Total Algebra

Verbal reasoning cannot always get us all the way to the answer. But verbal reasoning can be an important component to the solution to a great many problems. When engineers and physicists approach a problem that requires mathematics, they don't just set up the problem using the language of algebra--and then begin start thinking about how to solve the problem. For engineers and physicists, setting up the problem is part of the solution process. They try to set the problem up in a way that will make the rest of the solution as easy as possible. This is what I call total algebra.

Recall the cookie problem from the post on Verbal Algebra. Large cookies cost 30 cents, small ones cost 10 cents, and we bought 12 cookies for $2.00 or 200 cents. How many of each types did we buy?

The most straightforward way to set this problem up algebraically would be to use two equations and two unknowns. If x represents large cookies, and y represents small ones, then the fact that we have 12 cookies means (1) x + y = 12, and the $2 cost gives us our next equation (2) 30x + 10 y = 200. At this point, one way to proceed would be to solve (1) for y which gives us y = 12 - x, and then we could substitute this value for x in (2) which will give us (3) 30x + 10(12 - x) = 200.

The alternative approach we looked at in the Verbal Algebra post used just one variable. We let x be the number of large cookies, and verbal reasoning told us that (12 - x) was the number of small ones. This approach allowed us to set up our problem in the form 30x + 10(12 - x) = 200 instead of taking several steps to reach that point.

Algebra teachers usually teach the one variable approach to problems like this. This is done at a time when students do not yet know how to solve problems with two variables and two unknowns. Basically, verbal reasoning is used to stretch what we can accomplish with one variable algebra. A little creativity in setting up our problem allows us to make a one variable problem out of something that really should be a two variable problem. In the process we are able to eliminate some of the steps we would have to take if we took the more straightforward two variable approach.

But the emphasis in math education is not to teach students to make the most effective use of simple mathematical tools. Instead we try to teach students more advanced mathematical concepts even if we make rather trivial use of those concepts. We teach students to set up problems in the language of algebra, but do rather little to help them explore the different ways they could do so. We will now look at how that might be done.

Two numbers have a sum of 14 and a product of 45. What are the two numbers? We could call one number x, the other y and get two equations: (1) x + y = 14, and (2) xy = 45. Solving the first equation for y we would get y = 14 - x, and substituting in the second equation we get x(14 - x) = 45. In practice we usually call our second number x - 14, and then set up x(14 - x) = 45 as our original equation. Multiplying out we get 14x - x^2 = 45 which becomes 14x = x^2 + 45 or 0 = x^2 - 14x + 45. There are a number of ways to solve this quadratic equation. In one method, factoring, we do almost the opposite of what we did when we set up this particular problem.

To solve 0 = x^2 - 14x + 45 by factoring we look for two numbers whose sum is 14 and whose product is 45. 5 and 9 are two such numbers. We can now use this information to factor the expression x^2 - 14x + 45, and when we do we get 0 = x^2 - 14x + 45 = (x - 5)(x - 9). Since this product is zero one of the factors must be zero. Either x - 5 = 0, in which case x=5 or x - 9 = 0, in which case x = 9. In the first case our two numbers are x = 5, and 14 - x = 9 while in the second case they are x = 9, and 14 - x = 5. In either case 5 and 9 are the numbers. Therefore 5 and 9 are not only one possible solution, but in one order or the other they are the only possible solution.

Our final result is a nice one--but is it really necessary to do all this work? No. We could have saved a lot of effort if we had put a little more thought into how we set up our problem.

Two numbers have a sum of 60 and a product of 851. What are the two numbers? Instead of just making one number x, so the other number will be 60 - x, let's think about what we are doing. Two numbers that add up to 60 could both be 30--or if one is more than 30, the other is the same amount less than 30. If one is 30 + u, the other is 30 - u. There is a nice kine of symmetry here that suggests this might make our problem easier to solve. And it does.

The product of our two numbers is (30 + u)(30 - u) = 851. Multiplying out we get 900 - u^2 = 851. This becomes 900 = 851 + u^2, and subtracting 851 we get u^2 = 49. Taking square roots we see that u = +7, or u = -7. In the first case our two numbers are 37 and 23, In the second case they are 23 and 37.

Now consider two numbers that add up to s and have a product of p. For convenience we little let m = s/2 be the mean of the two numbers. Our numbers will now be m + u and m - u for some u. Multiplying we get (m + u)(m - u) = p or m^2 - u^2 = p. This means m^2 = p + u^2. And we get m^2 - p = u^2, Taking square roots on both sides we find SQRT(m^2 - p) = u or - SQRT(m^2 - p) = u. Substituting the first value for u we get m + SQRT(m^2 - p) and m - SQRT(m^2 - p) as our two numbers. The second value for u gives us the same two numbers in the other order.

Now suppose we wish to find the roots of the quadratic equation x^2 - 2mx + p = 0. We can factor x^2 - 2mx + p if we can find two numbers that add up to 2m and multiply to p. But we have just done that. Calling these two numbers q and r we can factor x^2 - 2mx + p and we get (x - q)(x - r). This product will be zero if and only if either (x - q) or (x - r) equals zero. This means our solution must be either m + SQRT(m^2 -p) or else m - SQRT(m^2 - p).

We can now derive the quadratic formula, or the solution for the general quadratic equation Ax^2 + Bx + C = 0, where B and C can be any numbers, and A is any nonzero number. We divide by A and get the equation x^2 + (B/A)x + (C/A). Setting (B/A) = -2m and (C/A) = p we get m = -B/(2A) and p = C/A. We now need only apply the result from the last paragraph to get our solution.

Our solution will unfortunately involve some messy fractions. There is, however, a trick that will allow us to avoid any fractions until the last step. We want to find q and r whose mean m = -B/(2A) and whose product p = C/A. Multiplying both these numbers by 2A we get Q = 2Aq and R = 2Ar. The mean of Q and R is -B. The product QR = (2Aq)(2Ar) = 4AA(C/A) = 4AC. We have two numbers Q and R whose mean M is -B and whose product P is 4AC. By our previous result Q and R in one order or the other must be =B + U and - B - U where U = SQRT([-B]^2 - 4AC) or simply U = SQRT(B^2 - 4AC).

Dividing Q and R by 2A we get q and r, our solution to the quadratic equation, in the nice form in which it is usually presented. That is, our roots are [-B + SQRT(B^2 - 4AC)]/(2A) and [-B = SQRT(B^2 - 4AC]/(2A).

Combining verbal reasoning and algebraic notation we have found an alternate approach to the study of quadratic equations which may be more accessible to many students. Certainly if this approach seems easier to you, you should consider using greater flexibility in setting up algebraic problems since it could make problem solving easier for you. A little creativity in setting up problems can yield considerable benefits.

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.

Algebraic Notation on the Keyboard and Elsewhere

There are two types of algebraic notation. There is the traditional notation we use when we write with pen or pencil. And there is keyboard notation that we use when we key in math without a good math word processor. As long as we keep things simple, there is little difference between the two. And we should keep things simple when we introduce students to algebraic notation.

The best way to learn algebraic notation is by working with formulas. We could start with simple formulas like P = 4S as a shorthand way to describe the rule that the perimeter of a square is 4 times the length of a side. We can then proceed to slightly more complicated formulas like P = 2L + 2W as shorthand for the perimeter of a rectangle is twice the length plus twice the width. We could also write P = 2(L + W) as shorthand for an alternate way to find the perimeter of a rectangle--twice the sum of the length and the width. In these formulas it was convenient to write 4 times S as 4S, and to write 2 times L as 2L etc. If we generalize this idea and use it to express the multiplication of two variables in the same way we get A = LW as the formula for the area of a square. We might get A = (1/2)BH as the area of a triangle or A = BH/2 as an alternate formula.

Working with formulas like this to solve practical problems can help students become familiar with a simple version of algebraic notation. Students are able to work in a setting where they are actually using the notation to solve interesting problems. At this stage, students are only required to apply the formulas by substituting in the appropriate input variables such as S or L or W or B or H.

Mathematics is a deductive science, It is based upon rules that we follow when we do math. But humans learn inductively. We learn from examples, generalizing what we see in specific cases. The inductive nature of human learning can be seen from the way we learn language. We easily pick up words inductively from the conversations we hear and we find ourselves using them correctly in our own conversation, while dictionary definitions help very little if we want to actually use the words that we learn.

In mathematics we learn best from examples where we can see things used in context, rather than from rigorous definitions. Ultimately we will need to work with rigorous definitions, but these will make little sense to us until we have the experience that will make these definitions meaningful to us.

At a slightly more sophisticated level, algebraic notation includes the use of exponents to represent repeated multiplication. These are usually written as superscripts--but without some level of word processing we may not be able to keyboard in superscripts. Instead we use what I call the hat symbol ^ to write a to the n th power as a^n.

Unfortunately, the hat symbol ^ gives us a less readable way of writing exponents than the traditional superscript notation. This pretty much forces us to do something we really should do anyway. Instead of writing the really ugly A = S^2 as the formula for the area of a square, we will simply write it as A = SS. We can write the volume of a cube as V = SSS.

When we get to multiplying algebraic expressions (x + 5)x = xx + 5x has a simple clarity that would be lacking in (x + 5)x = x^2 + 5x even if we were
able to write our exponent as a superscript. It is important to use exponents to represent a large number of repeated multiplications. It is essential that we use exponents if we wish to represent repeated multiplication where the number of factors in the product could vary. But there is a lot of basic algebra where the use of exponents is optional, and we can achieve greater clarity for beginners by not using exponents.

Arithmetic in the Age of Calculators

In the past people actually needed to use paper and pencil arithmetic for practical everyday purchases. Adding machines and cash registers did a lot to reduce the amount of arithmetic people needed to do. But people still did a lot more arithmetic than they do today.

One nice example of what people did was making change. A cash register added up the total--but did not tell you how to make change. Instead of subtracting people counted up. If the total was 78 cents and I gave the clerk a dollar, the clerk would give me two pennies and then two dimes, counting up 79 cents, 80 cents, 90 cents, and finally one dollar. The actual process of counting out change in this way requires no more effort than it would require today to count out 22 cents after a modern cash register did the subtraction and told us what change to give.

Not only is there less need for traditional arithmetic today--there are also fewer situations in the real world where these skills are reinforced. We cannot expect the students of today to be as strong in this area as students were in the past. But there still is a place for traditional arithmetic.

Suppose two movie tickets and a $3 bag of popcorn costs $13. How much does a movie ticket cost? Without the popcorn you pay $10 for two movie tickets. Dividing by 2 we get $5 as the price of one movie ticket.


This solution is as clear as it is because we can do the arithmetic easily in our heads. If we needed a calculator to find 13 - 3 and 10/2 it would interrupt the flow of the argument. That would be fine if we already understood the process. But if we had never seen a solution like this before, that added distraction would make things harder for us.


One benefit to traditional arithmetic background is this instant access to certain simple basic facts which can be incorporated into when a certain type of problem is first being learned. This allows us to place the greatest possible focus on the algebraic parts of the problem with minimum distraction from the arithmetic part.


A second benefit comes from the fact that algebra in many ways generalizes what we do in arithmetic. Familiarity working with numbers expressed in powers of 10 provides a background that makes it easier to understand working with polynomials expressed in powers of x or some other variable.


Carrying and borrowing are major problems for many people working with whole number arithmetic. These problems are as bad as they are because we rush students into doing problems that require carrying and borrowing, and we do not treat the subject in a sufficiently flexible manner. This is particularly true today when students do not have the mastery of basic facts that students once had.


What we need to do is work with problems that allow students to place nearly all their focus on carrying or borrowing with as little distraction as possible. This can be done if we add or subtract relatively small two digit numbers ending in either 5 or 0. The addition or subtraction in the ones place is very simple. Addition or subtraction in the tens place is fairly simple. Students can place a maximum of focus on the carrying or borrowing part until they become familiar with the process. With problems this simple, students can learn to carry or borrow without ever writing down little carry or borrow numbers. The process is learned with problems simple enough that writing such numbers is not necessary--and once the process is learned in a simple setting such as the one I described, slightly more complicated settings can be considered, such as two digit numbers ending in 2, 4, 6, 8, or 0. It is easier to never learn to write these little numbers than it is to first learn to use them, and then learn to stop using them.


But in a age of calculators, mental math may be more important than paper and pencil techniques. There are different methods for carrying and borrowing in your head. When I add 365 and 366 in my head, I don't start with the ones place--I start with the hundreds place, 3 hundred and 3 hundred make 6 hundred. Sixty and sixty make one hundred twenty for a total of 7 hundred twenty, And finally 5 and 6 make 11 bringing the total to 731.


Subtraction is similar. To find 82 minus 38 I start with eighty minus thirty and I get fifty. Then I try to do 2 - 8 but I am 6 short so I take 6 away from fifty to get 44. Even if our accuracy is less than perfect, being able to visualize simple arithmetic problems in this fashion can help follow algebraic examples as long as the arithmetic isn't too messy.


Another area of difficulty for many students is working with fractions. The rules for adding fractions can be made clearer if we write out the denominators in words. We cannot directly add 1 half and 1 fourth because we would be adding different things--apples and oranges. But if we re-express 1 half as 2 fourths, then we can add 2 fourths and 1 fourth and clearly the answer is 3 fourths.


If we are asked to add two unlike fractions, how can we re-express the problem in a way where we are no longer trying to add different things? How can we make it into a problem where we have like fractions? We can start working with halves, quarters, and eighths--which are the fraction we usually encounter. In this case, the largest denominator will always work--and there are just a few conversions we need to learn. Once we become comfortable with the need to re-express a problem using like fractions, we can worry about the situation where doing so is more of a challenge.


Too many people are misinformed by teachers and textbooks that tell them they need to find the lowest common denominator--and then they are given a complicated procedure involving factor trees and prime factorization they are supposed to use to find the necessary lowest common denominator.


You don't really need to find a lowest common denominator. Any common denominator will work--and you can always find a common denominator by multiplying the two given denominators together. Lower denominators are usually easier to work with--but experience will usually help you find a lower common denominator when there is one, and it is not really that important if you don't use the lowest possible common denominator.


In conclusion I would like to mention one neat little trick that you can do when you subtract mixed numbers. What is 5 and one third minus 1 and two thirds? 5 minus 1 is 4. One third minus two thirds leaves us short one third. This leaves us one third short so we wind up with one third less than 4 or 3 and two thirds.

Background for Algebra

Algebra is built upon a foundation of arithmetic. This much is obvious. But how much of a background in arithmetic is needed before it makes sense to begin the study of algebra?

At one time little if anything was done to introduce students to algebra prior to high school, and in high school algebra was something that was only taught to the more academically inclined students. In an effort to upgrade mathematics education in America around the time of Sputnik, we began teaching algebra more widely, and we began introducing students to algebraic concepts at an earlier age.

Fifty years later we have students who are starting to study algebra earlier. They are spending more years in high school studying algebra. And they are taking dumbed down SATs to make the comparison with students who took the test before Sputnik look less embarrassing than it would otherwise. It is true that more students are now going to college, and more students are now taking the SATs--but that only explains the decline if we assume there has been no real improvement in mathematics education despite massive time and effort devoted to that goal.

Before Sputnik we did drag our feet, so to speak, delaying the teaching of algebra far more than we needed. But in many ways the reforms in math education have only made things worse. Students need to have a math background that will help them to make sense of algebraic concepts. These concepts can be confusing to many students. And like people, algebra has only one chance to make a first impression. When we teach math, it is better to wait a little longer before introducing students to algebra, than rush to introduce algebra, and make the first impression that students get of algebra a confusing one.

What if algebra has already made its first impression, and that first impression has not been a good one? In that case we will want to go back and reintroduce you to algebra. Just as in human relationships we can overcome a bad first impression--but it does take some effort, and it does take a recognition that something of this sort needs to be done.

EZ Algebra rests upon a foundation consisting of three pillars. The first pillar is basic arithmetic: counting, addition, subtraction, multiplication, division, and straightforward problems that require the use of these basic arithmetic operations.

A second pillar is algebraic notation. The best introduction to algebraic notation is one where we use formulas such as p = 4s as a shorthand notation for a verbal rule such as the perimeter is 4 times the length of the side of a square.

The third pillar is non-straightforward arithmetic or what I call verbal algebra. The more challenging percentage problems are an example. If adding a 20% tip to a bill makes the total $60, what was the cost of the meal without the tip? A common mistake would be to subtract 20% of $60 and get $60 - $12 = $48. But this would be wrong. The tip isn't 20% of the total including the tip. It is only 20% of the total before the tip--and that total is something we don't know. But if we think of the problem in a different way, we can solve the problem. Adding 20% to our basic 100% gives us 120% of the basic bill or 1.2 times the basic bill. Dividing $60 by 1.2 will give us our answer of $50. The reasoning we have used here is quite similar to what we do in algebra, and having a background that includes working with problems like this one will give us a leg up when we begin our study of algebra.