The Lowly Mathematician

In High schools across the country, kids learn algebra. I know quite a few parents who have some big problems with algebra. For some, it’s just because it’s been a while since they’ve needed to use it, for others, it’s always been a hard thing, but it doesn’t have to be. In this post, I’ll take you through some of the basic concepts of algebra as if you had never seen it before, it’ll be a bit of a whirlwind review of the idea, but (hopefully) written well enough so that someone who has never fully understood algebra. Oh, yes, by the way, there will be word problems.

Here’s the solution from last time.

Recall that the problem was to determine what score you would need on the new SATs to match your older brothers score of 1460 on the old SATs.

The first step is to realize that the old score is really 1460 out of 1600, or 1460/1600 = 146/160 = 73/80. We’ll call that number B, for “Brothers score.” Next, we need to realize that your score will be something out of 2400, we can write that as m = x/2400. The next step is to convert B to a percentage, and then we have the problem in a form we know, that of a “x percent of y” problem. Using the definition, we find that 73/80 is 91.25% of 80. So now we want to find 91.25% of 2400, using a calculator we find that this is: 2190.0 even, so you’d need to score 2190 to match your older brothers score.

So what is algebra? Most people would answer, “That math with the letters in it.” But really, this is just what algebra looks like. Really, Algebra is about equalities, and specificly about equalities involving unknown numbers. To understand equalities, we need to name the two types, what I call “resolved equalities”, eg:

“x = 2″, “y = z”, or “x = 3/4 + 1″.

and “unresolved equalities”, eg:

“(x + y)*(x - y) = 2″ or “x^2 + 2 = 4″

The basic idea is that, when you read the equality, if it looks like “<some letter> = <any old expression>” then it’s a resolved equality.

The neat thing about equalities is that you can substitute with them, in math, this is called the transitive property, it just means that if I know: “x = y” and “y = z”, then I can substitute ‘z’ for ‘y’ to get “x = z”. Simple enough.

The other idea that gives rise to algebra is the idea of unknowns, we’ve used them a bit already, but let me explain the motivation for them a little more.

When you do normal arithmetic, you typically do problems like “12 x 23 = something”, or “12 + 23 = something”, or some combination of these. These things are really algebraic expressions, in the language of algebra, we’d write them as “x = 12*23″ or “x = 12 + 23″, when we do arithmetic we’re really solving resolved equalities in algebra. So then, algebra is the next logical jump, it’s a set of rules for solving unresolved equalities.

So, how do we come up with rules to do that? Really, it only requires one rule, we’ll call it the “fairness” rule, specificly it says that, given any equality, we can take some operation and do it to both sides, and get an equality back. Later we’ll learn how to write this in symbols, but for now, we’ll just remember it as the following adage, “Anything you do to one side, you have to do to the other.” This rule, along with the transitivity rule, allows us to transform just about any unresolved equality into a resolved one. Pretty powerful stuff.

So lets do a couple of problems with unresolved equalities, solve for x in the following:

• “3x = 9″
• “3x - 1 + y = 2x + y”
• “3x^2 = 27″

The first one is pretty easy, we just need to find an operation to get rid of the 3 on the left side, the obvious choice is to divide by three on both sides, and in fact, thats the right choice. Doing so, we get “3x/3 = 9/3.” So now how do we simplify that to get a resolved equality? Well, realize that saying “3x” is just saying that you have a number with a factor of three in it, a factor which you can remove by dividing, and in general saying “n*x” is the same as saying that you have a number with a factor of n in it. So what do you have left from a multiple of three when you divide by three? Simply, “x” — the other factor of the equation. So our answer is then “x = 9/3″, or more simply “x = 3″. That was easy.

What we did with multiplication, also works with addition, or any other operation. So lets try it out on the second equation. We won’t do as much explaining, I’m just going to list the steps:

1. 3x - 1 + y = 2x + y
2. 3x - 1 + y - y = 2x + y - y
3. 3x - 1 = 2x
4. 3x - 2x - 1 = 2x - 2x
5. x - 1 = 0
6. x - 1 + 1 = 0 + 1
7. x = 1

The important things to note here are in steps 4 and 5. In step 4, we’re subtracting a multiple of an unknown from a multiple of an unknown, this implicitly[1] uses a notion called “factoring,” which you may have heard of in arithmetic, where you would factor numbers into individual components. Here, it means the following, If you have something like: k*x + n*x, then you can write it as x * (k + n). This is because when we expand multiplication on an expression in parenthesis, we multiply the expression on the outside of the parenthesis with each item in the parenthesis. We can see this with numbers by looking at the following example:

• x = 9 + 27
• x = (3*3) + (3*9)
• x = 3(3 + 9) — you see how we removed the common factor of three, and now we have two simpler problems
• x = 3*12
• x = 36

The other step I noted, step 5, underlies a really important thing in algebra, the existence of the number zero, and specifically, the fact that you can have expressions equal to it. This comes up when dealing with quadratic equations (we’ll talk about those later), Calculus, and virtually all of the rest of mathematics.

The third equation I listed is solved as follows:

1. 3x^2 = 27
2. 3x^2 / 3 = 27/3
3. x^2 = 9
4. sqrt(x^2) = sqrt(9)
5. x = +3 or -3

The important, and possibly confusing step here is step 5, after taking the square root (written as sqrt(x^2) here) — the inverse operation of x^2, which means “x multiplied by itself twice”, and in general x^n means “x times itself n times” — I wrote that x is equal to either +3 or -3. It’s easy to see if we plug either answer in to the equation in step 3, both +3*+3 = 9, and (-3) * (-3) = 9. [2]

So, with what I’ve armed you with so far, you should be able to master a few more problems, since algebra is such a big subject, I can’t teach it in just one post, so this is as far as we’ll go this time, I’ll leave a few more algebra problems for you to try out. It’ll work like this, I’ll give you the original unresolved equality, and a variable to solve for, and your job is to find the list of manipulations, one at a time, to go from the unresolved equality to a resolved equality. Here they are.

In all three, solve for x, they rank from easiest to hardest:

1. 8x + 1 = 17
2. 2(9x + 4y) = (18 + 16y)/2
3. 10x^2 + 6 = 1006

[1] In case you haven’t seen that term before, it means that we’re using some rule without explicitly stating it.

[2] It may seem counterintuitive now, but you’ll have to take my word on it, a negative times a negative is a positive. We’ll get to this idea later.