Verbal Algebra

There are a variety of problems that cannot be solved with straightforward arithmetic. The trick to solving such problems is looking at the problems in the right way. We can use standard algebra for this purpose--but the methods of standard algebra are not the only way to solve these problems. Long before an efficient algebraic notation was ever developed, techniques were found that helped people solve problems that lie just beyond what we would call straightforward arithmetic.

A number plus a seventh of the number adds up to 17. What is the number? This problem was found in the Rhind Papyrus dated about 1650 BC. To solve the problem pick a convenient number such as 7. Adding a seventh we get 8. Our actual total is 17 which is 17/8 times 8, so our original number needs to be 17/8 times 7 or 119/8. This is also 14 and 7/8 in our notation or 14 and 1/2 and 1/4 and 1/8 as the ancient Egyptians would have written it.

In 19th Century American similar techniques were used to solve son-father-grandfather problems. A father is twice as old as his son, and the grandfather is twice as old as the father. If the sum of the ages is each person. We might use1 in place of the son's age. The father would then be 2, and the grandfather 4 so the sum would be 7. In fact the sum was 20 times this much--so we multiply all these ages by 20 to get 20, 40, and 80 as our answers. This method was called false position. We use a false value in place of the son's age to get a preliminary result which we then correct to get our final answer. In this form the method only works with problems that turn out to be ratio and proportion problems.

A family of 6 spends $33 at the movies for tickets and $3 in popcorn. How much does each ticket cost? For free tickets the cost would be $3, since they would only pay for the popcorn. With $1 tickets the cost would be an extra $6. Since the extra cost of $30 was 5 times this much, the tickets must be $5 each. This is a very simple example of what was called double false position. We first need to check one particular value, and then look at the rate at which the output changes with changing input.

For a more interesting example, suppose we spend $2.00 on a dozen cookies, with small cookies 10 cents and large ones 30 cents. How many of each kind did we buy? All small cookies would cost $1.20. Since a large cookie costs an extra 20 cents and we spent an extra 80 cents, we must have bought 4 large cookies, and the other 8 must be small ones.

Verbal solutions like these are very useful if we want to solve problems in our heads. Solving simple problems in our heads is an excellent way to develop a feel for what we do in a particular area of mathematics. Standard algebra is more rigorous--but that rigor can be quite confusing for beginners. Learning verbal algebra first is like trying to write the final draft of a paper as our first draft. That would not be a good idea for a beginning writer. and trying to start with all the complexities of standard algebra is not a good idea for a beginning algebra student.

But we do want to get to standard algebra. Let's look at how our verbal algebra solutions resemble solutions we might find using standard algebra. A number and a seventh add up to 17. The Egyptians used 7 as a test value for the problem and got an output of 8. Since they wanted 17 which is 17/8 times this much they multiplied their input of 7 by 17/8 to get their answer. We might call our input 7x. Adding one seventh or x we get an output of 8x. Since we want an output of 17, x must be 17/8 and our desired input of 7x must be 7 times 17/8. The solution is essentially the same.

For the cookie problem we might let x be the number of large cookies. The number of small cookies would be 12-x. The total cost would be 30x + 10(12-x) and this must equal 200. But 30x + 10(12-x) = 30x + 120 - 10x. For each large cookie we pay 30 cents, but save the 10 cents we would spend on a small cookies. Our cost is an extra 20 cents for every large cookie we buy, or 20x + 120 which must equal 200. Our extra cost is 20x cents, but it is also 80 cents. Dividing by 20 we get x = 4 for the number of large cookies, and 12 - x = 8 for the number of small ones.

We will want to reach the stage where we can solve a problem expressed algebraically without attaching concrete meaning to every step in the solution process--but attaching concrete meaning in this fashion can help students make the transition from a more concrete to a more abstract way of looking at math.

Sometimes verbal algebra can be used to solve problems that could be a struggle even using standard algebra. The first year of the high school IMP program includes a problems they call the Perils of Pauline. A girl is 3/8ths of the way through a railroad tunnel when she hears the whistle of an approaching train coming from behind. She just has time to make it to either the beginning or the end of the tunnel the same time the train gets to there. If the train goes 60 mph, how fast does Pauline run?

You might want to try solving this problem with standard algebra. It isn't easy.

But we can solve the problem very easily using verbal algebra. We only have to ask the right question. There are three important points in time:
(1) When she hears the whistle.
(2) When the train reaches the beginning of the tunnel.
(3) When the train reaches the end of the tunnel.

We know ask where Pauline is at each of thses times.
(1) She is 3/8 ths of the way thrught the tunnel.

(2) She just reaches the beginning of the tunnel when the train gets there, if she runs back the way she came. To get there she runs through 3/8 ths of the tunnel. If she continues to run forward she should also run through 3/8 ths of the tunnel, and she would get to the 3/8 + 3/8 = 6/8 or 3/4 point in the tunnel.

(3) If she runs forward she will run from the 3/4 point to the end of the tunnel in the time it takes the train to go fron beginning to end. She covers 1/4 the distance the train does in this time interval, so she must be running at 1/4 of the 60 mph speed of the train.