The Secant Method". In the next blog I'll tell you why.